Systems of Equations — Free Algebra 1 Review Games.

Substitute: x + (x+1) = 5, 2x = 4, x = 2, y = 3.

Parallel lines never intersect, so there is no solution.

Add equations: 2x = 14, x = 7. Then y = 3.

One intersection point means exactly one solution: consistent and independent.

2x = 8, x = 4. So (4, 8).

Add: 3x = 9, x = 3. Then y = 7-6 = 1.

3(2) + 2y = 12, 6 + 2y = 12, 2y = 6, y = 3.

Same slope (2) but different y-intercepts means parallel lines, no solution.

From second: y = 7-2x. Substitute: x + 3(7-2x) = 11, x+21-6x=11, -5x=-10, x=2, y=3.

Infinitely many solutions means both equations represent the same line.

Add: 6x = 18, x = 3. Then 2(3)+3y=8, 3y=2, y=2/3.

Multiply first by 2: 10x+4y=2. Add to second: 13x=13, x=1. Then 5+2y=1, y=-2.

Let p=pens, n=notebooks. p+n=4, 2p+5n=14. p=4-n, 2(4-n)+5n=14, 8+3n=14, n=2, p=2.

From second: x=y+3. Substitute: (y+3)/2+y/3=4. Multiply by 6: 3(y+3)+2y=24, 5y+9=24, y=3, x=6.

No solution when parallel: slopes equal. 2/k = 1/3 fails; rewrite: slope of first is -2/k, second is -1/3. Equal when 2/k=1/3, k=6.

The solution to a system of two linear equations is the ordered pair (x, y) that satisfies both equations simultaneously. On a graph, this corresponds to the point where both lines cross. The y-intercept (choice A) and slope (choice C) describe individual line properties, not the shared solution.

When both equations represent the same line, every point on that line is a solution, giving infinitely many solutions. This is a consistent and dependent system. An inconsistent system (choice A) has no solution because the lines are parallel but distinct. An independent system (choice B) has exactly one solution.

Substitute y = 3x into x + y = 8: x + 3x = 8, so 4x = 8 and x = 2. Then y = 3(2) = 6. The solution is (2, 6). Choice B, (3, 9), satisfies y = 3x but gives 3 + 9 = 12, not 8, so it fails the second equation.

The elimination method involves adding or subtracting the two equations (sometimes after multiplying by constants) so that one variable cancels out, leaving one equation in one unknown. Choice B describes the substitution method, and choice A describes the graphing method.

Since y = 3, substitute into the first equation: 2x + 3 = 9, so 2x = 6 and x = 3. The solution is (3, 3). Choice A gives 2(2) + 3 = 7, not 9, so it is incorrect.

Exactly one solution means one ordered pair satisfies both equations, which occurs when two distinct lines cross at one point. Parallel lines (choice A) never intersect, giving no solution. Identical lines (choice B) overlap completely, giving infinitely many solutions.

Set the right sides equal: x + 2 = -x + 4. Adding x to both sides gives 2x + 2 = 4, so 2x = 2 and x = 1. Then y = 1 + 2 = 3. The solution is (1, 3). Choice B, (2, 4), satisfies y = x + 2 but gives -(2) + 4 = 2, not 4, for the second equation.

Add the two equations: (3x + 2y) + (x - 2y) = 16 + 0, giving 4x = 16, so x = 4. Substitute into x - 2y = 0: 4 - 2y = 0, so y = 2. The solution is (4, 2). Choice B, (3, 4), gives 3(3) + 2(4) = 17, not 16, failing the first equation.

Substitute y = 2x - 3 into 3x + y = 7: 3x + (2x - 3) = 7, so 5x = 10 and x = 2. Then y = 2(2) - 3 = 1. The solution is (2, 1). Choice B, (1, -1), satisfies y = 2(1) - 3 = -1 but gives 3(1) + (-1) = 2, not 7.

Divide the first equation by 2: x + 2y = 4. This is identical to the second equation, so both represent the same line. Every point on that line is a solution, giving infinitely many solutions. This is a consistent and dependent system. Choice A would apply only if the lines were parallel but distinct.

Add the two equations: (4x - y) + (2x + y) = 9 + 3, giving 6x = 12, so x = 2. Substitute into 2x + y = 3: 4 + y = 3, so y = -1. The solution is (2, -1). Choice B, (1, -5), satisfies 4(1) - (-5) = 9 but fails the second equation: 2(1) + (-5) = -3, not 3.

Set the savings equal: 30 + 10w = 60 + 5w. Subtract 5w from both sides: 30 + 5w = 60, so 5w = 30 and w = 6. After 6 weeks both have $90. Choice A (4 weeks) gives Friend A $70 and Friend B $80, which are not equal.

Parallel lines share the same slope but have different y-intercepts. In choice A, both lines have slope 2 but y-intercepts of 3 and -1, so they never intersect. Choice B has slopes of 3 and -3, meaning the lines cross at one point. Choice C has slopes of 1 and 2, which are different, so those lines also intersect.

Multiply the second equation by 2: 2x + 4y = 14. Subtract the first equation: (2x + 4y) - (2x + 3y) = 14 - 11, giving y = 3. Substitute into x + 2y = 7: x + 6 = 7, so x = 1. The solution is (1, 3). Choice C, (1, 4), gives 2(1) + 3(4) = 14, not 11.

Let a = adult tickets and s = student tickets. The system is: a + s = 100 and 8a + 5s = 620. Substitute s = 100 - a: 8a + 5(100 - a) = 620, so 3a + 500 = 620, giving 3a = 120 and a = 40. Choice C (50 adult tickets) would yield 8(50) + 5(50) = $650, not $620.

Substitute x = 2y + 1 into 3x - y = 13: 3(2y + 1) - y = 13, so 6y + 3 - y = 13, giving 5y = 10 and y = 2. Then x = 2(2) + 1 = 5. The solution is (5, 2). Choice C, (7, 3), satisfies x = 2y + 1 since 7 = 2(3) + 1, but 3(7) - 3 = 18, not 13.

The goal of elimination is to add or subtract equations so one variable disappears. Multiplying by a constant adjusts a coefficient so it matches or becomes the opposite of the corresponding coefficient in the other equation, allowing cancellation. For example, if one equation has 3y and another has 2y, you multiply to create equal magnitudes. Choice D describes a verification step, not a solving step.

Subtract the second equation from the first: (5x - 2y) - (3x - 2y) = 4 - 0, giving 2x = 4, so x = 2. Substitute into 3x - 2y = 0: 6 - 2y = 0, so y = 3. The solution is (2, 3). Choice B, (1, 1.5), satisfies 3(1) - 2(1.5) = 0 but gives 5(1) - 2(1.5) = 2, not 4.

Multiply the first equation by 6 to clear fractions: 2x + 3y = 12. Multiply the second by 4: 2x - y = 4. Subtract the second from the first: 4y = 8, so y = 2. Substitute into 2x - y = 4: 2x - 2 = 4, so x = 3. The solution is (3, 2). Choice B, (6, 0), satisfies 2(6) + 3(0) = 12 but gives 2(6) - 0 = 12, not 4.

Let x = liters of 20% solution and y = liters of 50% solution. The system is: x + y = 60 and 0.20x + 0.50y = 0.30(60) = 18. Substitute y = 60 - x: 0.20x + 0.50(60 - x) = 18, so -0.30x = -12 and x = 40. Choice B (30 liters) gives 0.20(30) + 0.50(30) = 6 + 15 = 21, not 18.

Let a = larger and b = smaller. The system is: a + b = 20 and a - b = 8. Adding the equations: 2a = 28, so a = 14. Then b = 20 - 14 = 6. Finally, 3a - 2b = 3(14) - 2(6) = 42 - 12 = 30. A common error is stopping after finding a and b and computing 14 + 6 + 14 - 6 = 28 (choice B), which skips the final expression.

Multiply the first equation by 2: 6x - 8y = 4. The second equation states 6x - 8y = 5. The left sides are identical but 4 does not equal 5, a contradiction. The lines are parallel and distinct, so there is no solution. Choice C (infinitely many) would apply only if multiplying one equation produced the other exactly, which would require the constants to match as well.

Multiply the first equation by 2: 4x + 10y = -2. Multiply the second by 5: 15x - 10y = 40. Add: 19x = 38, so x = 2. Substitute into 2x + 5y = -1: 4 + 5y = -1, giving 5y = -5 and y = -1. The solution is (2, -1). Choice D, (2, 1), gives 2(2) + 5(1) = 9, not -1.

Downstream speed = 24 miles / 2 hours = 12 mph; the current aids the boat, so b + c = 12. Upstream speed = 24 miles / 3 hours = 8 mph; the current opposes the boat, so b - c = 8. Solving gives b = 10 mph and c = 2 mph. Choice C incorrectly uses the travel times (2 and 3) as the speeds rather than dividing distance by time.

Multiply the first equation by 5 to clear decimals: 2x + y = 8. Multiply the second by 10: 3x - y = 7. Add: 5x = 15, so x = 3. Substitute into 2x + y = 8: 6 + y = 8, giving y = 2. The solution is (3, 2). Choice B, (2, 4), satisfies the first equation since 0.4(2) + 0.2(4) = 1.6, but gives 0.3(2) - 0.1(4) = 0.2, not 0.7.

The point where two lines cross is where both equations are satisfied simultaneously, making it the solution to the system. The slope and y-intercept are properties of each individual line, not the intersection point.

A system has no solution only when the lines never intersect. Parallel lines have the same slope but different y-intercepts, so they run alongside each other without ever crossing. If they shared the same y-intercept and slope, they would be the same line with infinitely many solutions.

Substitute y = 3x into the second equation: 2x + 3x = 10, so 5x = 10 and x = 2. Then y = 3(2) = 6. The solution is (2, 6). Choice A gives 2(1) + 3(1) = 5, which does not equal 10.

A solution must satisfy both equations. Test (3, 4): 3+4=7 and 2(3)-4=2. Both check out. Choice A (2, 5) satisfies the first equation (2+5=7) but fails the second: 2(2)-5=-1, not 2.

When two equations represent the same line, every point on that line satisfies both equations. Since a line contains infinitely many points, the system has infinitely many solutions. This is called a dependent system, as opposed to an inconsistent system (no solution) or a consistent independent system (one solution).

Add the two equations to eliminate y: (x+y) + (x-y) = 8+2, giving 2x = 10, so x = 5. Substitute back: 5+y = 8, so y = 3. The solution is (5, 3). Choice B (4, 4) gives 4-4 = 0, not 2, in the second equation.

Both lines have slope 4 but different y-intercepts (-3 and 1), making them parallel. Parallel lines never intersect, so this system is inconsistent and has no solution. If the y-intercepts were equal, the lines would be identical, creating a dependent system with infinitely many solutions.

Set the two expressions equal: x+2 = 3x-4. Subtract x: 2 = 2x-4. Add 4: 6 = 2x, so x = 3 and y = 3+2 = 5. The solution is (3, 5). Choice B gives y = 2+2 = 4 from the first equation but y = 3(2)-4 = 2 from the second — the y-values differ, so it is not the intersection.

Substitute y = x-4 into the second equation: 2x + (x-4) = 14. Combine like terms: 3x-4 = 14, so 3x = 18 and x = 6. Then y = 6-4 = 2. The solution is (6, 2). Choice A (5, 1) gives 2(5)+1 = 11, not 14.

Let x be the larger number and y the smaller. Set up: x+y=18 and x=2y. Substitute: 2y+y=18, so 3y=18 and y=6, giving x=12. Choice C (9 and 9) satisfies the sum but neither number is twice the other.

Set the expressions equal: -2x+6 = x-3. Add 2x: 6 = 3x-3. Add 3: 9 = 3x, so x = 3, and y = 3-3 = 0. The intersection is (3, 0). Choice A (1, 4) gives y = -2(1)+6 = 4 from the first equation but y = 1-3 = -2 from the second — these are not equal.

The y-coefficients (+2y and -2y) are already opposites, so add the equations directly: 4x = 16, x = 4. Substitute into the second equation: 4-2y = 0, so y = 2. The solution is (4, 2). Choice A (2, 5) satisfies the first equation but gives 2-10 = -8, not 0, in the second.

Let n = nickels and d = dimes. System: n+d=25 and 5n+10d=175 (in cents). Substitute n=25-d: 5(25-d)+10d=175, giving 125+5d=175, 5d=50, d=10. Check: 15 nickels and 10 dimes give 75+100=175 cents. Choice C (15 dimes) would give 10(5)+15(10)=200 cents, which exceeds $1.75.

The phrase 'three times the first plus twice the second equals 24' translates to 3x+2y=24. The phrase 'the first minus the second equals 1' translates to x-y=1. Choice A uses addition (x+y) instead of subtraction for the second equation, and Choice C reverses the coefficients of x and y in the first equation.

Substitute y=2x-4 into the first equation: 3x+(2x-4)=11. Combine: 5x-4=11, 5x=15, x=3. Then y=2(3)-4=2. The solution is (3, 2). Choice A (2, 0) gives 3(2)+0=6, not 11, in the first equation.

The y-coefficients are +3 and -3, which are already opposites. Adding the equations directly eliminates y: (2x+4x)+(3y-3y)=7+5, giving 6x=12 and x=2. Choice A would match the x-coefficients instead, which also works but requires an extra multiplication step before subtracting.

Let s = small candles and l = large candles. System: s+l=10 and 3s+7l=42. Substitute s=10-l: 3(10-l)+7l=42, giving 30+4l=42, 4l=12, l=3. There are 3 large and 7 small candles. Choice C (4 large) would cost 3(6)+7(4)=18+28=$46, which exceeds $42.

Multiply the first equation by 3 to get 9x+15y=3, and the second by 5 to get 10x+15y=0. Subtract the first result from the second: x=-3. Substitute: 2(-3)+3y=0, 3y=6, y=2. The solution is (-3, 2). Choice C (2, -1) satisfies the first equation (6-5=1) but gives 2(2)+3(-1)=1, not 0, in the second.

Let t = hours after the second car departs. The first car travels for t+1 hours, the second for t hours. Set distances equal: 50(t+1)=75t. Expand: 50t+50=75t, so 50=25t, t=2 hours. Choice C (3 hours) would put the first car at 50(4)=200 miles and the second at 75(3)=225 miles, meaning the second car has already passed the first.

For infinitely many solutions, both equations must represent the same line. Divide the second equation by 3: 2x+y=8. Comparing with 2x+ky=8, the coefficients match when k=1. If k=3, the first equation becomes 2x+3y=8, which has a different slope than 2x+y=8, so the lines would intersect at exactly one point instead.

Clear fractions by multiplying each equation by 6: the first becomes 3x+2y=24 and the second becomes 2x+3y=21. Multiply the new first equation by 3 (9x+6y=72) and the second by 2 (4x+6y=42), then subtract: 5x=30, x=6. Substitute: 3(6)+2y=24, 2y=6, y=3. Choice A (3, 6) fails the first original equation: 3/2+6/3=1.5+2=3.5, not 4.

Let c = Colombian pounds and b = Brazilian pounds. System: c+b=20 and 12c+8b=9.60(20)=192. Substitute b=20-c: 12c+8(20-c)=192, giving 4c+160=192, 4c=32, c=8. Choice D (10 pounds) gives 12(10)+8(10)=120+80=200, not 192.

For no solution, the lines must be parallel: equal slopes but different y-intercepts. The first equation rewrites as y=2x-3 (slope 2). The second equation rewrites as y=(2/k)x-(4/k). Setting 2/k=2 gives k=1. With k=1: the second equation becomes 2x-y=4, or y=2x-4, which is parallel to y=2x-3 but not identical, confirming no solution. Choice k=2 gives slope 1 in the second equation, so the lines would intersect at one point.

Multiply the first equation by 2: 10x+4y=6. Subtract the second equation: 7x=-7, so x=-1. Substitute: 3(-1)+4y=13, 4y=16, y=4. The solution is (-1, 4). Choice A (1, -1) satisfies 5(1)+2(-1)=3 in the first equation but gives 3(1)+4(-1)=-1, not 13, in the second.

Substitute (2, 1) into each equation. First: 1=m(2)+5, so 2m=-4, m=-2. Second: 1=-(2)+n, so n=3. The values are m=-2 and n=3. The solution is unique because line 1 has slope -2 and line 2 has slope -1, confirming they are not parallel. Choice B (m=2, n=-3) would place line 1 as y=2x+5, which passes through (2, 9), not (2, 1).

Set the expressions equal: 2x - 1 = -x + 5, so 3x = 6 and x = 2. Substitute back: y = 2(2) - 1 = 3. The solution is (2, 3). Choice (1, 2) fails the first equation since 2(1) - 1 = 1, not 2.

Infinitely many solutions occur when both equations describe the exact same line, so every point on the line satisfies both equations. Parallel lines (choice A) represent a system with no solution, not infinitely many.

Adding the two equations gives 2x = 8, so x = 4. Substituting into x + y = 7 gives y = 3. The solution is (4, 3). Choice (3, 4) fails the second equation: 3 - 4 = -1, not 1.

Substitution requires isolating one variable in one equation first, then substituting that expression into the other equation. Multiplying equations by a constant is the first step in elimination, not substitution.

No solution means the lines never meet, which happens when they are parallel — same slope but different y-intercepts. If they were the same line (choice B), the system would have infinitely many solutions.

In x + 2y = 8 and x - 2y = 4, the y-coefficients are +2 and -2, which are opposites. Adding the equations immediately eliminates y. Choice D has coefficients that are multiples, not opposites, so adding would not eliminate a variable.

A system with exactly one solution means the two lines cross at a single point. If the equations had the same slope (choice A), the lines would be parallel (no solution) or the same line (infinitely many solutions), not a unique solution.

Substitute y = 3x - 2 into 2x + y = 8: 2x + (3x - 2) = 8, so 5x - 2 = 8, giving 5x = 10 and x = 2. Choosing x = 3 is a common error from adding instead of substituting correctly: 5(3) - 2 = 13, not 8.

Adding the two equations eliminates y: 3x = 12, so x = 4. Substituting into x - y = 3 gives 4 - y = 3, so y = 1. The solution is (4, 1). Choice (3, 3) fails both equations: 2(3) + 3 = 9 checks out, but 3 - 3 = 0, not 3.

Parallel lines never intersect and have the same slope but different y-intercepts, giving no solution. If lines have different slopes (choice C), they will always intersect at exactly one point. Perpendicular lines also intersect, just at a right angle.

Substitute x = 2y + 1 into 3x - 4y = 7: 3(2y + 1) - 4y = 7, so 6y + 3 - 4y = 7, giving 2y = 4 and y = 2. Choosing y = 1 is incorrect: 2(1) = 2 and 3(3) - 4(1) = 5, not 7.

The y-coefficients are -1 and +1, which are opposites. Adding the equations gives 6x = 12, so x = 2. Substituting into 2x + y = 7 gives 4 + y = 7, so y = 3. The solution is (2, 3). Choice (2, 1) fails the second equation: 2(2) + 1 = 5, not 7.

Let a = adult tickets and c = child tickets. Set up: a + c = 60 and 8a + 5c = 378. Substituting c = 60 - a gives 8a + 5(60 - a) = 378, so 3a + 300 = 378 and a = 26. Choosing 20 would give revenue of 8(20) + 5(40) = 160 + 200 = 360, which is less than $378.

Multiplying 2x + y = 4 by 2 gives 4x + 2y = 8, which is identical to the second equation — so both equations describe the same line, giving infinitely many solutions. Choice B is parallel lines (no solution), since doubling the first equation gives 4x + 2y = 8, not 6.

Line 1 has slope (0-5)/(5-0) = -1 and equation y = -x + 5. Line 2 has slope (1-(-1))/(1-0) = 2 and equation y = 2x - 1. Setting equal: -x + 5 = 2x - 1, so 6 = 3x and x = 2, y = 3. The intersection is (2, 3). Choice (3, 2) fails Line 1: -3 + 5 = 2, not 2 — wait actually 2 = 2 checks, but Line 2: 2(3) - 1 = 5, not 2.

Substitute y = x + 1 into 2x - y = 4: 2x - (x + 1) = 4, so x - 1 = 4 and x = 5. Then y = 5 + 1 = 6. Choosing x = 3 is incorrect: 2(3) - (3+1) = 6 - 4 = 2, not 4.

Let q = quarters and n = nickels. Set up: q + n = 26 and 25q + 5n = 450. Substituting n = 26 - q gives 25q + 5(26 - q) = 450, so 20q + 130 = 450 and 20q = 320, giving q = 16. Choosing 18 quarters would total 18(25) + 8(5) = 450 + 40 = 490 cents, which exceeds $4.50.

Adding x + 2y = 10 and 3x - 2y = 6 eliminates y: 4x = 16, so x = 4. Back-substituting gives 4 + 2y = 10, so y = 3. Choosing x = 5 would require 4(5) = 20 after adding, not 16.

Multiply the first equation by 3: 9x + 12y = 3. Multiply the second by 4: 8x - 12y = 48. Adding gives 17x = 51, so x = 3. Substituting: 3(3) + 4y = 1, giving 4y = -8 and y = -2. The solution is (3, -2). Choice (2, -3) fails both: 3(2) + 4(-3) = 6 - 12 = -6, not 1.

Let x be the larger and y the smaller. From x + y = 20 and x - y = 8, adding gives 2x = 28 so x = 14, and y = 6. Then 3(14) + 2(6) = 42 + 12 = 54. Choice 48 is a common error from computing 3(12) + 2(6) = 36 + 12, using the wrong value of x.

For infinitely many solutions, the two equations must be proportional. Dividing the second equation by 2 gives 3x + 2y = 6. Comparing with 3x + ky = 6, the coefficients and constant are identical when k = 2. Choosing k = 4 would give 3x + 4y = 6 and 6x + 4y = 12, which are not proportional (the first doubled becomes 6x + 8y = 12, not 6x + 4y = 12).

Let b = boat speed in still water and c = current speed. Downstream: b + c = 60/3 = 20. Upstream: b - c = 60/4 = 15. Adding the equations gives 2b = 35, so b = 17.5. Subtracting gives 2c = 5, so c = 2.5 mph. Choosing 5 mph would make b = 15, but then downstream speed would be 20 and 15 + 5 = 20 checks, yet upstream would be 15 - 5 = 10 mph covering 60 miles in 6 hours, not 4.

Clear fractions in equation 1 by multiplying by 3: 2x + 1 + 3y = 15, giving 2x + 3y = 14. Clear fractions in equation 2 by multiplying by 2: 2x - (y - 2) = 8, giving 2x - y = 6. Subtracting the second from the first: 4y = 8, so y = 2. Then 2x = 8, so x = 4. Thus x + y = 6. Choice 8 is the error of adding x and y before fully solving: confusing x = 4 with x + y = 4 + 4.

Let x = liters of 20% solution and y = liters of 50% solution. Set up: x + y = 12 and 0.20x + 0.50y = 3.6. Substituting x = 12 - y gives 0.20(12 - y) + 0.50y = 3.6, so 2.4 + 0.30y = 3.6 and y = 4. Using 6 liters of each (choice C) would give 0.20(6) + 0.50(6) = 1.2 + 3.0 = 4.2 liters of acid, which is 35%, not 30%.

The equation 0 = 2k - 8 is a condition on k, not on x or y. When k = 4, the equation becomes 0 = 0, which is always true — the equations are dependent and have infinitely many solutions. When k is not equal to 4, the equation becomes a false statement like 0 = 2, meaning the system is inconsistent with no solution. Choice D is wrong because k must equal exactly 4 for infinitely many solutions to exist.

Set the expressions equal: 3x - 1 = x + 3, so 2x = 4 and x = 2. Then y = 2 + 3 = 5, giving the solution (2, 5). Choice A, (1, 2), satisfies y = 3x - 1 since 3(1) - 1 = 2, but fails y = x + 3 since 1 + 3 = 4, not 2. A solution must satisfy both equations.

The solution to a system is the ordered pair (x, y) that satisfies both equations simultaneously. Graphically, that is the point where both lines cross — the intersection. The y-intercept or x-intercept of a single line is a different concept and is not the solution to the system unless the lines happen to cross at those specific points.

Both lines have slope 4 but different y-intercepts (3 and -5), so they are parallel and never intersect. A system of parallel lines has no solution. If the slopes were different, the lines would cross at exactly one point. If the equations described the same line, there would be infinitely many solutions.

Divide the second equation by 4 to get y = 2x + 1, which is identical to the first equation. Both equations represent the same line, so every point on that line is a solution, giving infinitely many solutions. A student might mistakenly think two different-looking equations always produce exactly one solution, but here they simplify to the same equation.

Substitute y = 5 into 3x + y = 14: 3x + 5 = 14, so 3x = 9 and x = 3. Choice A (x = 2) is a common error from dividing 9 by the wrong number. The coefficient of x is 3, so divide both sides by 3 to get x = 3.

Add the equations: (x + y) + (2x - y) = 7 + 2 gives 3x = 9, so x = 3 and y = 7 - 3 = 4. Check: 3 + 4 = 7 and 2(3) - 4 = 2. Both equations check out, so (3, 4) is the solution. Choice A, (2, 5), satisfies the first equation but 2(2) - 5 = -1, not 2, so it fails the second.

An inconsistent system has no solution because the equations contradict each other — graphically this occurs when the two lines are parallel. A system with exactly one solution is called consistent and independent. A system with infinitely many solutions is consistent and dependent. Linear systems cannot have exactly two solutions.

The goal of elimination is to add the equations so that one variable drops out. To do this, the coefficients of that variable must be equal in magnitude but opposite in sign so they sum to zero. Multiplying by a constant scales the equation to achieve this. Making y-intercepts or slopes equal is not the goal of elimination.

Subtract the second equation from the first: (2x + 3y) - (2x - y) = 12 - 4 gives 4y = 8, so y = 2. Substituting back: 2x - 2 = 4, so x = 3. Choice D (y = 4) is a common error from adding the equations instead of subtracting, which gives 2y = 8 if the y-terms are handled incorrectly.

Let p = pens and n = notebooks. Set up: p + n = 8 and 2p + 5n = 25. From the first, n = 8 - p. Substitute: 2p + 5(8 - p) = 25, giving 2p + 40 - 5p = 25, so -3p = -15 and p = 5. Choice A (3) is the number of notebooks, not pens — a student who solves for n first and stops there gets the wrong quantity.

Set the equations equal: 2x - 3 = -x + 6, so 3x = 9 and x = 3. Substitute back into either equation: y = 2(3) - 3 = 3. The y-coordinate of the intersection is 3. Choice D (y = 6) is the y-intercept of the second line, not the intersection value. Always compute y after finding x.

Multiply every term in 3x + 2y = 10 by 2: 2(3x) + 2(2y) = 2(10) gives 6x + 4y = 20. Subtracting the second original equation then eliminates y and gives x = 4. Choice C (6x + 2y = 20) is wrong because the y-coefficient must also be multiplied by 2 — every term in the equation must be scaled by the same factor.

Substitute y = 2x - 5 into 3x + y = 10: 3x + (2x - 5) = 10, so 5x - 5 = 10, giving 5x = 15 and x = 3. Then y = 2(3) - 5 = 1. Choice A (x = 2) gives y = -1, but checking: 3(2) + (-1) = 5, not 10, so it fails the second equation.

Rewrite the second equation in slope-intercept form: 6x - 2y = 5 becomes y = 3x - 5/2. Both lines have slope 3 but different y-intercepts (1 and -5/2), so they are parallel and never intersect. The system has no solution. A student might choose 'exactly one solution' by assuming different-looking equations always intersect, but slope determines parallelism.

Let w = width and l = length. Set up: 2l + 2w = 40 and l = w + 4. Substitute: 2(w + 4) + 2w = 40, so 2w + 8 + 2w = 40, giving 4w = 32 and w = 8 cm. The length is 12 cm. Choice A (6 cm) is wrong: a 6 cm width with a 10 cm length gives perimeter 32 cm, not 40 cm.

Choice C has lines with slopes -1 and 2, which are different, so the lines cross at exactly one point. Choice A has two parallel lines (both slope 2, y-intercepts 3 and -1), giving no solution. Choice B: dividing the second by 2 gives y = x + 2, identical to the first, so infinitely many solutions. Choice D: dividing the first by 3 gives y = 2x + 3, identical to the second, again infinitely many solutions.

Sum means addition and difference means subtraction, so the system is x + y = 30 and x - y = 10. Solving by adding the equations: 2x = 40, so x = 20 and y = 10. Choice B reverses which condition equals 30 and which equals 10, producing an incorrect pair. Choices C and D introduce multiplication, which does not match the meaning of sum or difference.

From the second equation, y = x - 1. Substitute into the first: 2x + (x - 1)/2 = 7. Multiply every term by 2 to clear the fraction: 4x + x - 1 = 14, so 5x = 15 and x = 3. Then y = 3 - 1 = 2. Choice C (4, 3) results from forgetting to multiply the -1 constant when clearing the fraction, giving 5x = 16 instead of 15.

For infinitely many solutions, one equation must be a scalar multiple of the other. Divide the first by 2: x + 2y = 4. Divide the second by 3: x + (b/3)y = 4. For these to be identical, b/3 = 2, so b = 6. Alternatively, using ratios: 2/3 = 4/b = 8/12 = 2/3 confirms b = 6. Choosing b = 4 creates a line with a different slope, yielding exactly one intersection.

Let A and B represent each car's speed. Set up: A = B + 10 and 3A + 3B = 210. Divide the second by 3: A + B = 70. Substitute A = B + 10: (B + 10) + B = 70, so 2B = 60 and B = 30 mph. Choice A (25 mph) gives Car A at 35 mph; total distance = 3(25 + 35) = 180 miles, not 210 — verifying 30 is correct.

Let x = liters of 20% solution and y = liters of 50% solution. Set up: x + y = 12 and 0.20x + 0.50y = 0.35(12) = 4.2. From the first equation, x = 12 - y. Substitute: 0.20(12 - y) + 0.50y = 4.2, giving 2.4 + 0.30y = 4.2, so y = 6 liters. Choice A (4 liters) yields only 0.20(8) + 0.50(4) = 3.6 liters of alcohol, falling short of the required 4.2 liters.

Rewrite in slope-intercept form. The first gives y = (-m/3)x + 2 and the second gives y = -2x + 4. For no solution, the lines must be parallel: equal slopes and different y-intercepts. Set slopes equal: -m/3 = -2, so m = 6. The y-intercepts 2 and 4 are different, confirming the lines are parallel. With m = 2, the slope of the first line would be -2/3, which differs from -2, so the lines would intersect at one point.

Distribute the first equation: 3x - 3 + 2y = 9, simplifying to 3x + 2y = 12. From the second equation, y = 2x - 1. Substitute: 3x + 2(2x - 1) = 12, giving 3x + 4x - 2 = 12, so 7x = 14 and x = 2. Then y = 2(2) - 1 = 3. Choice A (x = 1) often results from failing to distribute fully, for example treating 3(x - 1) as 3x - 1 instead of 3x - 3.

Multiply 2x - y = 5 by 2 to get 4x - 2y = 10. Subtracting from 4x - 2y = 13 gives 0 = 3, which is a false statement. This contradiction shows the two lines are parallel — the left sides are identical but the right sides are different — so the system has no solution. By contrast, if the result were 0 = 0 (always true), the system would have infinitely many solutions.

Line 1 slope = (7 - 3)/(3 - 1) = 2, giving y = 2x + 1. Line 2 slope = (7 - 1)/(2 - 0) = 3, giving y = 3x + 1. Set equal: 2x + 1 = 3x + 1, so x = 0 and y = 1. The intersection is (0, 1). Choice B, (1, 3), lies on Line 1 but checking Line 2: y = 3(1) + 1 = 4, not 3. Choice D, (3, 7), lies on Line 1 but y = 3(3) + 1 = 10 for Line 2, not 7.

The solution to a system of two linear equations is the ordered pair (x, y) that satisfies both equations simultaneously. On a graph, this is the point where the two lines cross, or intersect. The y-intercept is where a single line crosses the y-axis, which is not the same as the solution to the system.

Substitute x = 2 and y = 3 into both equations. For y = x + 1: 3 = 2 + 1 = 3 (true). For y = 2x - 1: 3 = 2(2) - 1 = 3 (true). Since (2, 3) satisfies both equations, it is the solution to the system. A point must satisfy every equation in the system to count as a solution.

Both lines have the same slope (3) but different y-intercepts (5 and -2), so they are parallel and never intersect. Parallel lines produce a system with no solution. Algebraically, setting 3x + 5 = 3x - 2 simplifies to 5 = -2, a contradiction, which confirms there is no solution.

In the substitution method, you replace a variable in one equation with an equivalent expression from the other. Since the first equation is already solved for y, substitute 4x + 1 for y in the second equation: 2x + (4x + 1) = 7. Adding or multiplying equations is part of the elimination method, not substitution.

A linear equation has variables raised only to the first power with no radicals or variables in denominators. Only y = 2x + 1 and y = -x + 4 are both linear. The first option contains x², making it quadratic; the third contains a square root; and the fourth has a variable in the denominator, making those equations nonlinear.

When elimination produces 0 = 0, a true statement with no variables, it means the two equations are equivalent and describe the same line. Every point on that line is a solution, so there are infinitely many solutions. By contrast, if elimination produces 0 = 5 (a false statement), there is no solution. The result 0 = 0 is not an error.

Set the expressions equal: x + 2 = 3x - 4. Subtract x: 2 = 2x - 4. Add 4: 6 = 2x, so x = 3. Substitute back: y = 3 + 2 = 5. The solution is (3, 5) and the y-coordinate is 5. A common error is stopping at x = 3 and selecting that as the final answer instead of continuing to find y.

Substitute x = 3y - 1 into 2x + y = 12: 2(3y - 1) + y = 12 → 6y - 2 + y = 12 → 7y = 14 → y = 2. Then x = 3(2) - 1 = 5. The solution is (5, 2). A common error is reporting y = 2 as the answer for x since y is solved first.

To eliminate y by addition, the y-terms must be opposites. The first equation has +2y. Multiplying the second equation by 2 gives 4x - 2y = 2. Adding to x + 2y = 8 yields 5x = 10, so x = 2. Multiplying by -2 would give -4x + 2y = -2, and adding would eliminate x instead of y.

Let p = number of pens and n = number of notebooks. Set up the system: p + n = 8 and 2p + 5n = 25. From the first equation: p = 8 - n. Substitute: 2(8 - n) + 5n = 25 → 16 - 2n + 5n = 25 → 3n = 9 → n = 3. Then p = 8 - 3 = 5 pens. The answer 3 is the number of notebooks, not pens.

Sum means addition: x + y = 45. Difference means subtraction: x - y = 13. Choice A incorrectly uses multiplication for the difference. Choices B and D swap the 45 and 13. The correct system is x + y = 45 and x - y = 13, which can be solved to find x = 29 and y = 16.

Set the expressions equal: -2x + 6 = x - 3. Add 2x to both sides: 6 = 3x - 3. Add 3: 9 = 3x, so x = 3. Substitute: y = 3 - 3 = 0. The intersection is (3, 0). Choice D results from correctly finding x = 3 but then substituting into the wrong equation and getting y = -2(3) + 6 = 0, or making a sign error.

Notice that 4x - 2y = 10 is exactly 2 times 2x - y = 5. The two equations are equivalent and represent the same line. Every point on the line 2x - y = 5 is a solution, giving infinitely many solutions. The result 0 = 0 means the system is dependent. If the result were 0 = nonzero (like 0 = 3), the system would be inconsistent with no solution.

Since the y-coefficients are already opposites, add the equations directly: 6x = 18 → x = 3. Substitute into the first equation: 2(3) + 3y = 12 → 6 + 3y = 12 → 3y = 6 → y = 2. Therefore x + y = 3 + 2 = 5. Choice A gives only x and choice D gives only y — the question asks for x + y.

Let d = distance between cities. Time going = d/60 and time returning = d/40. Set up: d/60 + d/40 = 5. The LCD is 120: 2d/120 + 3d/120 = 5 → 5d = 600 → d = 120 miles. A common error is averaging the speeds (50 mph) and multiplying by 5, giving 250 miles, which is incorrect because the train spends more time at the slower speed.

A system has no solution when the lines are parallel: same slope but different y-intercepts. The second line has slope 2. Setting m = 2 gives y = 2x + 4 and y = 2x - 1, which are parallel (same slope, y-intercepts 4 and -1 differ). For any other value of m, the slopes differ and the lines will intersect at exactly one point.

Substitute y = 2x - 4 into 3x - y = 7: 3x - (2x - 4) = 7 → 3x - 2x + 4 = 7 → x + 4 = 7 → x = 3. Then y = 2(3) - 4 = 2. The solution is (3, 2). Choice B reverses x and y. Forgetting to distribute the negative sign — writing 3x - 2x - 4 instead of 3x - 2x + 4 — is a common error that gives an incorrect answer.

Infinitely many solutions means every point on one line is also on the other — the two equations describe the same line. This appears graphically as two coincident lines (lines on top of each other). Lines with equal slopes that never touch are parallel, giving no solution. Lines crossing at the origin or at right angles intersect at exactly one point.

Multiply each equation by 10 to clear decimals: 5x + 3y = 21 and 2x - y = 4. From the second: y = 2x - 4. Substitute: 5x + 3(2x - 4) = 21 → 5x + 6x - 12 = 21 → 11x = 33 → x = 3. Then y = 2(3) - 4 = 2. Clearing decimals first is a critical strategy because working with them directly greatly increases the chance of arithmetic errors.

For no solution, the lines must be parallel: equal slopes but different y-intercepts. Rewrite the first: y = (-a/2)x + 3, so its slope is -a/2. Rewrite the second: y = 3x - 4, so its slope is 3. Set slopes equal: -a/2 = 3 → a = -6. Check: y-intercepts are 3 and -4, which differ, confirming no solution. If a = 6, the slope would be -3, not 3, and the lines would intersect.

Let t = hours until they meet. Their combined distances must equal 15 miles: 3t + 2t = 15 → 5t = 15 → t = 3 hours. Alex walks 3 mph for 3 hours = 9 miles. Jordan walks 2 mph for 3 hours = 6 miles. Check: 9 + 6 = 15 miles total. Choice A (6 miles) is Jordan's distance, not Alex's — a common error when the question asks about a specific person.

Multiply the first equation by 2: x + y = 10. Multiply the second equation by 3: x - y = 3. Add the two results: 2x = 13 → x = 6.5. Then y = 10 - 6.5 = 3.5. Check: (6.5 - 3.5)/3 = 3/3 = 1 and (6.5 + 3.5)/2 = 10/2 = 5. Failing to clear the fractions first is the most common source of error in this type of problem.

Line 1: y = 2x - 3. For Line 2, find the slope: (1 - 5)/(3 - 1) = -4/2 = -2. Use point (3, 1): y - 1 = -2(x - 3) → y = -2x + 7. Set equal: 2x - 3 = -2x + 7 → 4x = 10 → x = 2.5. A common error is assuming Line 2 passes through the origin or using only one point to write its equation without computing the slope first.

Let x = mL of 30% solution and y = mL of 60% solution. System: x + y = 300 and 0.30x + 0.60y = 0.40(300) = 120. From the first equation: x = 300 - y. Substitute: 0.30(300 - y) + 0.60y = 120 → 90 - 0.30y + 0.60y = 120 → 0.30y = 30 → y = 100 mL. Since 40% is closer to 30% than to 60%, you need more of the 30% solution, which confirms that 100 mL (not 200 mL) of the 60% solution is correct.

Clear fractions by multiplying the first equation by 6: 3x + 2y = 24. Multiply the second equation by 3: 3x - y = 6. Subtract the second from the first: (3x + 2y) - (3x - y) = 24 - 6 → 3y = 18 → y = 6. Substitute back: 3x - 6 = 6 → 3x = 12 → x = 4. Check: 4/2 + 6/3 = 2 + 2 = 4 and 4 - 6/3 = 4 - 2 = 2. Both equations check out.

The solution to a system of equations is the ordered pair (x, y) that satisfies both equations simultaneously. Graphically, this corresponds to the point where the two lines cross. Choice A is incorrect because y-intercepts show where each individual line meets the y-axis, not where the two lines meet each other.

A consistent and independent system has lines that intersect at exactly one point, giving exactly one solution. Choice A (no solution) describes an inconsistent system of parallel lines. Choice D describes a consistent dependent system where both equations represent the same line. Two distinct lines cannot cross at two separate points, so Choice C is impossible.

Test (2, 5): y = 3(2) - 1 = 5 and y = -(2) + 7 = 5. Both equations yield y = 5, confirming (2, 5) is correct. Choice A, (1, 2): y = 3(1) - 1 = 2 satisfies the first equation, but y = -(1) + 7 = 6 does not equal 2, so it fails the second equation. A solution must satisfy both equations.

The substitution method starts by isolating one variable in one equation, creating an expression to substitute into the other equation. Choice A describes the addition step used in the elimination method. Choice D describes the graphing method. Choice B is a technique used in elimination when coefficients need to be matched.

If two equations share the same slope and the same y-intercept, they represent the identical line. Every point on that line satisfies both equations, producing infinitely many solutions. Choice A (no solution) applies when lines have the same slope but different y-intercepts, making them parallel rather than identical.

Test (4, 3): x + y = 4 + 3 = 7 and x - y = 4 - 3 = 1. Both equations are satisfied. Choice A, (3, 4): x + y = 7 checks out, but x - y = 3 - 4 = -1, which does not equal 1. A solution must satisfy every equation in the system, not just one.

Lines with different slopes are neither parallel nor identical, so they must intersect at exactly one point, giving exactly one solution. Choice A applies to parallel lines, which share the same slope but have different y-intercepts. Choice C applies when both equations describe the exact same line, which requires both the same slope and the same y-intercept.

An inconsistent system has no solution. Choice B has both lines with slope 4 but y-intercepts of -1 and 5, making them parallel — they never intersect. Choice A has different slopes (2 and 3), so those lines cross at one point. Choice D appears to be two different equations but dividing the second by 2 gives 2x + y = 5, which is the same as the first — making it dependent with infinitely many solutions.

Substitute y = 2x - 3 into the second equation: 4x + (2x - 3) = 21, which simplifies to 6x - 3 = 21, then 6x = 24, giving x = 4. To confirm, y = 2(4) - 3 = 5, and 4(4) + 5 = 21. Choice A (x = 3) results from a common error of writing 6x = 18 instead of correctly adding 3 to both sides to get 6x = 24.

Set the equations equal: (1/2)x + 3 = -2x + 8. Multiply every term by 2 to clear the fraction: x + 6 = -4x + 16, so 5x = 10 and x = 2. Then y = (1/2)(2) + 3 = 4. The intersection is (2, 4). Choice C, (3, 4.5): checking the second equation gives y = -2(3) + 8 = 2, not 4.5, so that point is not on both lines.

Adding the equations eliminates y: (3x + 2y) + (3x - 2y) = 14 + 4, giving 6x = 18 and x = 3. Substituting back into the first equation: 3(3) + 2y = 14, so 9 + 2y = 14, meaning 2y = 5 and y = 2.5. Choice B (y = 2) is a common error from forgetting to divide 5 by 2, stopping at 2y = 5 and writing y = 2 instead of 2.5.

The total item count gives n + p = 10 and the total cost gives 3n + 1.5p = 24. Choice A reverses the constants — claiming Maria bought 24 items at a total cost of $10 contradicts the problem. Choice D has the prices swapped, charging $1.50 per notebook and $3 per pen, which reverses the given price information.

Substitute x = y + 1 into 3x + 2y = 13: 3(y + 1) + 2y = 13. Distributing gives 3y + 3 + 2y = 13, which combines to 5y + 3 = 13. Choice B has a sign error — the constant should be +3, not -3. Choice C results from not combining 3y and 2y correctly, leaving only 3y instead of 5y.

Choice B has both equations with the identical left side (3x - y) but different right sides (7 and 4). This is a direct contradiction — the same expression cannot equal two different values, so there is no solution. Choice A is actually a dependent system: multiplying the first equation by -2 produces the second, meaning they represent the same line and have infinitely many solutions.

From the first equation, x = 8 - 2y. Substitute into the second: 2(8 - 2y) - y = 6, giving 16 - 4y - y = 6, so -5y = -10 and y = 2. Then x = 8 - 2(2) = 4. The solution is (4, 2). Choice A, (2, 3): checking the second equation gives 2(2) - 3 = 1, which does not equal 6, so this point is not on the second line.

For (4, -1) to be the solution, it must satisfy both equations. Choice C: y = -(4) + 3 = -1 and y = 3(4) - 13 = -1. Both check out. Choice B: the first equation gives y = 4 - 5 = -1, which passes, but the second gives y = 2(4) + 1 = 9, not -1. A solution must satisfy every equation in the system simultaneously.

Divide the first equation by 2: 3x - y = 5. For infinitely many solutions, both equations must represent the same line, so 3x - y = b must match 3x - y = 5, giving b = 5. Choice D (b = 10) is a common error — students sometimes use the constant from the original equation before simplifying, forgetting that dividing both sides by 2 changes the constant from 10 to 5.

Multiply the first equation by 3 and the second by 4 to get 9x - 12y = 30 and 8x + 12y = 4. Adding eliminates y: 17x = 34, so x = 2. Substitute back: 3(2) - 4y = 10, giving -4y = 4 and y = -1. Choice C, (3, 0): checking the second equation gives 2(3) + 3(0) = 6, which does not equal 1.

Upstream the effective speed is b - c, downstream it is b + c. The system is 3(b - c) = 36 and 2(b + c) = 36, which simplifies to b - c = 12 and b + c = 18. Adding these equations gives 2b = 30, so b = 15 mph. Choice D (18 mph) is the downstream speed b + c, not the boat's speed in still water — a common error that confuses the combined speed with the boat's own speed.

Multiply the first equation by 2 and the second by 3: 8x + 6y = -2 and 15x - 6y = 48. Adding eliminates y: 23x = 46, so x = 2. Substitute back into the first equation: 4(2) + 3y = -1, giving 8 + 3y = -1, so 3y = -9 and y = -3. Choice B (y = -2) results from a subtraction error when solving 3y = -9, incorrectly dividing to get -2 instead of -3.

Let c = chickens and k = cows. The system is c + k = 30 (total animals) and 2c + 4k = 88 (total legs, since chickens have 2 legs and cows have 4). From the first equation, c = 30 - k. Substituting: 2(30 - k) + 4k = 88, giving 60 + 2k = 88, so k = 14. Choice B (12 cows): verifying gives 2(18) + 4(12) = 36 + 48 = 84 legs, not 88.

Line 1 has slope (7 - 3)/(3 - 1) = 2 and passes through (1, 3), giving equation y = 2x + 1. Line 2 has slope (9 - 5)/(2 - 0) = 2 and y-intercept 5, giving y = 2x + 5. Both lines have slope 2 but different y-intercepts, so they are parallel and never intersect. Choice C is a common misconception — same slope alone means parallel, not the same line. Infinitely many solutions require both the same slope and the same y-intercept.

Divide the second equation by 2 to get 2x + 3y = 6. For infinitely many solutions, both equations must be identical, so 2x + ky = 6 must match 2x + 3y = 6, requiring k = 3. Choice D (k = 6) is a common error — students use the coefficient 6 directly from the original second equation before dividing, forgetting that dividing by 2 reduces the y-coefficient from 6 to 3.

Substitute x = 2 into the first equation: 2 + 2y = 8, so 2y = 6 and y = 3. Now substitute x = 2 and y = 3 into the second equation: k = 3(2) - 3 = 6 - 3 = 3. Choice D (k = 9) comes from incorrectly adding instead of subtracting: 3(2) + 3 = 9 uses the wrong operation. Choice C (k = 5) results from using only x in the second equation and ignoring y: k = 3(2) - 1 = 5 is not valid.

Let a = pounds of Variety A and b = pounds of Variety B. The system is a + b = 30 and 9a + 15b = 12(30) = 360. Substituting a = 30 - b into the cost equation: 9(30 - b) + 15b = 360, giving 270 - 9b + 15b = 360, so 6b = 90 and b = 15. Choice A (10 pounds) would give a blend cost of 9(20) + 15(10) = 180 + 150 = 330, which is only $11 per pound rather than the target $12.

Substitute (1, 2) into each equation. For y = x + 1: 2 = 1 + 1 = 2, which is true. For y = 3x - 1: 2 = 3(1) - 1 = 2, which is also true. Since (1, 2) satisfies both equations, it is the solution. A common error is checking only one equation — a solution must satisfy every equation in the system.

Parallel lines never intersect, so there is no point that lies on both lines at the same time. A solution to a system must satisfy both equations, which corresponds graphically to a point of intersection. Since parallel lines have no intersection, the system has no solution. This is called an inconsistent system.

The solution to a system of equations is an ordered pair (x, y) that satisfies both equations simultaneously. Graphically, every point on a line satisfies that line's equation. Therefore, a point that lies on both lines — the intersection point — is the only point satisfying both equations at once. The y-intercepts and x-intercepts are specific points on individual lines, not necessarily the solution.

A system has infinitely many solutions when both equations describe the same line. Dividing the second equation, 4x + 2y = 12, by 2 gives 2x + y = 6 — identical to the first. Choice A has the same slope but different y-intercepts, making them parallel with no solution. Choices B and D have lines with different slopes that intersect at exactly one point.

Substitute y = 2x into x + y = 9: x + 2x = 9, so 3x = 9 and x = 3. Then y = 2(3) = 6. The solution is (3, 6). Note that choices A, B, and D all satisfy y = 2x, making them plausible distractors — but only (3, 6) also satisfies x + y = 9, since 3 + 6 = 9.

The substitution method works by isolating one variable in one equation and then replacing (substituting) that variable in the other equation with the equivalent expression. This reduces the system to a single equation with one unknown, which can be solved directly. Choice B describes the addition step of the elimination method, not substitution.

A consistent system has at least one solution; an independent system means the equations are not multiples of each other (the lines are not identical). Together, 'consistent and independent' means the two lines intersect at exactly one point, giving exactly one solution. A system with no solution is inconsistent. A system with infinitely many solutions is consistent but dependent.

Since the first equation already expresses x in terms of y, substitute x = 2y + 1 into 3x - y = 8: 3(2y + 1) - y = 8, which gives 6y + 3 - y = 8, then 5y = 5, so y = 1. Substitute back: x = 2(1) + 1 = 3. A common mistake is forgetting to distribute the 3 across (2y + 1), which would incorrectly give 6y + 1 instead of 6y + 3.

The coefficients of y are +2 and -2, so adding the equations eliminates y: (5x + 2y) + (3x - 2y) = 12 + 4, giving 8x = 16, so x = 2. Substituting back: 5(2) + 2y = 12 gives y = 1. Choosing x = 3 is a common error that results from adding incorrectly or misreading the signs on the y-terms.

Let x be the larger number and y the smaller. The system is x + y = 30 and x - y = 8. Adding both equations eliminates y: 2x = 38, so x = 19. Then y = 30 - 19 = 11. Verify: 19 + 11 = 30 and 19 - 11 = 8. Choice 15 results from simply halving the sum without using the difference equation.

Multiply the second equation by 3: 3(x - 2y) = 3(3) gives 3x - 6y = 9, which is identical to the first equation. Both equations represent the same line. Every point on this line satisfies both equations, so there are infinitely many solutions. This is a dependent system. A student might incorrectly expect one solution by treating these as distinct equations.

Let a = adult tickets and s = student tickets. The system is a + s = 200 and 6a + 4s = 1,000. Substitute s = 200 - a into the second equation: 6a + 4(200 - a) = 1,000, which gives 2a + 800 = 1,000, so 2a = 200 and a = 100. Check: 100 + 100 = 200 tickets and 6(100) + 4(100) = 1,000. Choosing 90 or 75 results from arithmetic errors in distributing or combining terms.

Substitute y = 3x - 7 into 2x - y = 5: 2x - (3x - 7) = 5. Distribute the negative sign carefully: 2x - 3x + 7 = 5, giving -x + 7 = 5, so -x = -2 and x = 2. Then y = 3(2) - 7 = -1. A very common error is failing to distribute the negative sign, giving 2x - 3x - 7 = 5 and a wrong answer of x = -12.

Subtract the second equation from the first: (x + 4y) - (x - 2y) = 11 - (-1). The x terms cancel: 6y = 12, so y = 2. Substituting back into x - 2y = -1: x - 4 = -1, giving x = 3. The distractor y = 4 comes from incorrectly computing 11 - (-1) = 10 instead of 12, or from forgetting to subtract the -2y correctly.

Let w = width and l = length. The system is 2l + 2w = 38 and l = w + 5. Substitute into the perimeter equation: 2(w + 5) + 2w = 38, giving 2w + 10 + 2w = 38, so 4w = 28 and w = 7. The length is l = 12. Check: 2(12) + 2(7) = 24 + 14 = 38. Choosing 9 is tempting if a student divides 38 by 4 without accounting for the 5 cm difference.

Add the two equations: (2x + y) + (x - y) = 9 + 3, giving 3x = 12 and x = 4. Substitute into x - y = 3: 4 - y = 3, so y = 1. The solution is (4, 1). Choices A, B, and D are plausible because (1, 7) and (3, 3) both satisfy 2x + y = 9, and (5, -1) satisfies x - y = 3 — but none of those satisfy both equations simultaneously.

Lines with the same slope are parallel — they run in the same direction and never cross. If they also have different y-intercepts, they are distinct parallel lines with no intersection point, so the system has no solution. Choice A describes the same line (infinitely many solutions). Choices C and D both involve different slopes, meaning the lines will always intersect at exactly one point.

Substitute y = x + 1 into 4x + 2y = 14: 4x + 2(x + 1) = 14, which gives 4x + 2x + 2 = 14, so 6x = 12 and x = 2. Then y = 2 + 1 = 3. Check: 4(2) + 2(3) = 8 + 6 = 14. A student who forgets to multiply both terms inside the parentheses may write 4x + 2x + 1 = 14, getting x = 13/6 — a sign that the distribution step was skipped.

Let x be the larger number and y the smaller. The system is x + y = 50 and x - y = 14. Adding gives 2x = 64, so x = 32, and y = 18. The question asks for 2x + 3y = 2(32) + 3(18) = 64 + 54 = 118. Choosing 112 results from computing 2x + 2y = 100 and adding 12 incorrectly. This problem requires two steps: first solving the system, then applying a second calculation to the result.

First, solve Saturday's system: a + c = 150 and 12a + 8c = 1,560. Substituting c = 150 - a gives 12a + 8(150 - a) = 1,560, so 4a = 360 and a = 90, c = 60. The adult-to-child ratio is 90:60 = 3:2. For Sunday's 200 tickets at the same ratio, adults = 200 times (3/5) = 120 and children = 80. Revenue = 12(120) + 8(80) = 1,440 + 640 = $2,080. This problem requires using a system to find a ratio, then applying that ratio to a new scenario.

For infinitely many solutions, both equations must represent the same line — one must be a scalar multiple of the other. Multiply x - 2y = 2 by 3: 3x - 6y = 6. Comparing with 3x - ky = 6, the coefficients must match, so k = 6. If k = 3, the second equation multiplied by 3 gives 3x - 6y = 6, which still requires k = 6, not 3. Choosing k = 2 confuses the coefficient in the original second equation with the needed value.

To eliminate y, multiply the first equation by 5 and the second by 2: 15x + 10y = 35 and 4x - 10y = 22. Add: 19x = 57, so x = 3. Substitute into the first equation: 9 + 2y = 7, giving 2y = -2 and y = -1. Therefore x + y = 3 + (-1) = 2. Choosing 4 is a common error if students find x = 3 and then add y = 1 (using an absolute value or sign error when solving for y).

Let c = car speed and t = train speed. The system is t = 2c and 3t + 3c = 270 (distance apart equals sum of distances traveled in opposite directions). Substitute t = 2c into the second equation: 3(2c) + 3c = 270, giving 9c = 270 and c = 30 mph. The train travels at 60 mph. Choosing 45 mph is a common error from dividing 270 by 3 and then halving, which ignores that the speeds are summed, not equated.

Solve by substitution: y = 7 - 2x. Substitute into x - 3y = -14: x - 3(7 - 2x) = -14, giving x - 21 + 6x = -14, so 7x = 7 and x = 1. Then y = 7 - 2(1) = 5. So a = 1, b = 5. The value a squared - b squared = 1 - 25 = -24. Choosing 24 results from computing b squared - a squared instead. Choosing 16 may come from computing (a - b) squared = (1 - 5) squared = 16, which is not the same expression.

Let x = liters of 20% solution. The total acid before and after must be equal: 0.20x + 0.50(12) = 0.30(x + 12). Expanding: 0.20x + 6 = 0.30x + 3.6. Subtract 0.20x from both sides: 6 = 0.10x + 3.6, so 0.10x = 2.4 and x = 24. Choosing 12 liters comes from setting the two solution amounts equal rather than equating acid amounts. Choosing 18 liters results from incorrectly setting up 0.20x + 0.50(12) = 0.30(12) instead of 0.30(x + 12).

A system has exactly one solution when the two lines have different slopes and therefore intersect at exactly one point. In choice C, the slopes are -1 and 2 — different slopes, so the lines cross once. Choice A has two lines with the same slope (4) but different y-intercepts, making them parallel with no solution. Choice B: dividing the second equation by 2 gives the first equation exactly, so they are the same line with infinitely many solutions. Choice D: rewriting 2y = x + 8 as y = (1/2)x + 4 shows the same slope as the first line but a different y-intercept, so those lines are also parallel with no solution.

Substitution means replacing y in the second equation with the expression that equals y from the first equation. Since y = 3x - 5, you substitute into 2x + y = 9 to get 2x + (3x - 5) = 9. Choice B incorrectly subtracts the expression instead of adding it. Choice C has a sign error, writing +5 instead of -5. Choice D still contains y and therefore is not a completed substitution — the whole point is to eliminate one variable.

When two lines are identical, every single point on one line is also on the other. Every such point satisfies both equations simultaneously, producing infinitely many solutions. Choice C is a very common error: identical equations do NOT produce no solution — they produce infinitely many. A system has no solution only when the lines are parallel and distinct (same slope, different y-intercept). The number of intersection points determines the number of solutions: none, one, or infinitely many.

To eliminate y, multiply the first equation by 2 to get 4x + 2y = 14. Adding this to the second equation 5x - 2y = 4 gives 9x = 18, so x = 2. Substituting back: 2(2) + y = 7 gives y = 3. You can verify: 5(2) - 2(3) = 10 - 6 = 4. A common error is multiplying only the y-term rather than the entire equation, which leads to incorrect coefficients and answers like x = 1 or x = 3.

Isolate y in the first equation: y = 3x - 7. Substitute into the second equation: x + 2(3x - 7) = 14, giving x + 6x - 14 = 14, so 7x = 28 and x = 4. Then y = 3(4) - 7 = 5, and the solution is (4, 5). Note that choices A, B, and D each satisfy the first equation 3x - y = 7 (check: 6-(-1)=7, 9-2=7, 15-8=7), but all three fail the second equation. Students who only verify one equation will be misled by these plausible distractors.

Let d = number of dimes and q = number of quarters. Set up the system: d + q = 20 and 10d + 25q = 305 (working in cents to avoid decimals). From the first equation, d = 20 - q. Substituting: 10(20 - q) + 25q = 305 gives 200 - 10q + 25q = 305, so 15q = 105 and q = 7. Then d = 13. Verify: 13(10) + 7(25) = 130 + 175 = 305 cents = $3.05. Students who substitute incorrectly or forget to distribute the 10 may arrive at 5 or 9.

Downstream, the current adds to the boat's speed: distance = (b + c) x time, so (b + c)(2) = 36, which simplifies to b + c = 18. Upstream, the current subtracts: (b - c)(3) = 24, which simplifies to b - c = 8. Choice A skips dividing by time, incorrectly equating speed directly to distance. Choice C sets up the downstream equation correctly but uses b + c (downstream formula) for the upstream trip instead of b - c. Choice D swaps the current's effect entirely, applying b - c to the downstream trip and b + c to the upstream trip.

Multiply the second equation by 2 to get 10x - 4y = 16. Adding this to the first equation eliminates y: 13x = 26, so x = 2. Substituting back into the first equation: 3(2) + 4y = 10, giving 4y = 4 and y = 1. The solution is (2, 1). Choice D satisfies the first equation (3(0) + 4(2.5) = 10) but fails the second (5(0) - 2(2.5) = -5, not 8). Choice A reverses the coordinates — a common careless error when writing the ordered pair after solving.

Let a = older age and b = younger age. The system is: a + b = 48 and (a + 6) = 2(b + 6). Expanding the second equation: a + 6 = 2b + 12, so a = 2b + 6. Substituting: (2b + 6) + b = 48 gives 3b = 42, so b = 14 and a = 34. Verify: in 6 years they are 40 and 20, and 40 = 2(20). Choice A (30) results from only adding 6 to the older friend's side: writing a + 6 = 2b instead of 2(b + 6). Choice B (32) comes from ignoring the 6-year shift entirely and writing a = 2b directly. Choice D (36) comes from incorrectly simplifying a + 6 = 2b + 12 as a = 2b + 12 instead of a = 2b + 6.

Let t = tens digit and u = units digit. The two conditions give the system: t + u = 11 and (10u + t) - (10t + u) = 27. Simplifying the second equation: 9u - 9t = 27, or u - t = 3. Adding t + u = 11 and u - t = 3 gives 2u = 14, so u = 7 and t = 4, making the original number 47. All four answer choices have a digit sum of 11, so that condition alone cannot distinguish the answer. You must also check the reversed-number condition: 74 - 47 = 27, while 83 - 38 = 45, 65 - 56 = 9, and 92 - 29 = 63 — none of which equal 27.