Writing Linear Equations — Free Algebra 1 Review Games.

Slope-intercept form: y = mx + b = 2x - 3.

y = mx + b is the slope-intercept form where m is slope and b is y-intercept.

y-intercept is 4 (the point is on y-axis), so y = x + 4.

Move terms: -2x + y = -6, multiply by -1: 2x - y = 6.

Horizontal lines have equation y = constant. Here y = -2.

Point-slope form: y - y1 = m(x - x1) = y - 5 = 3(x - 2).

Slope = (7-3)/(3-1) = 2. Using (1,3): y-3 = 2(x-1), y = 2x+1.

Same slope -1: y - 1 = -1(x - 2), y = -x + 3.

Perpendicular slope = -1/2. y - 0 = -1/2(x - 4), y = -x/2 + 2.

Fixed cost $30 is the y-intercept, $0.10 per text is the slope.

Slope = (-5-4)/(2-(-1)) = -9/3 = -3. y-4 = -3(x+1), y = -3x+1.

-2y = -3x + 8, y = 3x/2 - 4.

Points (4,0) and (0,-2). Slope = (-2-0)/(0-4) = 1/2. y = x/2 - 2.

A line parallel to the x-axis is horizontal: y = 5.

Starts at 20000, decreases 1500 per year: V = -1500t + 20000.

Slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept. With slope m = -3 and y-intercept b = 5, the equation is y = -3x + 5. Choice A swaps the slope and y-intercept values, a very common error.

In slope-intercept form y = mx + b, the value b is the y-intercept. Here b = -7. The value 4 is the slope, not the y-intercept. The sign matters: the equation subtracts 7, so the y-intercept is -7, not 7.

Slope is calculated as rise divided by run. A vertical line has zero change in x (run = 0), making division by zero undefined. Vertical lines are written as x = constant, so x = -2 has an undefined slope. The line y = 3 is horizontal with slope 0 — defined, but equal to zero.

Point-slope form is y - y1 = m(x - x1), where (x1, y1) is a known point and m is the slope. This equation uses the point (1, 3) and slope 2. Standard form would look like Ax + By = C, and slope-intercept form would be y = mx + b.

To find the slope, rewrite in slope-intercept form: subtract 2x from both sides to get y = -2x + 8. The coefficient of x is the slope, which is -2. A common mistake is reading the coefficient of x directly from standard form (giving 2) without first isolating y.

A line has no y-intercept only when it never crosses the y-axis. Vertical lines (x = constant) run parallel to the y-axis and never intersect it (when the constant is nonzero). The line y = 2x passes through the origin, so its y-intercept is 0. The line y = -5 has y-intercept -5.

Passing through the origin means the y-intercept is 0. With slope -1, the equation is y = -1x + 0, which simplifies to y = -x. Choice B has slope -1 but a y-intercept of 1, meaning it passes through (0, 1), not the origin.

Point-slope form: y - 3 = 4(x - (-1)) = 4(x + 1). Distribute: y - 3 = 4x + 4. Add 3 to both sides: y = 4x + 7. A frequent error is writing 4(x - 1) instead of 4(x + 1), which ignores the negative sign in the x-coordinate -1.

First find the slope: m = (5 - (-3)) / (6 - 2) = 8 / 4 = 2. Then use point-slope with (2, -3): y - (-3) = 2(x - 2), so y + 3 = 2x - 4, giving y = 2x - 7. Choice C incorrectly uses the y-coordinate -3 as the y-intercept without completing the slope calculation.

Subtract 3x from both sides: 4y = -3x + 12. Divide every term by 4: y = -3/4 x + 3. Choice B drops the negative sign on the slope. Choice C inverts the fraction incorrectly. Dividing only the constant (giving y = -3x + 3) is another common error.

Parallel lines have the same slope, so m = 2/3. Using point-slope: y - 4 = 2/3(x - 6) = 2/3 x - 4. Adding 4: y = 2/3 x. The y-intercept is 0, so the line passes through the origin. Choice C uses the negative reciprocal slope (-3/2), which would be perpendicular, not parallel.

The base fare is a fixed charge regardless of distance, so it is the y-intercept. The per-mile rate is the slope. The model is C = 1.75m + 2.50. Choice A swaps the rate and the base fare. Choice D adds them into a single rate, which incorrectly implies no base fare when m = 0.

Perpendicular lines have slopes that are negative reciprocals. The slope of y = 3x - 2 is 3, so the perpendicular slope is -1/3. Using (3, 1): y - 1 = -1/3(x - 3) = -1/3 x + 1. Adding 1: y = -1/3 x + 2. Choice A keeps the original slope 3, which gives a parallel line rather than a perpendicular one.

Slope = (11 - 7) / (4 - 2) = 4 / 2 = 2. Using (2, 7): y - 7 = 2(x - 2) = 2x - 4. Adding 7: y = 2x + 3. Choice A results from the error of subtracting 2 from 7 (giving 5) instead of setting up the full equation properly.

Subtract 5x from both sides: -2y = -5x - 10. Divide by -2 (reversing both signs): y = 5/2 x + 5. Choice C results from forgetting to flip the sign on the x-coefficient when dividing by -2. Choice B forgets to flip the sign on the constant term.

Start with slope-intercept: y - 5 = -2(x - 3) = -2x + 6, so y = -2x + 11. Rearrange to standard form by adding 2x: 2x + y = 11. Standard form typically requires the coefficient of x to be positive, which rules out choice C. Choice D results from an arithmetic error in computing the constant.

When C = 0, F = 32, so 32 is the y-intercept. The rate of change (slope) is 1.8 degrees F per degree C. This gives F = 1.8C + 32. Choice A subtracts 32 instead of adding it, which would incorrectly give F = -32 when C = 0.

The y-intercept is -3 from point (0, -3). The x-intercept gives the point (5, 0). Slope = (0 - (-3)) / (5 - 0) = 3/5. The equation is y = 3/5 x - 3. Choice B has the correct slope but the wrong sign on the y-intercept. Choice A inverts the slope fraction.

Slope = (-5 - 7) / (2 - (-4)) = -12 / 6 = -2. Using (-4, 7): y - 7 = -2(x + 4) = -2x - 8, so y = -2x - 1. Rearranging with a positive x-coefficient: 2x + y = -1. Choice C incorrectly places a positive 1 on the right side. Choice D keeps the negative sign on x, violating standard form convention.

Rewrite 4x - 3y = 9 as y = 4/3 x - 3, giving slope 4/3. The perpendicular slope is the negative reciprocal: -3/4. Using (4, -2): y + 2 = -3/4(x - 4) = -3/4 x + 3. Subtracting 2: y = -3/4 x + 1. Choice A uses the original slope 4/3, which is the slope of the given line, not the perpendicular.

Slope = (500 - 350) / (5 - 2) = 150 / 3 = $50 per month. Using point (2, 350): B - 350 = 50(m - 2) = 50m - 100, so B = 50m + 250. At m = 0, B = $250 (initial deposit). Choice B incorrectly uses 350 directly as the y-intercept without accounting for the slope over those 2 months.

The slope of the line through (-3, 5) and (1, -3) is (-3 - 5) / (1 - (-3)) = -8 / 4 = -2. The perpendicular slope is the negative reciprocal: 1/2. Using (1, 2): y - 2 = 1/2(x - 1) = 1/2 x - 1/2. Adding 2: y = 1/2 x + 3/2. Choice B uses slope -2, the slope of the original line, which would give a parallel line instead.

Slope = (0 - 3) / (5 - 2) = -3 / 3 = -1. Using (5, 0): y - 0 = -1(x - 5), so y = -x + 5. Rearranging: x + y = 5. Here A = 1, B = 1, C = 5, and gcd(1, 1, 5) = 1 with A > 0. Therefore A + B + C = 1 + 1 + 5 = 7. Choice A (5) confuses the value of C alone with the requested sum.

Set up the slope equation: (3k - 1 - k) / (4 - (-2)) = 2, which gives (2k - 1) / 6 = 2. Solving: 2k - 1 = 12, so k = 6.5. The point (-2, 6.5) is on the line: y - 6.5 = 2(x + 2) = 2x + 4. Adding 6.5: y = 2x + 10.5. Choice B incorrectly uses k = 6.5 as the y-intercept without substituting back through the point-slope equation.

Slope = (90 - 45) / (25 - 10) = 45 / 15 = 3 thousand dollars per unit. Using point (10, 45): R - 45 = 3(n - 10) = 3n - 30, so R = 3n + 15. At n = 0, R = 15, meaning $15,000. Choice B incorrectly uses 45 as the y-intercept, which would mean R = 75 at n = 10, contradicting the given data.

Point-slope form is y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is a known point. Choice A is slope-intercept form and Choice B is standard form. Choice D is just a rearrangement of slope-intercept form, not point-slope.

Solve for y: 6y = -3x + 12, so y = -½x + 2. The slope is the coefficient of x, which is -1/2. A common mistake is to use the coefficient 3 or 6 directly from standard form without isolating y first.

Slope-intercept form is y = mx + b where m is the slope and b is the y-intercept. Since the line passes through (0, 4), the y-intercept is 4 and the slope is 3, giving y = 3x + 4. Choice B incorrectly swaps the slope and intercept values.

Standard form is Ax + By = C, where A, B, and C are integers and A is non-negative. Only 3x - 4y = 12 fits this pattern. Choice A is slope-intercept form, Choice B is point-slope form, and Choice D has a fraction.

Point-slope form is y - y₁ = m(x - x₁). Substituting m = 5, x₁ = 1, and y₁ = 2 gives y - 2 = 5(x - 1). Choice A incorrectly swaps the x and y values from the point. Choice C uses the wrong signs.

To find the y-intercept, substitute x = 0: 5y = 20, so y = 4. The y-intercept is 4. Choice C (10) results from dividing 20 by 2, and Choice D (20) is the constant without dividing by the coefficient of y.

A line through the origin has a y-intercept of 0, so b = 0 in y = mx + b. With slope -3, the equation is y = -3x. Choice A has slope -3 but a non-zero y-intercept, so it does not pass through the origin.

Slope equals rise divided by run. With rise = 4 and run = 2, slope = 4/2 = 2. Choice A (1/2) reverses the rise and run. Choice B just states the rise without dividing, and Choice C multiplies instead of divides.

Start with point-slope form: y - 1 = -4(x - 3), which gives y - 1 = -4x + 12, so y = -4x + 13. Rearranging to standard form: 4x + y = 13. Choice B results from sign errors when rearranging the equation.

The y-intercept is -2 (the point (0, -2)). The slope is (0 - (-2))/(5 - 0) = 2/5. Substituting into y = mx + b gives y = (2/5)x - 2. Choice B uses a negative slope, but the line rises from left to right, so the slope must be positive.

The flat fee of $75 is a fixed starting value (y-intercept), and $50 per hour is the rate of change (slope). This gives C = 50h + 75. Choice A incorrectly uses 75 as the rate and 50 as the flat fee. Choice D would only be correct if there were no flat fee.

Distribute: y - 4 = -(1/3)x - 2. Add 4 to both sides: y = -(1/3)x + 2. Choice B results from forgetting to distribute the negative sign to the 6. Choice D uses a positive slope, which contradicts the original equation.

First, find the slope of 3x - y = 7: rewrite as y = 3x - 7, so slope = 3. Parallel lines share the same slope. Using point-slope with slope 3 through (2, 1): y - 1 = 3(x - 2), giving y = 3x - 5. Choice D uses the reciprocal slope instead.

The cost is $0.10 per text (variable rate) plus a fixed $30 monthly fee. This gives C = 0.10t + 30. Choice B incorrectly uses 30 as the per-text rate. Choices A and C have incorrect operations or order that do not match the described pricing structure.

The y-intercept is 6 (the point (0, 6)). The slope is (6 - 0)/(0 - (-3)) = 6/3 = 2. Substituting into slope-intercept form: y = 2x + 6. Choice A uses a negative slope, but the line rises from left to right, making the slope positive.

To find the x-intercept, substitute y = 0: 5x = 15, so x = 3. The x-intercept is (3, 0). Choice A (x = 5) results from dividing the constant by the coefficient of y rather than x.

First, find slope: (3 - (-5))/(4 - 0) = 8/4 = 2. Using point-slope form with (4, 3): y - 3 = 2(x - 4). Choice B incorrectly uses the opposite signs for both coordinates. Choice C uses slope 1/2, which reverses the rise and run.

Rewrite 2x + 5y = 10 as y = -(2/5)x + 2, so its slope is -2/5. A perpendicular slope is the negative reciprocal: 5/2. Using point-slope through (1, 4): y - 4 = (5/2)(x - 1) gives y = (5/2)x - 5/2 + 4 = (5/2)x + 3/2. Choice A results from a sign error when combining the constant terms.

For collinear points, the slope between any two must be equal. Slope AB = (9 - 3)/(2 - (-1)) = 6/3 = 2. For BC: (k - 9)/(5 - 2) = 2, so k - 9 = 6, giving k = 15. Choice A results from adding 4 (the run from B to C) twice to 9 instead of computing slope correctly.

The slope is (6 - 0)/(0 - (-4)) = 6/4 = 3/2. With y-intercept 6: y = (3/2)x + 6. Multiply through by 2: 2y = 3x + 12. Rearranging: 3x - 2y = -12. Choice D has the wrong sign on the constant. Choice C is equivalent but not fully reduced (not in simplest integer form).

Slope = (27 - 15)/(5 - 2) = 12/3 = 4 cm per week. Using point-slope with (2, 15): h - 15 = 4(w - 2), giving h = 4w + 7. At w = 10: h = 40 + 7 = 47 cm. Choice A (39 cm) results from incorrectly multiplying 4 by 10 and subtracting 1 instead of using the full equation.

Rewrite the first line: y = (3/2)x - 3, so slope = 3/2. The parallel line has the same slope and y-intercept 5: y = (3/2)x + 5. Multiply by 2: 2y = 3x + 10, giving 3x - 2y = -10. Choice D uses +10, which would mean a y-intercept of -5, not +5.

Set up the slope equation: (2a + 1 - 3)/(5 - a) = 2. Simplify the numerator: (2a - 2)/(5 - a) = 2. Cross-multiply: 2a - 2 = 10 - 2a. Adding 2a to both sides: 4a - 2 = 10, so 4a = 12 and a = 3. Verify: slope from (3, 3) to (5, 7) is 4/2 = 2. Choice A (a = 1) fails verification: slope from (1, 3) to (5, 3) would be 0.

The slope is (17500 - 5500)/(500 - 100) = 12000/400 = 30. In a cost model C = mn + b, the slope m represents the variable cost per unit, meaning each additional unit costs $30 to produce. Choice A describes the y-intercept (the fixed cost), not the slope. Choice D describes average cost, which changes with n and differs from the marginal cost.

Slope of Line 1 = (-1 - 5)/(1 - (-2)) = -6/3 = -2. The perpendicular slope is the negative reciprocal: 1/2. Since Line 2 passes through (0, 4), its y-intercept is 4. So Line 2 is y = (1/2)x + 4. Choice B has the same slope as Line 1, making it parallel rather than perpendicular. Choice C uses slope 2, which is not the negative reciprocal of -2.

Point-slope form is y - y1 = m(x - x1). With m = 3 and the point (2, 5), substitute to get y - 5 = 3(x - 2). Choice D reverses the coordinates, using the point (5, 2) instead of (2, 5).

Standard form is Ax + By = C, where A, B, and C are integers and A is non-negative. Only 4x - 3y = 12 fits this structure. Choice A is slope-intercept form. Choices B and D are written in point-slope form.

Slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept. With m = -2 and b = 7, the equation is y = -2x + 7. Choice A incorrectly switches the roles of slope and y-intercept.

Solve for y: subtract 6x to get -2y = -6x + 10, then divide by -2 to get y = 3x - 5. The slope is the coefficient of x, which is 3. Choice A incorrectly reads the coefficient of x before isolating y. Choice B uses the correct magnitude but the wrong sign.

Distribute the right side: y - 1 = 4x - 12. Then add 1 to both sides: y = 4x - 11. Choice B makes a sign error when distributing 4 times -3. Choice D incorrectly adds 1 and 12 instead of subtracting: -12 + 1 = -11, not +13.

Using slope-intercept form y = mx + b with slope m = 2 and y-intercept b = -3, the equation is y = 2x - 3. Choice A switches the slope and y-intercept values. Choice C uses the positive value of the y-intercept instead of -3.

The base fare of $3.00 is the starting cost regardless of distance (y-intercept), and $2.50 per mile is the rate of change (slope). The model is C = 2.50m + 3. Choice A incorrectly uses $3 as the per-mile rate and $2.50 as the flat fee.

Find the slope: m = (7 - (-1))/(6 - 2) = 8/4 = 2. Use point-slope with (2, -1): y + 1 = 2(x - 2), so y + 1 = 2x - 4 and y = 2x - 5. Choice A has the correct slope but an incorrect y-intercept. Choice C uses the rise of 8 (not divided by the run) as the slope.

Subtract 4x from both sides: 2y = -4x + 8. Divide every term by 2: y = -2x + 4. Choice A loses the negative sign on the slope. Choice C divides only part of the equation by 2, leaving -4x instead of -2x.

The slope is m = (2 - 0)/(0 - 4) = -1/2. In slope-intercept form: y = -(1/2)x + 2. Multiply every term by 2: 2y = -x + 4. Rearrange: x + 2y = 4. Choice B results from incorrectly computing slope as 2 instead of -1/2.

The y-value increases by 3 for each unit increase in x, so the slope is 3. When x = 0, y = 5, confirming the y-intercept is 5. The equation is y = 3x + 5. Choice A switches the slope and y-intercept. Choice D uses the y-value at x = 1 as the slope.

Distribute: y + 2 = (3/2)x - 6, so y = (3/2)x - 8. Multiply every term by 2: 2y = 3x - 16. Rearrange: 3x - 2y = 16. Choice B has the correct left side but a wrong sign for the constant. Choice D swaps the coefficients of x and y.

The pool starts at 2,400 gallons (the y-intercept) and decreases by 60 gallons each hour (negative rate of change). The equation is W = 2400 - 60t. Choice C adds water over time instead of removing it. Choice A produces negative values immediately, which is not physically meaningful here.

Substitute into slope-intercept form: 11 = 5(3) + b, so 11 = 15 + b and b = -4. The y-intercept is -4. Choice B makes a sign error, computing b = 4 instead of -4. Choice C results from subtracting in the wrong order: 11 - 15 = -4, not -26.

Substitute x = 4 and y = k: 2(4) - 3k = 2, so 8 - 3k = 2, giving 3k = 6 and k = 2. Choice D finds 3k = 6 correctly but forgets to divide by 3. Choice B makes a sign error when solving 8 - 3k = 2.

The rate of change is (12 - 14)/(50 - 0) = -2/50 = -1/25 gallons per mile. The starting value (at m = 0) is 14 gallons. The model is G = -(1/25)m + 14. Choice A uses the reciprocal 25 as the rate. Choice D uses 12 as the initial value rather than 14.

With slope 2/3 and y-intercept 0, the equation is y = (2/3)x. Multiply every term by 3: 3y = 2x. Rearrange: 2x - 3y = 0. Choice B reverses the coefficients of x and y. Choice A sets C = 1, but a line through the origin must have C = 0.

Rewrite in slope-intercept form: y - 3 = -2x - 2, so y = -2x + 1. Check each point: (2, 5): y = -4 + 1 = -3, not 5. (0, 1): y = 0 + 1 = 1, which matches. (1, 3): y = -2 + 1 = -1, not 3. (-1, -1): y = 2 + 1 = 3, not -1. Only (0, 1) works.

Rewrite 2x - y = -4 as y = 2x + 4, giving a y-intercept of 4. Line L passes through (1, 5) and (0, 4), so its slope is (5 - 4)/(1 - 0) = 1. The equation is y = x + 4. Choice B uses y = 5 (the y-coordinate of the given point) as the y-intercept rather than 4.

For Line 1: substitute (2, -1): -1 = 2m + 5, so 2m = -6 and m = -3. For Line 2: substitute (2, -1): -1 = -3(2) + b = -6 + b, so b = 5. Therefore m + b = -3 + 5 = 2. Choice A results from computing b as -5 instead of +5.

Set the slope equal to 3: (10 - 4)/(2k - k) = 6/k = 3, so k = 2. The points become (2, 4) and (4, 10). Use point-slope with (2, 4): y - 4 = 3(x - 2), giving y = 3x - 2. Choice D incorrectly uses 6 (the rise) as the slope instead of 3.

Find the slope: m = (13 - 7)/(5 - 2) = 6/3 = 2. Find b using f(2) = 7: 7 = 2(2) + b, so b = 3 and f(x) = 2x + 3. Then f(0) = 3 and f(10) = 23, giving f(0) + f(10) = 26. Choice A gives only f(10) = 23 without adding f(0) = 3.

Line A has slope (4 - 0)/(0 - (-2)) = 2 and equation y = 2x + 4. Line B has perpendicular slope -1/2 and equation y = -(1/2)x - 1. Set equal: 2x + 4 = -(1/2)x - 1, so (5/2)x = -5 and x = -2. Then y = 2(-2) + 4 = 0. The intersection is (-2, 0). Choice A is a point on Line B but not the intersection of both lines.

The monthly rate is the slope: (12400 - 5400)/(8 - 3) = 7000/5 = $1,400/month. Find the setup fee: 5400 = 1400(3) + b gives b = 5400 - 4200 = $1,200. For 12 months: C = 1400(12) + 1200 = 16800 + 1200 = $18,000. Choice B omits the $1,200 setup fee when computing the 12-month total.

If both intercepts equal a, the line passes through (a, 0) and (0, a). The intercept form x/a + y/a = 1 simplifies to x + y = a. Substitute (3, 5): 3 + 5 = 8, so a = 8 and the equation is x + y = 8. Choice C incorrectly multiplies the coordinates (3 times 5 = 15) instead of adding them.

The point-slope form is y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is a known point on the line. Choice A is slope-intercept form, C is standard form, and D incorrectly swaps the coordinates in the formula.

Standard form is Ax + By = C, where A, B, and C are integers. Only 4x + 3y = 12 fits this structure. Choices A and D are in slope-intercept form, and Choice B is in point-slope form.

Point-slope form is y - y₁ = m(x - x₁). With m = -3 and the point (2, 5): y - 5 = -3(x - 2). Choice A incorrectly writes (x + 2) instead of (x - 2), which would represent a point at x = -2. Choice D swaps the x and y coordinate values.

Slope equals rise over run. The rise is 4 and the run is 3, so slope = 4/3. Choice A reverses the ratio (run over rise). Negative slopes would describe a line falling from left to right, but this line rises, so the slope must be positive.

If both A and B were zero, the equation would reduce to 0 = C, which is either always true or always false and does not define a line in two variables. A does not need to be positive by definition, C can be any value, and there is no requirement comparing B to A.

In point-slope form y - y₁ = m(x - x₁), the point is (x₁, y₁). From y - 7 = 4(x - 3), we read x₁ = 3 and y₁ = 7, giving the point (3, 7). Choice A reverses the coordinates. Choice D incorrectly uses the slope value 4 as a coordinate.

Slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept. With m = -2 and b = 6, the equation is y = -2x + 6. Choice A incorrectly uses 6 as the slope and -2 as the intercept, swapping the two values.

Starting with y = 2x + 4, subtract y from both sides and subtract 4 from both sides: 2x - y = -4. This has leading coefficient A = 2 > 0. Choice D gives 2x - y = 4, which rearranges to y = 2x - 4, using a y-intercept of -4 instead of 4. Choice B gives y = -2x + 4, which has the wrong slope sign.

Using point-slope form: y - 1 = 2(x - (-3)) → y - 1 = 2(x + 3) → y - 1 = 2x + 6 → y = 2x + 7. The y-intercept is 7. A common error producing Choice A is computing 1 + 3 + 1 = 5 without properly distributing the slope of 2 through the parentheses.

Find slope: m = (11 - 3)/(3 - (-1)) = 8/4 = 2. Use point-slope with (-1, 3): y - 3 = 2(x + 1) → y - 3 = 2x + 2 → y = 2x + 5. Choice D (y = 2x + 1) results from not distributing correctly when substituting: writing y - 3 = 2x - 1 instead of 2x + 2.

To find the x-intercept, set y = 0: 3x = 24, so x = 8. To find the y-intercept, set x = 0: 4y = 24, so y = 6. Choice B reverses these values. Choice C mistakenly uses the coefficients A = 3 and B = 4 as the intercepts rather than dividing C by each coefficient.

The flat fee is a fixed starting value (the y-intercept, $3.50) and the per-mile rate is the slope ($1.75 per mile). So C = 1.75m + 3.50. Choice B swaps the two values, incorrectly making $3.50 the per-mile rate. Choice D incorrectly shows cost decreasing as miles increase.

Point-slope form: y - y₁ = m(x - x₁). With (x₁, y₁) = (2, -1) and m = 3: y - (-1) = 3(x - 2), which simplifies to y + 1 = 3(x - 2). Choice B incorrectly writes y - 1 instead of y + 1 and changes the sign inside the parentheses. Choice C swaps the x and y coordinates of the given point.

The intercepts give points (5, 0) and (0, -3). Slope = (-3 - 0)/(0 - 5) = 3/5. Using slope-intercept: y = (3/5)x - 3. Multiply by 5: 5y = 3x - 15, then rearrange: 3x - 5y = 15. Verify: 3(5) - 5(0) = 15 and 3(0) - 5(-3) = 15, both check out. Choice B adds instead of subtracts the y-term.

The hourly rate is the slope (rate of change): (285 - 150)/(5 - 2) = 135/3 = $45 per hour. To verify the fixed fee: 150 = 45(2) + b → b = $60. Then 45(5) + 60 = 225 + 60 = 285, which confirms the model. Choice B ($50/hr) gives a fixed fee of $50, but then 50(5) + 50 = 300 ≠ 285.

Slope = (y₂ - y₁)/(x₂ - x₁) = (0 - 6)/(4 - 0) = -6/4 = -3/2. The negative value makes sense because the line falls from left to right (y decreases as x increases). Choice A (-2/3) results from inverting the fraction: using 4/6 reduced to 2/3, then applying the negative sign.

Solve for y: 5x - 2y = 10 → -2y = -5x + 10 → y = (5/2)x - 5. Choice B has the wrong sign on the slope; dividing -5x by -2 gives positive 5/2, not negative. Choice C inverts the slope fraction. Choice D has the correct slope but an incorrect positive y-intercept.

Rewrite Line A in slope-intercept form: 4y = -3x + 12, so y = (-3/4)x + 3; slope = -3/4. A perpendicular line has slope 4/3 (negative reciprocal). Line B through (0, 1) with slope 4/3: y = (4/3)x + 1. Multiply by 3: 3y = 4x + 3, then rearrange: 4x - 3y = -3. Choice D (4x - 3y = 3) implies y-intercept of -1, not 1.

Slope = (3 - a)/(a - (-1)) = (3 - a)/(a + 1) = 1. Cross-multiplying: 3 - a = a + 1 → 2 = 2a → a = 1. Check: points become (1, 3) and (-1, 1); slope = (3 - 1)/(1 - (-1)) = 2/2 = 1. Since a = 1 does not equal 3, the condition is satisfied. Choice C (a = 2) gives slope = (3 - 2)/(2 + 1) = 1/3, not 1.

Hiker A position: 2 + 4t. Hiker B position: 20 - 2t. Set equal and solve: 2 + 4t = 20 - 2t → 6t = 18 → t = 3 hours. At t = 3: Hiker A is at 2 + 12 = 14 and Hiker B is at 20 - 6 = 14. Choice A (2 hours) is wrong: at t = 2, A is at 10 and B is at 16, they have not yet met.

Substituting (2, 8) into y = mx + b gives 8 = 2m + b. Using the constraint m + b = 6, solve b = 6 - m and substitute: 8 = 2m + (6 - m) = m + 6 → m = 2, and b = 4. So y = 2x + 4. Choice B satisfies m + b = 6 but fails the point check: 4(2) + 2 = 10, not 8. Choice C satisfies m + b = 6 but 3(2) + 3 = 9, not 8.

Let the y-intercept be b, so the x-intercept is 2b. The line passes through (2b, 0) and (0, b). Slope = (b - 0)/(0 - 2b) = -1/2. So y = (-1/2)x + b. Substituting (-2, 5): 5 = (-1/2)(-2) + b = 1 + b → b = 4. The equation is y = (-1/2)x + 4. Choice B has y-intercept 2 and x-intercept 4, but gives (-1/2)(-2) + 2 = 3, not 5.

For two equations to represent the same line, their coefficients must be proportional. Using the constants: 4/12 = 1/3. So each coefficient ratio must equal 1/3: a/6 = 1/3 gives a = 2, and 2/b = 1/3 gives b = 6. Verify: multiply ax - 2y = 4 by 3 to get 6x - 6y = 12, which matches 6x - by = 12 when b = 6. Choice A gives coefficient ratios 3/6 = 1/2 and 2/4 = 1/2, which are consistent with each other but not with the constant ratio of 1/3.

Substitute x = 12: 25(12) + 50y = 600 → 300 + 50y = 600 → 50y = 300 → y = 6. Choice B (y = 9) would give a total of 300 + 450 = 750, exceeding the $600 revenue. Choice D (y = 12) gives 300 + 600 = 900, also too high.

Slope of the first line: m = (-4 - 0)/(3 - 0) = -4/3. The perpendicular slope is 3/4 (negative reciprocal). The second line through (1, 5) with slope 3/4: y - 5 = (3/4)(x - 1) → 4(y - 5) = 3(x - 1) → 4y - 20 = 3x - 3 → 3x - 4y = -17. Verify: 3(1) - 4(5) = 3 - 20 = -17. Choice D passes through (1, 5) since 4(1) + 3(5) = 19, but its slope is -4/3, which is parallel to the original line, not perpendicular.

Point-slope form is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a known point on the line. Choice A is slope-intercept form. Choice C is standard form. Choice D incorrectly swaps the roles of m and b.

Point-slope form y - y1 = m(x - x1) reveals the point (x1, y1) directly. Matching y - 4 = 2(x - 1) to the template gives x1 = 1 and y1 = 4, so the point is (1, 4). A common error is switching the coordinates to get (4, 1), which is incorrect.

Standard form is Ax + By = C, where A, B, and C are integers and A is typically non-negative. Choice B is slope-intercept form, Choice C is point-slope form, and Choice D is intercept form — a less common but valid alternative that is not called standard form.

Substituting into y - y1 = m(x - x1) with (x1, y1) = (3, 5) and m = 2 gives y - 5 = 2(x - 3). Choice A incorrectly swaps x1 and y1. Choice C incorrectly uses addition instead of subtraction. Choice D uses the coordinates as slope values instead of the given slope of 2.

Standard form requires both variable terms on one side and the constant on the other: Ax + By = C. Only 4x - 3y = 12 fits this structure. Choice A is slope-intercept form. Choice B is point-slope form. Choice C has variables on different sides and is not written in Ax + By = C form.

Since the line passes through (0, 7), the y-intercept b = 7. Using slope-intercept form y = mx + b with m = -3 and b = 7 gives y = -3x + 7. Choice B uses the wrong sign for the slope. Choice C uses the wrong sign for the y-intercept. Choice D swaps the roles of slope and y-intercept.

Solve for y to get slope-intercept form: 4y = -2x + 8, so y = -1/2 x + 2. The slope is -1/2. Choice A (2) is the coefficient of x before isolating y, not the slope. Choice D (-2) comes from incorrectly dividing -2 by 1 instead of by 4.

First find the slope: m = (9 - 3)/(4 - 1) = 6/3 = 2. Then use point-slope with (1, 3): y - 3 = 2(x - 1), which simplifies to y = 2x + 1. Choice B results from a sign error when distributing: y - 3 = 2x - 2 gives y = 2x + 1, not y = 2x - 1. Verify: plugging (4, 9) into y = 2x + 1 gives 9 = 9. Correct.

The flat fee is a fixed starting value (y-intercept) of $20, and the per-text cost of $0.05 is the rate of change (slope). This gives C = 0.05t + 20. Choice A swaps the slope and y-intercept. Choice C incorrectly subtracts the flat fee. Choice D combines both values incorrectly into a single rate.

Distribute the right side: y - 2 = 3x + 3. Then add 2 to both sides: y = 3x + 5. A common mistake is to use the 1 from (x + 1) as the y-intercept, giving y = 3x + 1 — but this ignores the -2 that must be moved to the right side.

Slope = (-1 - 4)/(3 - (-2)) = -5/5 = -1. Using point-slope with (-2, 4): y - 4 = -1(x + 2) → y = -x + 2 → x + y = 2. Choice B gives x - y = 2, which would require slope +1. Choice C results from a sign error on the constant. Always verify with both points: for (-2, 4), -2 + 4 = 2. Correct.

The slope is m = (1 - 4)/(2 - 0) = -3/2. The slope-intercept form is y = -3/2 x + 4. Multiplying every term by 2: 2y = -3x + 8, then rearranging: 3x + 2y = 8. Choice A has the wrong sign on 2y. Choice C has the correct coefficients but an incorrect constant — forgetting to multiply the 4 by 2.

Using y - y1 = m(x - x1) with (x1, y1) = (2, -3) and m = -4: y - (-3) = -4(x - 2), which simplifies to y + 3 = -4(x - 2). Choice A incorrectly uses +3 and +2, reversing both signs. Choice C uses the correct point but the wrong sign for the slope. Choice D swaps x1 and y1.

In a linear model y = mx + b, the slope m represents the rate of change. Here m = 5 means the temperature increases by 5 degrees Fahrenheit for each additional hour. The 30 is the y-intercept, representing the starting temperature at midnight (h = 0), not the rate. A common error is confusing the slope with the y-intercept.

The y-intercept is 2 since the line crosses the y-axis at (0, 2). The slope is m = (2 - 0)/(0 - (-3)) = 2/3. So the equation is y = (2/3)x + 2. Choice A inverts the slope to 3/2. Choice D uses a negative slope, but moving right from (-3, 0) to (0, 2) shows the line is rising, so the slope must be positive.

When x = 0, y = -5, so the y-intercept is b = -5. The slope is m = (4 - (-5))/(3 - 0) = 9/3 = 3. Therefore y = 3x - 5. Choice B has the wrong sign on the y-intercept. Choice D incorrectly uses the y-intercept as the slope and vice versa.

Solve for y: -3y = -6x + 9, so y = 2x - 3. The y-intercept is -3. A common error is to set x = 0 in the original equation correctly (getting -3y = 9, y = -3) but then report y = 3 by forgetting the negative sign. Choice D is the original constant before dividing, not the y-intercept.

Total calories = (calories per minute jogging)(minutes jogging) + (calories per minute walking)(minutes walking) = 5j + 3w. Setting this equal to 200 gives 5j + 3w = 200. Choice B swaps the calorie rates for each activity. Choice D counts only total minutes, ignoring the different calorie rates for each activity.

The given line has slope 1/3. The perpendicular slope is the negative reciprocal: -3. Using point-slope with (2, -1): y - (-1) = -3(x - 2) → y + 1 = -3x + 6 → y = -3x + 5. Choice A uses the correct slope but wrong y-intercept (forgetting to add 1 to both sides). Choice B uses the original slope rather than its negative reciprocal. Choice D uses slope 3 instead of -3.

Set up the system: d + 100m = 65 and d + 250m = 110, where d is the daily fee and m is the per-mile rate. Subtracting the first from the second: 150m = 45, so m = 0.30. Substituting back: d + 100(0.30) = 65 → d + 30 = 65 → d = 35. Choice B ($30) is the per-mile cost for 100 miles, not the base fee. Verify: 35 + 0.30(250) = 35 + 75 = 110. Correct.

Step 1: Find the slope of line P: m = (8 - 2)/(3 - 1) = 6/2 = 3. Step 2: Find the y-intercept of 2x - y = -4 by rewriting as y = 2x + 4, giving y-intercept = 4. Step 3: Line Q is parallel to P (same slope, 3) with y-intercept 4: y = 3x + 4. Choice A uses slope 2 from the second line rather than slope 3 from line P. Choice D uses y-intercept 2 instead of 4.

The growth rate (slope) is (14 - 2)/(6 - 0) = 12/6 = 2 cm per week. The linear model is h = 2w + 2. Setting h = 20: 20 = 2w + 2 → 2w = 18 → w = 9. Week 9 is correct. Choice A (Week 10) results from forgetting the initial height of 2 cm: solving 20 = 2w gives w = 10, but the starting height must be included in the model.

For two equations to represent the same line, one must be a scalar multiple of the other. Multiplying 2x - y = 5 by 2 gives 4x - 2y = 10. Therefore c = 10. Choice A (5) is the constant from the original equation before scaling. Choice B (20) comes from multiplying only the constant by 4 instead of 2. The equations must be proportional across all terms.

Using the slope formula: (k - 1)/(2 - k) = 3. Cross-multiply: k - 1 = 3(2 - k) = 6 - 3k. Adding 3k to both sides and adding 1: 4k = 7, so k = 7/4. Verify: the two points are (7/4, 1) and (2, 7/4). Slope = (7/4 - 1)/(2 - 7/4) = (3/4)/(1/4) = 3. Correct. Choice D (2) is excluded by the given condition k is not equal to 2, which would make the denominator zero.

Find the drain rate (slope): (270 - 450)/(30 - 10) = -180/20 = -9 gallons per minute. Using point (0, 540) as the starting volume, the model is V = -9t + 540. Set V = 0: 0 = -9t + 540 → t = 540/9 = 60 minutes. Verify: at t = 10, V = -90 + 540 = 450. At t = 30, V = -270 + 540 = 270. Both check out. Choice A (54) results from incorrectly using the drain rate as 10 gallons per minute instead of 9.

The point-slope form is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a known point on the line. Choice A is slope-intercept form, and Choice C is standard form. Choice D is simply slope-intercept form with terms rearranged, not point-slope.

Standard form is Ax + By = C. Choice C, 4x - y = 7, fits this format with A = 4, B = -1, and C = 7. Choice A is slope-intercept form, and Choice B is point-slope form. Choice D is slope-intercept form with b = 0.

Point-slope form is y - y1 = m(x - x1). Substituting m = 3, x1 = 2, y1 = 5 gives y - 5 = 3(x - 2). Choice A adds 2 inside the parentheses instead of subtracting, and Choice D incorrectly swaps the roles of the x and y coordinates.

To find the slope, solve for y: 2y = -4x + 8, so y = -2x + 4. The slope is -2. A common error is reading the coefficient of x directly from standard form as the slope without first converting to slope-intercept form, which would incorrectly give a slope of 4.

Distribute the right side: y - 3 = 2x - 2. Then add 3 to both sides: y = 2x + 1. Choice B, y = 2x - 1, results from forgetting to add 3 after distributing, effectively only completing part of the simplification.

Slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept. With m = -3 and b = 7, the equation is y = -3x + 7. Choice A swaps the positions of the slope and y-intercept. Choice C uses the correct slope but applies a negative sign to the y-intercept.

To find the x-intercept, substitute y = 0: 3x + 5(0) = 15, so 3x = 15 and x = 5. The x-intercept is (5, 0). Choice A, (0, 3), is the y-intercept (set x = 0: 5y = 15, y = 3). Students often confuse which variable to set equal to zero.

Slope is defined as rise over run. Here, rise = 6 and run = 3, so slope = 6/3 = 2. Choice D, 1/2, results from dividing run by rise instead of rise by run, a common reversal error. Choice B incorrectly uses only the run value as the slope.

Slope = (4 - (-2))/(3 - 1) = 6/2 = 3. Using point-slope with (1, -2): y + 2 = 3(x - 1) → y = 3x - 5. Rearranging: 3x - y = 5. Choice D uses -5 on the right side, a sign error that occurs when incorrectly moving the constant term across the equal sign.

The slope is (0 - 6)/(3 - 0) = -6/3 = -2. Since the line passes through (0, 6), the y-intercept is 6. The equation is y = -2x + 6. Choice A uses a positive slope, which would describe a line rising from left to right and would not pass through (3, 0) with y-intercept 6.

The flat fee of $75 is the starting value (y-intercept) and $50 per hour is the rate of change (slope). So C = 50h + 75. Choice B incorrectly uses 75 as the per-hour rate and 50 as the flat fee. Choice D adds both dollar amounts together as a single per-hour rate, ignoring that the flat fee is charged once regardless of hours.

Solving for y: subtract 3x to get -4y = -3x + 12, then divide by -4: y = (3/4)x - 3. Choice D, y = (3/4)x + 3, results from dividing 12 by -4 incorrectly and getting +3 instead of -3. Choice B applies a negative sign to the slope without justification.

Point-slope form is y - y1 = m(x - x1). With y1 = 5, x1 = -1, and m = -2: y - 5 = -2(x - (-1)) = -2(x + 1). Choice A incorrectly writes y + 5 and x - 1, flipping both signs. Choice B uses a positive slope instead of -2.

To find the y-intercept, set x = 0: 2y = -10, so y = -5. The y-intercept is (0, -5). Choice A, (0, 5), results from dropping the negative sign. Choice B is found by setting y = 0, which gives the x-intercept, not the y-intercept.

The cyclist starts 10 miles from home (initial value) and increases distance at 12 miles per hour, so d = 12t + 10. Choice C, d = 10 - 12t, would model a cyclist riding toward home. Choice A subtracts the initial 10 miles, incorrectly treating the starting distance as a negative offset.

Parallel lines have the same slope. Since y = 4x + 1 has slope 4, the new line also has slope 4. The line passes through (0, -3), so the y-intercept is -3. The equation is y = 4x - 3. Choice A uses slope -4, which is the negative of the original slope, not the same slope. Choice B uses the reciprocal of the slope.

Substitute each point: (8, 0): 8 + 0 = 8 ✓. (0, 4): 0 + 8 = 8 ✓. (2, 3): 2 + 6 = 8 ✓. (4, 3): 4 + 6 = 10 ≠ 8. So (4, 3) does not lie on the line. A common error is forgetting to multiply the y-value by the coefficient 2 before checking the sum.

Rewrite 3x - y = 6 as y = 3x - 6, so the slope is 3. The perpendicular slope is -1/3 (negative reciprocal). Using point-slope with (3, 1): y - 1 = -(1/3)(x - 3) → y = -(1/3)x + 1 + 1 = -(1/3)x + 2. Choice B uses slope 3, which would produce a parallel line. Choice D uses the correct slope but makes a sign error in computing the y-intercept.

Slope = (-5 - 7)/(2 - (-4)) = -12/6 = -2. Using point-slope with (-4, 7): y - 7 = -2(x + 4) → y = -2x - 1. At x = 5: y = -2(5) - 1 = -11. Choice B (-9) results from computing the y-intercept as 1 instead of -1. Choice D (9) likely comes from using a positive slope.

The production rate is 0.5 grams per 4 minutes = 0.125 grams per minute. Using the data point (8, 3): 3 = 0.125(8) + b → 3 = 1 + b → b = 2. So A = 0.125t + 2. Choice A uses 0.5 as the slope directly without converting to a per-minute rate. Choice C uses the known data point value (3) as the y-intercept without solving for it.

Using point-slope: y - 3 = (2/3)(x - 1). Multiply both sides by 3: 3y - 9 = 2x - 2. Rearranging: -2x + 3y = 7, or equivalently 2x - 3y = -7. Choice D, 2x - 3y = 7, results from a sign error when moving the constant term, forgetting that multiplying both sides of -2x + 3y = 7 by -1 changes the right side to -7.

Rewrite 4x - 2y = k as y = 2x - k/2. For this to equal y = 2x + 5, we need -k/2 = 5, which gives k = -10. Choice A (10) results from setting k/2 = 5 and forgetting the negative sign introduced when dividing -2y = -4x + k by -2.

The slope of the line through (1, 1) and (4, 7) is (7 - 1)/(4 - 1) = 6/3 = 2. The perpendicular slope is -1/2. Using point-slope with (-2, 3): y - 3 = -(1/2)(x + 2) → y - 3 = -(1/2)x - 1 → y = -(1/2)x + 2. The y-intercept is 2. Choice B (4) results from a sign error when distributing -(1/2) through (x + 2).

The slope is (1 - (-3))/(2 - 0) = 4/2 = 2, and the y-intercept is -3 (from the point (0, -3)), giving y = 2x - 3. At x = 10: y = 2(10) - 3 = 17. Choice A (13) likely results from using a slope of 1 instead of 2. Choice C (15) may come from computing 2(10) - 5, using an incorrect y-intercept.

The value after t years is V = 25,000 - 1,500t. Setting V < 10,000: 25,000 - 1,500t < 10,000 → 1,500t > 15,000 → t > 10. Since t must be a whole number of complete years, the first integer satisfying t > 10 is t = 11. After exactly 10 years, V = 25,000 - 15,000 = $10,000, which equals the threshold and is not strictly less than it.

Standard form is Ax + By = C, where A, B, and C are integers and A is non-negative. Choice A is slope-intercept form, Choice C is point-slope form, and Choice D is a rearrangement of slope-intercept form — all different named forms of a linear equation.

In slope-intercept form y = mx + b, the coefficient of x is the slope m. Here m = -4. The value 7 is the y-intercept, not the slope. A common error is confusing the sign of the slope or swapping slope and y-intercept.

Since the line passes through (0, -2), that is the y-intercept, so b = -2. With slope m = 3, the slope-intercept form gives y = 3x - 2. Choice A swaps the slope and y-intercept. Choice C uses the wrong sign for the y-intercept.

Using slope-intercept form y = mx + b with m = -5 and b = 1 gives y = -5x + 1. Choice B uses the positive slope 5 instead of -5. Choice D uses the correct slope but changes the y-intercept to -1.

Point-slope form is y - y₁ = m(x - x₁). Choice B incorrectly uses addition instead of subtraction on both sides — that would give a different line. Choice A is slope-intercept form and requires knowing the y-intercept b. Choice D is standard form.

Slope = (y₂ - y₁)/(x₂ - x₁) = (11 - 5)/(4 - 2) = 6/2 = 3. A common mistake is forgetting to divide and reporting the rise alone (6), which gives choice D. Another error is computing only the run (2) instead of rise over run.

A horizontal line has a slope of 0 and takes the form y = constant. Here y = 4 is a horizontal line at height 4. Choice A (x = 4) is a vertical line. Choice C has slope 4 and is not horizontal. Choice D can be rewritten as y = -4x, which also has a nonzero slope.

Slope = (5 - (-1))/(6 - 3) = 6/3 = 2. Using point-slope with (3, -1): y - (-1) = 2(x - 3) → y + 1 = 2x - 6 → y = 2x - 7. Choice B incorrectly adds 6 instead of subtracting. Choice D gives y-intercept -5, which would require different arithmetic.

Using point-slope: y - 3 = -1/2(x - 2). Multiply both sides by 2: 2y - 6 = -(x - 2) = -x + 2. Then x + 2y = 8. Choice D gives 7 on the right side — an arithmetic error. Choice C gives 2x + y = 7, which corresponds to slope -2, not -1/2.

The x-intercept is where y = 0. Substituting: 2x + 5(0) = 20 → x = 10, so the x-intercept is (10, 0). Choice A is the y-intercept (set x = 0: 5y = 20, y = 4). Choices C and D swap the intercept values.

The monthly charge is the rate of change, so the slope is 15. The one-time setup fee is the starting value, so the y-intercept is 20. The model is C = 15m + 20. Choice B switches slope and y-intercept. Choice D incorrectly adds the two rates together as if both were monthly.

Parallel lines have equal slopes. The given line has slope 3/4, so the new line also has slope 3/4. Since it passes through (0, 5), the y-intercept is 5, giving y = (3/4)x + 5. Choice A uses the negative reciprocal slope, which would be perpendicular. Choice C uses the original y-intercept -2 instead of 5.

The two intercept points are (0, -4) and (2, 0). Slope = (0 - (-4))/(2 - 0) = 4/2 = 2. Choice A results from computing (0 - 2)/(-4 - 0) = -2/(-4), an error in assigning x and y coordinates. Choice B inverts the rise and run.

Expand and simplify: y - 6 = -3x - 6 → y = -3x - 6 + 6 → y = -3x. The slope is -3 and the y-intercept is 0. Choice A reads the 6 from the original form without converting to slope-intercept. Choice D gives y-intercept -6, ignoring that the +6 cancels the -6.

The slope = (10 - 4)/(3 - 1) = 6/2 = 3. Using point (1, 4): y - 4 = 3(x - 1) → y = 3x + 1. Verify with (5, 16): 3(5) + 1 = 16 ✓. Choice C gives 3(1) - 1 = 2, not 4, so it fails the check. Choice A uses slope 2, which does not match the table.

Using points (0, 32) and (100, 212): slope = (212 - 32)/(100 - 0) = 180/100 = 9/5. The y-intercept is 32 (from the freeze point). So F = (9/5)C + 32. Choice A uses the inverted slope 5/9, which is the formula to convert Fahrenheit to Celsius. Choice B has the correct slope but subtracts 32.

Solve for y: -3y = -4x - 12 → y = (4/3)x + 4. Dividing both sides by -3 flips the signs on both terms. Choice A keeps the y-intercept negative, an error when dividing -12 by -3. Choice B negates the slope incorrectly. Choice D uses the reciprocal slope 3/4.

The line x = 3 is vertical (undefined slope). A line perpendicular to a vertical line must be horizontal (slope = 0). A horizontal line through (0, 6) has equation y = 6. Choice B would be a vertical line, which is parallel to x = 3, not perpendicular. Choices C and D both have nonzero slopes.

Set up the slope equation: (6 - 2)/(-3 - a) = -2 → 4/(-3 - a) = -2 → 4 = 6 + 2a → -2 = 2a → a = -1. Verify: slope through (-1, 2) and (-3, 6) = (6-2)/(-3-(-1)) = 4/(-2) = -2 ✓. Choice A (a = 1) gives slope 4/(-4) = -1, not -2.

Rewrite the second equation in slope-intercept form: ky = -3x + 9 → y = (-3/k)x + 9/k. For parallel lines, slopes must be equal: -3/k = 1/3 → k = -9. Choice A (k = 9) gives slope -3/9 = -1/3, which is not equal to 1/3. Choice D (k = -3) gives slope -3/(-3) = 1, also incorrect.

Rewrite line L: y = 2x - 5, so slope = 2. The perpendicular slope is -1/2. Using point-slope through (4, 1): y - 1 = -1/2(x - 4) → 2y - 2 = -(x - 4) = -x + 4 → x + 2y = 6. Choice D gives x + 2y = 8, which would require a y-intercept of 4 instead of 3. Choice C uses slope -2, not -1/2.

Set the cost equations equal: 1.50m + 3 = 1.00m + 4.50 → 0.50m = 1.50 → m = 3 miles. Total cost = 1.50(3) + 3.00 = $7.50. Check: 1.00(3) + 4.50 = $7.50 ✓. Choice B (m = 2) gives taxi $6.00 but ride-share $6.50 — not equal. Choice D has the correct mileage but an incorrect cost.

Slope = (8 - k)/(k - 2) = 2 → 8 - k = 2(k - 2) = 2k - 4 → 12 = 3k → k = 4. Verify: points (2, 4) and (4, 8), slope = (8-4)/(4-2) = 4/2 = 2 ✓. Choice A (k = 2) makes the denominator 0, giving an undefined slope. Choice C gives slope (8-6)/(6-2) = 2/4 = 1/2, not 2.

Slope = (7 - (-1))/(-1 - 3) = 8/(-4) = -2. Using point-slope with (3, -1): y - (-1) = -2(x - 3) → y + 1 = -2x + 6 → y = -2x + 5. Verify: f(3) = -6+5 = -1 ✓ and f(-1) = 2+5 = 7 ✓. Choice C gives f(3) = -6+3 = -3, not -1. Choice D gives f(3) = -6+7 = 1, not -1.

The intercepts give points (-3, 0) and (0, 5). Slope = (5 - 0)/(0 - (-3)) = 5/3. Slope-intercept form: y = (5/3)x + 5. Multiply by 3: 3y = 5x + 15 → 5x - 3y = -15. Check: 5(-3) - 3(0) = -15 ✓ and 5(0) - 3(5) = -15 ✓. Choice B gives +15, which fails the x-intercept check. Choice A uses the wrong coefficient signs.

Point-slope form is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a known point. Choice B matches this template exactly with slope 2 and point (1, 5). Choice A is slope-intercept form. Choice C is standard form. Choice D moves the 2x term to the right side without factoring, so it is not written in point-slope form.

Slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept. With m = -2 and b = 7, the equation is y = -2x + 7. Choice A swaps the slope and y-intercept values. Choice B negates both values incorrectly. Choice D uses the correct numbers but applies the wrong signs to each.

Standard form requires Ax + By = C with integer coefficients and A greater than 0. Choice C satisfies all conditions: A = 5 (positive integer), B = 2, C = 10. Choice A is slope-intercept form. Choice B is point-slope form. Choice D has A = -2, which is negative, so it does not meet the standard form requirement even though it can be rewritten as one by multiplying through by -1.

Point-slope form is y - y1 = m(x - x1). With point (2, -3) and slope 4: y - (-3) = 4(x - 2), which simplifies to y + 3 = 4(x - 2). Choice B uses the wrong signs for both coordinates of the point. Choice C swaps the x and y values of the point. Choice D incorrectly places the slope value and point coordinates.

Isolate y by subtracting 3x from both sides: 4y = -3x + 20. Then divide everything by 4: y = -(3/4)x + 5. Choice A drops the negative sign on the slope, a common error when moving 3x to the other side. Choice C divides only by 3 instead of 4 and inverts the coefficient. Choice D inverts the slope ratio and changes the sign of the y-intercept.

The daily charges for 3 days equal 3 times 40 = $120. The flat fee is 145 - 120 = $25. Since the daily rate is $40 and the flat fee is $25, the equation is C = 40d + 25. Verify: when d = 3, C = 120 + 25 = 145. Choice B swaps the rate and the flat fee. Choice C incorrectly combines the values into a single rate. Choice D uses the total 3-day cost as the flat fee.

Find the slope: m = (7 - 1) / (4 - (-2)) = 6 / 6 = 1. Use point-slope form with (-2, 1): y - 1 = 1(x + 2), so y = x + 3. Check with (4, 7): 4 + 3 = 7. Choice A results from a sign error when solving for b: using 1 = -2 + b gives b = 3, not -3. Choice C doubles the numerator when computing slope. Choice D halves the slope by dividing the wrong values.

Convert 2x - 5y = 10 to slope-intercept form: y = (2/5)x - 2, so the slope is 2/5. A parallel line shares this slope. Using point (3, -2): y + 2 = (2/5)(x - 3). Multiply both sides by 5: 5y + 10 = 2x - 6, then rearrange: 2x - 5y = 16. Choice B results from using slope -2/5 (sign error on the slope). Choice C uses the perpendicular slope of 5/2 instead. Choice D comes from an arithmetic error when computing the constant after rearranging.

The candle starts at 12 inches when t = 0, so the y-intercept is 12. The rate of change is (9 - 12) / (3 - 0) = -3/3 = -1 inch per hour. The equation is h = -t + 12. Setting h = 0 gives t = 12 hours. Choice B incorrectly uses the total drop of 3 inches as the hourly burning rate rather than dividing by the 3 hours elapsed. Choice C inverts the rate. Choice D uses the 3-hour height of 9 as the starting value instead of 12.

Find the slope of the first line: m = (11 - 3) / (5 - 1) = 8/4 = 2. The perpendicular slope is the negative reciprocal: -1/2. The second line passes through (0, 6), so its y-intercept is 6. Its equation is y = -(1/2)x + 6. Choice A uses the original slope of 2 rather than the perpendicular slope. Choice C has the correct perpendicular slope but incorrectly negates the y-intercept. Choice D uses the original slope with an unrelated y-intercept.