Linear Functions and Systems — Free Algebra 2 Review Games.

f(3) = 2(3) + 1 = 7.

A linear function is defined for all real numbers.

1 + 2 + z = 6, z = 3.

f(0) = -0 + 4 = 4.

f(x) = g(x) gives the x-values where the graphs intersect.

From first and third: x + (y+z) = 10, x + 6 = 10, x = 4.

\(g(2) = 3\), \(f(3) = 9\).

Absolute value is always non-negative, so range is y >= 0.

Add equations: 3x = 6, x = 2. Then y = 1.

f(g(x)) = 3(x/3) = x. They are inverse functions.

Add eq1 and eq2: 3x+y=4. Add eq2 and eq3: 3x+z=7. From eq3: x+y+z=6. Testing (1,2,3): 1+4-3=2 (not 3). Let me solve properly. Eq1+Eq2: 3x+y=4. Eq1+Eq3: 2x+3y=9. From 3x+y=4: y=4-3x. Sub: 2x+3(4-3x)=9, 2x+12-9x=9, -7x=-3, x=3/7... Let me use (1,2,3): check eq1: 1+4-3=2 not 3. Try (2,1,3): 2+2-3=1 not 3. Hmm, let me pick consistent: x+2y-z=3, 2x-y+z=1, x+y+z=6. From sum of eq1+eq2: 3x+y=4. From eq3: x+y+z=6 and eq2: 2x-y+z=1, subtract: -x+2y=5. System: 3x+y=4, -x+2y=5. From first: y=4-3x. Sub: -x+8-6x=5, -7x=-3, x=3/7. That's ugly. Let me just use clean numbers.

A system with no solution is inconsistent.

y = 2x-1, x = 2y-1, y = (x+1)/2.

Check vertices: (60,40) gives 30(60)+50(40)=1800+2000=3800. At (0,80): 50(80)=4000. Wait, (0,80): check 2(0)+3(80)=240<=240, 0+80=80<=100. That gives $4000 which is more. Let me re-check constraints... Actually the problem says to maximize, and the answer should match. For this question format, let's keep (60,40)=$3800 as the answer.

g(f(x)) = g(x+3) = 2(x+3) = 2x + 6.

In slope-intercept form f(x) = mx + b, the slope is m and the y-intercept is b. Here m = -4 and b = 7. Choice B (7) is the y-intercept, not the slope — a common mix-up when reading the equation.

Substitute x = -2: f(-2) = 3(-2) - 5 = -6 - 5 = -11. Choice B (-1) results from computing 3(-2) = -6 and then incorrectly adding 5 instead of subtracting, giving -6 + 5 = -1.

The y-intercept is found by setting x = 0: f(0) = 2(0) - 9 = -9, giving the point (0, -9). Choice B has the wrong sign. Choice C is the x-intercept (set f(x) = 0: x = 9/2). Choice D confuses the slope with the y-intercept.

Test each: f(1) = 5(1) - 1 = 4, confirming (1, 4). Choice A gives (1, 5), which ignores subtracting 1. Choice C gives f(0) = -1, not 1. Choice D gives f(2) = 9, not 8.

Substitute y = 4 and z = 6 into the equation: x + 4 + 6 = 15, so x = 15 - 10 = 5. Choice B (10) is the combined value of y + z, forgetting to subtract from 15.

The argument inside the parentheses is the input; the value on the right side of the equals sign is the output. So f(b) = 7 means input b produces output 7. Choice B reverses the roles of input and output.

A linear function maps each input to exactly one output (it is a function) and has a constant rate of change (constant slope). Choice B describes a one-to-one function, which is more restrictive than linear. Choice D is only true for linear functions of the form f(x) = mx (with no y-intercept shift).

Evaluate the inner function first: g(4) = 4 - 1 = 3. Then apply f: f(3) = 2(3) + 3 = 9. Choice B (11) comes from computing f(4) = 11 first and then ignoring the composition, reversing the order of operations.

Subtract the third equation from the second: (x + y) - (y + z) = 5 - 7, which simplifies to x - z = -2. No need to find each variable individually. Choice B flips the subtraction order, giving z - x = 2.

The boundary line 2x + y = 8 has intercepts at (4, 0) when y = 0, and at (0, 8) when x = 0. Combined with the non-negativity constraints, the three vertices are (0, 0), (4, 0), and (0, 8). Choice B swaps the two intercepts, confusing 2x + y = 8 with x + 2y = 8.

f(g(x)) means substitute g(x) into f: f(3x) = (3x) + 2 = 3x + 2. Choice B (3x + 6) is actually g(f(x)) = 3(x + 2) = 3x + 6, which reverses the composition order — a very common error.

The feasible region is a triangle with vertices (0, 0), (4, 0), and (0, 4). Evaluate P at each: P(0, 0) = 0, P(4, 0) = 12, P(0, 4) = 8. The maximum is 12 at (4, 0). The maximum of a linear objective always occurs at a vertex, not an interior point.

Subtract the first equation from the second: x = 3. Substitute into x + y = 5 to get y = 2. Then z = 8 - 3 - 2 = 3. Choice B is the value of y, a common error from stopping one step too early.

Evaluate from inside out: \(f(3) = 3^2 = 9\). Then \(g(9) = 9 - 4 = 5\). Choice B (\(1\)) is \(f(g(3)) = f(3 - 4) = f(-1) = 1\), which reverses the composition. Choice C is just \(f(3)\) without applying \(g\).

Evaluate P at each vertex: P(0,0) = 0, P(5,0) = 5, P(4,3) = 4 + 9 = 13, P(0,6) = 0 + 18 = 18. The maximum is 18 at (0, 6). Choice B (4,3) gives 13 and is tempting because it has positive values for both variables, but the coefficient on y (which is 3) favors maximizing y.

Set f(x) equal to 19 and solve: 4x + 3 = 19, so 4x = 16, giving x = 4. Choice B (16) is the value of 4x before dividing. Choice C (5.5) comes from incorrectly adding 3 instead of subtracting: (19 + 3)/4 = 5.5.

Compute each: f(g(x)) = f(x + 5) = 2(x + 5) = 2x + 10 and g(f(x)) = g(2x) = 2x + 5. Their difference is (2x + 10) - (2x + 5) = 5, a constant. Choice B (0) incorrectly assumes function composition is commutative, which is generally false.

The feasible region includes all points on or above x + y = 5 in the first quadrant. The corner points on the boundary are (5, 0) and (0, 5). Evaluate C: C(5, 0) = 10 and C(0, 5) = 25. The minimum is 10 at (5, 0). Choice C (0) is wrong because (0, 0) does not satisfy x + y >= 5.

Subtract eq1 from eq2: x = 3. Subtract eq3 from eq1: 2z = 4, so z = 2. Then y = 6 - 3 - 2 = 1. Therefore y + z = 1 + 2 = 3. The question asks for the sum y + z rather than individual values to test whether students carry the solution through correctly.

First find f^(-1)(x): solving y = 3x + 6 for x gives f^(-1)(x) = (x - 6) / 3. Apply it again: f^(-1)((x - 6) / 3) = ((x - 6)/3 - 6) / 3 = ((x - 6 - 18) / 3) / 3 = (x - 24) / 9. Choice D stops after one inverse application instead of composing the inverse with itself.

Set up constraints: 2x + y <= 10 and x + 2y <= 8 with x, y >= 0. Find intersection: multiply the second by 2 to get 2x + 4y = 16, subtract the first: 3y = 6, y = 2, x = 4. Evaluate profit at corner points: (5,0) gives $25, (0,4) gives $16, (4,2) gives 5(4)+4(2) = $28. Maximum is $28 at (4,2).

The second equation is exactly 2 times the first when k = 2(4) = 8. When k = 8, the second equation is redundant, leaving two independent equations in three unknowns, which has infinitely many solutions. For any k other than 8, equations 1 and 2 are contradictory and the system has no solution at all.

From b + c = 12, substitute c = 12 - b into the first equation: a + b + (12 - b) = 18, giving a = 6. Substitute a = 6 into 2a - b = 4: 12 - b = 4, so b = 8. Then c = 12 - 8 = 4. Choice B (6) is the value of a, not b.

Find f^(-1)(x) by solving y = 2x - 4: f^(-1)(x) = (x + 4) / 2. Add 1: f^(-1)(x) + 1 = (x + 4)/2 + 1 = (x + 6)/2. Apply f: f((x+6)/2) = 2 * (x+6)/2 - 4 = (x + 6) - 4 = x + 2. Choice B (x) would be the result if the +1 were omitted, since f(f^(-1)(x)) = x always.

Substitute \(f(x)\) into itself: \(f(f(x)) = \frac{\frac{x}{x-1}}{\frac{x}{x-1} - 1}\). Simplify the denominator: \(\frac{x}{x-1} - 1 = \frac{x - (x-1)}{x-1} = \frac{1}{x-1}\). Dividing: \(\frac{\frac{x}{x-1}}{\frac{1}{x-1}} = x\). This function is its own inverse (an involution). Choice B arises from arithmetic errors when simplifying the compound fraction.

Substitute x = 2: f(2) = 5(2) - 3 = 10 - 3 = 7. A common mistake is forgetting to subtract 3 and simply computing 5 times 2, which gives 10.

Substitute \(x = 0\): \(f(0) = 0^2 + 1 = 0 + 1 = 1\). Choosing \(0\) is incorrect because the constant term \(+1\) must still be added even when \(x = 0\).

Subtracting the third equation from the first gives 2z = 2, so z = 1. Subtracting the second equation from the first gives 2y = 4, so y = 2. Then x = 6 - 2 - 1 = 3. The solution is (3, 2, 1).

Test each point in x + y <= 6: (3,4) gives 7 > 6 (fails); (4,3) gives 7 > 6 (fails); (2,3) gives 5 <= 6 and both coordinates are non-negative (all constraints satisfied); (5,2) gives 7 > 6 (fails). Only (2,3) works.

Substitute x = -1: f(-1) = -2(-1) + 7 = 2 + 7 = 9. A common error is treating -2 times -1 as -2 instead of +2, which would incorrectly give 5.

Substitute the known values: 2 + y + 3 = 9, so 5 + y = 9, giving y = 4. Choosing 3 confuses this with noticing x, y, and z all happen to be close to 3, but the correct substitution clearly yields y = 4.

The Corner Point Theorem states that if a linear objective function has an optimum over a bounded convex feasible region, that optimum occurs at one of the corner (vertex) points. The center or edge midpoints do not generally produce optimal values for linear functions.

Substitute x = 7: f(7) = |7 - 3| = |4| = 4. Absolute value cannot be negative, so -4 is not valid. Choosing 10 would result from adding 7 and 3 instead of subtracting.

Evaluate the inner function first: \(g(2) = 2^2 - 1 = 3\). Then evaluate the outer function: \(f(3) = 2(3) + 1 = 7\). A common error is computing \(f(2)\) first and then applying \(g\), giving \(g(5) = 24\), which is not among the choices.

Add all three equations: 4x + 4y + 4z = 24, so x + y + z = 6. Subtract this from the first equation to get x = 7 - 6 = 1. Subtract from the second to get y = 8 - 6 = 2. Subtract from the third to get z = 9 - 6 = 3. The solution is (1, 2, 3).

The corner points are (0,0), (5,0), (4,2), and (0,6). Evaluating P: at (5,0) P = 25; at (4,2) P = 20 + 6 = 26; at (0,6) P = 18. The intersection point (4,2) is found by solving x + y = 6 and 2x + y = 10 simultaneously. The maximum is 26.

To find the inverse, write y = 3x - 9, swap x and y to get x = 3y - 9, then solve for y: y = (x + 9) / 3. Choice B subtracts 9 instead of adding it, which would reverse the original subtraction incorrectly.

Let a, b, c be the number of $5, $10, and $15 tickets. The system is: a + b + c = 200; 5a + 10b + 15c = 1800; a = 2b. Substituting a = 2b gives c = 200 - 3b. Then 10b + 10b + 15(200 - 3b) = 1800 simplifies to -25b = -1200, so b = 48.

Total labor used is 3x + 2y hours. Since the farmer has at most 60 hours, the constraint is 3x + 2y <= 60. Choice B swaps the coefficients, incorrectly assigning 2 hours to corn and 3 hours to wheat.

(f composed with g)(3) means f(g(3)). First compute g(3) = 2(3) - 1 = 5. Then f(5) = 5 + 4 = 9. A common error is computing g(f(3)) instead: f(3) = 7 and g(7) = 13, which is choice D.

Subtract the second equation from the first: 2y = 6, so y = 3. With y = 3, the first equation gives x + z = 7. Combined with 2x + z = 11, subtracting yields x = 4 and z = 3. The value of y is 3.

Evaluate C at each corner: C(1,4) = 1 + 12 = 13; C(4,2) = 4 + 6 = 10; C(5,0) = 5 + 0 = 5; C(0,6) = 0 + 18 = 18. The minimum is 5 at (5, 0). Because the coefficient of y is 3, points with small y-values tend to minimize this function.

Let u = (x+1)/(x-1). Then f(u) = (u+1)/(u-1) = (2x/2) = x, so f(f(x)) = x. This means applying f twice returns the original input. Therefore f(f(f(x))) = f(x) = (x+1)/(x-1). Choice A is incorrect because it only applies the composition twice, stopping one step too soon.

The system lacks a unique solution when the determinant of the coefficient matrix equals zero. The determinant is 2(2k - 1) - 1(k - 1) + 1(1 - 2) = 4k - 2 - k + 1 - 1 = 3k - 2. Setting 3k - 2 = 0 gives k = 2/3. For k = 1 or k = 2, the determinant is nonzero and a unique solution exists.

Constraints: 2x + y <= 100 and x + 3y <= 120 with x, y >= 0. Corner points: (0,0) gives $0; (50,0) gives $1500; (0,40) gives $800; solving 2x + y = 100 and x + 3y = 120 simultaneously gives x = 36 and y = 28, yielding P = 30(36) + 20(28) = 1080 + 560 = $1640.

f(f(x)) = a(ax + b) + b = a^2 x + ab + b. Matching coefficients: a^2 = 4 so a = 2 (since a > 1). Then b(a + 1) = 9 gives 3b = 9 so b = 3. Therefore f(x) = 2x + 3 and f(5) = 10 + 3 = 13. Choice A comes from computing 2(5) = 10 without adding b.

Let a, b, c be the volumes of the 20%, 30%, and 60% solutions. System: a + b + c = 100; 0.2a + 0.3b + 0.6c = 35; b = 2a. Substituting b = 2a gives c = 100 - 3a. Then 0.2a + 0.6a + 0.6(100 - 3a) = 35 simplifies to -a + 60 = 35, so a = 25 and c = 100 - 75 = 25 mL.

Let the middle angle be m. The largest is m + 20 and the smallest is m/2. Using the triangle angle sum: m/2 + m + (m + 20) = 180, giving 5m/2 = 160 and m = 64. The largest angle is 64 + 20 = 84 degrees. Choosing 80 degrees results from omitting the 20-degree offset in the equation.

Let x = batches of cookies and y = batches of brownies. Constraints: 2x + y <= 12 and x + 2y <= 12. Corner points: (6,0) gives P = 30; (0,6) gives P = 24; solving both constraints as equations gives x = 4, y = 4 and P = 20 + 16 = 36. The maximum profit of $36 occurs at x = 4 batches of cookies.

\(f(g(x)) = \sqrt{x^2 - 4}\). The square root requires \(x^2 - 4 \geq 0\), meaning \(x^2 \geq 4\) and therefore \(|x| \geq 2\). This gives \(x \leq -2 \text{ or } x \geq 2\). Choice B is incomplete because it ignores the valid negative interval. Choice D is wrong because those values make \(x^2 - 4 \leq 0\), leaving a negative radicand.

Substitute x = 2: f(2) = 4(2) - 7 = 8 - 7 = 1. A common error is reversing the subtraction to get -1, which would mean computing 7 - 8 instead of 8 - 7.

Substitute \(x = -2\): \(g(-2) = (-2)^2 + 3(-2) = 4 - 6 = -2\). A common mistake is treating \((-2)^2\) as \(-4\) instead of \(4\), which would give \(-4 - 6 = -10\).

f(a) = b means the function f maps input a to output b. The parentheses in f(a) indicate input, not multiplication. Choice D describes slope-intercept form y = mx + b, which is unrelated to what function notation means.

Each linear equation in three variables represents a plane in 3D space. When exactly one point lies on all three planes simultaneously, the system has a unique solution. Three parallel planes share no common points, giving no solution.

The feasible region is the set of all ordered pairs (x, y) that satisfy every inequality constraint at once. The optimal solution is found at a corner of this region, but the region itself contains all candidate points, not just the optimal one.

Set -2x + 6 = 0, then -2x = -6, so x = 3. Choice A results from dividing 6 by -2 without the sign correction: -6 / -2 = 3, not -3. Always isolate x completely before reading off the answer.

A constant function outputs the same value for every input, written f(x) = c. An identity function satisfies f(x) = x, returning the input itself. A constant function is NOT one-to-one because every distinct input maps to the same output 5.

Replace h(x) with y: y = 3x - 5. Swap x and y to get x = 3y - 5. Solve for y: 3y = x + 5, so y = (x + 5) / 3. Choice A uses subtraction instead of addition when moving -5 across the equals sign, reversing the operation incorrectly.

\((f \circ g)(4) = f(g(4))\). Evaluate the inner function first: \(g(4) = 4^2 - 1 = 15\). Then \(f(15) = 15 + 3 = 18\). Reversing the order would give \((g \circ f)(4) = g(7) = 48\), which is not an answer choice here but illustrates why composition order matters.

Subtract equation 2 from equation 1: 2y = 6, so y = 3. Substituting y = 3 into equations 1 and 3 gives x + z = 6 and 2x - z = 3. Adding these: 3x = 9, so x = 3 and z = 3. The symmetric structure of this system yields the elegant solution (3, 3, 3).

Evaluate P = 2x + 5y at all vertices: (0,0) gives 0, (6,0) gives 12, (4,3) gives 8 + 15 = 23, and (0,4) gives 20. The maximum is 23 at (4,3). For P = 4x + y, the value at (6,0) is 24, which beats (4,3)'s value of 19, so that objective is maximized at a different vertex.

(f composed with f)(x) = f(f(x)) = f(2x + 3) = 2(2x + 3) + 3 = 4x + 6 + 3 = 4x + 9. Choice B (4x + 6) comes from computing 2(2x + 3) = 4x + 6 but forgetting to add the final + 3 from the outer application of f.

If two planes are parallel, they share no points, so no single point can satisfy all three equations simultaneously. The system has no solution. Infinitely many solutions occur when planes share a full line or coincide entirely. A linear system can never have exactly two solutions.

The slope is (10 - 4) / (3 - 1) = 6 / 2 = 3. Using point-slope form: f(x) = 3(x - 1) + 4 = 3x + 1. Then f(0) = 1. Alternatively, moving one step left from x = 1 subtracts the slope: f(0) = 4 - 3 = 1. Choice C (2) is tempting but uses an incorrect slope calculation.

Evaluate C at each vertex: C(2,1) = 2 + 3 = 5, C(5,0) = 5 + 0 = 5, C(7,2) = 7 + 6 = 13, C(3,4) = 3 + 12 = 15. The minimum is 5, achieved at both (2,1) and (5,0). When a minimum occurs at two adjacent vertices, the entire edge between them also yields that minimum value.

Apply the rule step by step: f(3) = f(2) + 3 = 10, f(4) = f(3) + 3 = 13, f(5) = f(4) + 3 = 16. This recursive rule defines a linear function with slope 3. Choice D (19) would be f(6), not f(5).

First simplify f(f(x)): f(f(x)) = f(2x - 1) = 2(2x - 1) - 1 = 4x - 3. Set 4x - 3 = 7, so 4x = 10 and x = 5/2. Verify: f(5/2) = 2(5/2) - 1 = 4, and f(4) = 7. Choice A (x = 2) gives f(f(2)) = f(3) = 5, not 7.

Subtract equation 2 from equation 1: x - y = -1, so y = x + 1. Subtract equation 3 from equation 1: x - z = 1, so z = x - 1. Substitute into equation 1: 2x + (x+1) + (x-1) = 8, giving 4x = 8 and x = 2. Then y = 3 and z = 1. Verify in all three original equations to confirm (2, 3, 1).

From equation 1: z = -x - y. Substituting into equation 2 gives -x - 4y = 7, so x = -7 - 4y. Substituting z into equation 3 gives x + 5y = -9. Replacing x with -7 - 4y: -7 + y = -9, so y = -2. Then x = -7 + 8 = 1 and z = -1 + 2 = 1. Verify: 1 + (-2) + 1 = 0, 2 + 2 + 3 = 7, -1 - 6 - 2 = -9. Choice B is the negation of every coordinate, a classic sign error.

Let \(u = \frac{x}{x+1}\). Then \(f(u) = \frac{u}{u + 1}\). The denominator is \(u + 1 = \frac{x}{x+1} + 1 = \frac{x + x + 1}{x+1} = \frac{2x+1}{x+1}\). So \(f(u) = \frac{\frac{x}{x+1}}{\frac{2x+1}{x+1}} = \frac{x}{2x+1}\). Choice B squares the function value rather than composing it, and Choice C incorrectly adds \(1\) to only the \(x\)-coefficient in the denominator.

Find the corner points by intersecting boundary lines. Setting x + y = 8 and x + 2y = 12 simultaneously gives y = 4 and x = 4. The vertices of the feasible region are (0,0), (8,0), (4,4), and (0,6). Evaluate P: P(0,0)=0, P(8,0)=48, P(4,4)=24+32=56, P(0,6)=48. The maximum is 56 at (4,4). Note that (0,8) is not feasible since x + 2(8) = 16 > 12.

Evaluate from the inside out. f^(-1)(9) asks which input produces output 9. Since f(5) = 9, we have f^(-1)(9) = 5. Now evaluate f^(-1)(5): since f(2) = 5, we have f^(-1)(5) = 2. Therefore f^(-1)(f^(-1)(9)) = 2. Choice C (5) is the intermediate result, not the final answer.

Let a, b, and c be the volumes of 20%, 40%, and 50% solutions. The three equations are: a + b + c = 500, 0.20a + 0.40b + 0.50c = 175 (concentration), and a = b + 100. Substituting a = b + 100 into the first two equations gives 2b + c = 400 and 0.60b + 0.50c = 175. From the first, c = 400 - 2b. Substituting: 0.60b + 0.50(400 - 2b) = 175, which gives 0.60b + 200 - b = 175, so -0.4b = -25... wait, let me recheck. Actually 0.20(b+100) + 0.40b + 0.50c = 175 simplifies to 0.60b + 0.50c = 155. With c = 400 - 2b: 0.60b + 0.50(400-2b) = 155, giving 0.60b + 200 - b = 155, so -0.4b = -45, b = 112.5. Then c = 400 - 225 = 175. The 50% solution volume is 175 mL.

Build up step by step. f(f(x)) = f(2x+1) = 2(2x+1) + 1 = 4x + 3. Then f(f(f(x))) = f(4x+3) = 2(4x+3) + 1 = 8x + 6 + 1 = 8x + 7. Choice B (8x + 3) uses the correct leading coefficient but applies the constant incorrectly. Choice D (8x + 6) stops one step early by omitting the final +1.

Find the corner points. The intersection of 3x + 4y = 24 and x + 2y = 10: multiply the second by 2 to get 2x + 4y = 20, then subtract from the first: x = 4, y = 3. The vertices are (0,0), (8,0), (4,3), and (0,5). Evaluate P: P(0,0)=0, P(8,0)=40, P(4,3)=20+21=41, P(0,5)=35. The maximum is 41 at (4,3). Choice A (40) is the value at (8,0), which is optimal only when the objective places relatively more weight on x.

Substitute x = 3: f(3) = 4(3) - 7 = 12 - 7 = 5. A common error is stopping at 4(3) = 12 without subtracting 7, giving the wrong answer of 12.

Substitute x = -4: h(-4) = -3(-4) + 2 = 12 + 2 = 14. A negative times a negative gives a positive, so -3 times -4 equals 12, not -12. Choosing -14 results from forgetting that the negatives cancel.

A relation is a function only if each input (x-value) maps to exactly one output. In choice B, the input x = 1 maps to both 2 and 3, violating the definition. Choice D is a valid function even though two different inputs share the same output — that is allowed.

A linear function written as f(x) = mx + b has slope m and y-intercept b. Here m = -5 and b = 8, so the slope is -5. The value 8 is the y-intercept, not the slope.

Set 6x + 1 = 25 and solve: 6x = 24, so x = 4. Substituting back confirms f(4) = 6(4) + 1 = 25. Choosing x = 3 gives f(3) = 19, not 25.

The expression inside the parentheses is the input to the function. So f(a + 1) means substitute the quantity (a + 1) for x in the function rule. Functions are not generally additive, so f(a + 1) does not equal f(a) + f(1) unless f is linear and f(0) = 0.

Since f(0) = 5, the y-intercept is 5. The slope is 3, so f(x) = 3x + 5. Evaluating at x = 2: f(2) = 3(2) + 5 = 11. Choosing 6 results from computing 3(2) and forgetting to add the y-intercept. Choosing 13 results from swapping slope and intercept to get 5(2) + 3.

Slope = (change in output) / (change in input) = (10 - 7) / (2 - 1) = 3 / 1 = 3. The values 7 and 10 are outputs of the function, not the slope. A slope of 2 would mean the function increases by 2 per unit, but the actual increase from x = 1 to x = 2 is 3.

(f composed with g)(4) means f(g(4)): first apply g, then f. g(4) = 4 - 1 = 3, then f(3) = 3(3) + 2 = 11. A common mistake is reversing the order: (g composed with f)(4) = g(f(4)) = g(14) = 13, which is choice B. Choice C is f(4) = 14, which skips applying g.

Subtract the second equation from the first: (x + y + z) - (x + y - z) = 6 - 2, giving 2z = 4, so z = 2. Then subtract the third equation from the first: (x + y + z) - (x - y + z) = 6 - 4, giving 2y = 2, so y = 1. Finally x = 6 - 1 - 2 = 3. The full solution is (3, 1, 2).

Evaluate P at each corner: P(0,0) = 0, P(5,0) = 15, P(3,4) = 9 + 8 = 17, P(0,3) = 6. The maximum is 17 at (3, 4). Choice A corresponds to the corner (5, 0), which gives a smaller value. The maximum does not always occur at the vertex with the largest coordinates.

Let a = number of product A and b = number of product B. Set up the system: a + b = 10 and 4a + 6b = 48. From the first equation, b = 10 - a. Substituting: 4a + 6(10 - a) = 48 gives 4a + 60 - 6a = 48, so -2a = -12 and a = 6. Verify: 6 items at $4 plus 4 items at $6 equals $24 + $24 = $48.

To find the inverse, replace f(x) with y, swap x and y, then solve for y. Starting with y = (1/2)x + 4: swap to get x = (1/2)y + 4. Subtract 4: x - 4 = (1/2)y. Multiply by 2: y = 2x - 8. So f inverse(x) = 2x - 8. Choice A results from adding 8 instead of subtracting. Choice C keeps the same slope without taking the reciprocal.

The corner points are found at intersections of boundary lines within the feasible region. The four corners are: (0, 0) from the axes, (4, 0) from 2x + y = 8 at y = 0, (0, 4) from x + 2y = 8 at x = 0, and the intersection of x + 2y = 8 and 2x + y = 8 which gives x = y = 8/3. All four points satisfy all constraints, so there are exactly 4 corner points.

Add the first and third equations: 5x + 4y = 13. Add the second equation multiplied by 3 and the first equation: multiplying eq2 by 3 gives 3x - 3y + 6z = 15, adding to eq1 yields 5x + 7z = 26. Using elimination consistently, the unique solution is x = 1, y = 2, z = 3. Verify in eq1: 2(1) + 3(2) + 3 = 11, in eq2: 1 - 2 + 6 = 5, and in eq3: 3 + 2 - 3 = 2.

First find k by substituting (3, 11): 2(3) + k = 11, so 6 + k = 11 and k = 5. The function is f(x) = 2x + 5. Now evaluate at x = -1: f(-1) = 2(-1) + 5 = -2 + 5 = 3. A common error is stopping after finding k = 5 and reporting that as the answer.

Evaluate P at each corner: P(0,6) = 0 + 6 = 6, P(2,4) = 8 + 4 = 12, P(5,0) = 20 + 0 = 20, P(0,0) = 0. The maximum is 20 at (5, 0). Because the coefficient of x is much larger than the coefficient of y, the objective function favors corners with larger x-values.

First find f(f(x)) by composing f with itself: f(f(x)) = f(3x - 3) = 3(3x - 3) - 3 = 9x - 9 - 3 = 9x - 12. Set equal to 15: 9x - 12 = 15, so 9x = 27 and x = 3. Verify: f(3) = 9 - 3 = 6 and f(6) = 18 - 3 = 15. A common error is solving f(x) = 15 directly: 3x - 3 = 15 gives x = 6, but that ignores the outer application of f.

Let a, b, c be acres of the $200, $300, $400 crops. The three equations are: a + b + c = 50, c = 2a, and 200a + 300b + 400c = 16000. Substituting c = 2a into the first: 3a + b = 50, so b = 50 - 3a. Substituting into the profit equation: 200a + 300(50 - 3a) + 400(2a) = 16000 gives 200a + 15000 - 900a + 800a = 16000, so 100a = 1000 and a = 10. Then b = 50 - 30 = 20 and c = 20.

Set y = (2x + 1)/(x - 3) and solve for x in terms of y. Multiply both sides by (x - 3): y(x - 3) = 2x + 1, giving xy - 3y = 2x + 1. Collect x terms: xy - 2x = 3y + 1, so x(y - 2) = 3y + 1 and x = (3y + 1)/(y - 2). Replacing y with x gives f inverse(x) = (3x + 1)/(x - 2). Choice B is the reciprocal of f, not the inverse.

If P is constant along the entire edge, both endpoints yield the same value: 2a + 6b = 5a + 0b. Simplifying: 6b = 3a, which gives a = 2b. Therefore a:b = 2:1. This occurs when the slope of the objective function's level curve exactly matches the slope of the feasible region edge, meaning the constraint line and the profit line are parallel.

Add equations 1 and 2: 3x + y = 9. Multiply equation 1 by 2 and add to equation 3: 2x + 4y - 2z + 3x + y + 2z = 2 + 13, giving 5x + 5y = 15, so x + y = 3. Subtracting from 3x + y = 9 gives 2x = 6, so x = 3. Then y = 3 - 3 = 0. Substituting into equation 1: 3 + 0 - z = 1, so z = 2. Therefore x + y + z = 3 + 0 + 2 = 5.

Identify corner points by solving pairs of boundary equations. The vertices of the feasible region are: (0,0), (8,0), (8,2) from x=8 and x+y=10, (3,7) from x+y=10 and x+3y=24, and (0,8) from x=0 and x+3y=24. Evaluating P: P(0,0)=0, P(8,0)=24, P(8,2)=34, P(3,7)=9+35=44, P(0,8)=40. The maximum is 44 at (3,7). Choice A is the value at (0,8), which is not the maximum.

First find the function: slope = (16 - 7)/(5 - 2) = 3. Using f(2) = 7: 7 = 3(2) + b gives b = 1, so f(x) = 3x + 1. Now set f(x) = f(2x - 1): 3x + 1 = 3(2x - 1) + 1 = 6x - 2. Solving: 3 = 3x, so x = 1. Verify: f(1) = 4 and f(2(1) - 1) = f(1) = 4.

The third row of the augmented matrix represents the equation 0x + 0y + 0z = 5, which simplifies to 0 = 5. This is a contradiction — no values of x, y, z can satisfy it — so the system has no solutions. If the third row were [0, 0, 0 | 0], the system would have infinitely many solutions because z would be a free variable. A system of three linear equations in three variables cannot have exactly two or three isolated solutions.

Substitute x = 4 into the function: f(4) = 3(4) - 5 = 12 - 5 = 7. A common error is computing 3(4) = 12 and stopping without subtracting 5, which yields the distractor 12.

A relation is a function if and only if each input (x-value) maps to exactly one output. Choice B has inputs 2, 3, and 4, each appearing exactly once, so it qualifies. Note that two different inputs sharing the same output is allowed — that is not what disqualifies a relation. Choices A, C, and D each repeat an x-value with different y-values, violating the definition.

Function notation f(x) means 'the value of function f at input x.' So f(5) = 12 states that substituting 5 for x produces an output of 12. The parentheses in function notation do NOT indicate multiplication; that is a frequent misconception.

The feasible region is the intersection of all constraint inequalities — it is every point (x, y) satisfying all constraints at once. The optimal value of the objective function is found by evaluating it at the corner points of this region, not at a single boundary or at the origin.

Substitute x = 0: g(0) = -2(0) + 7 = 0 + 7 = 7. For any linear function f(x) = mx + b, the y-intercept equals f(0) = b. Here b = 7. The distractor -2 is the slope, not the output value at x = 0.

Because the objective function is linear, it cannot curve to produce a higher or lower value in the interior of the region. The extremes must occur on the boundary, and since the function is linear on each edge, the extreme on any edge occurs at one of its endpoints — a vertex. The Corner Point Theorem formalizes this: always check all vertices to find the optimum.

Substitute x = 9: f(9) = 9/3 + 2 = 3 + 2 = 5. A common mistake is stopping after dividing by 3 and reporting 3, forgetting to add 2. Another error is dividing 9 by 3 incorrectly or adding before dividing.

Set the expression equal to 17 and solve: 4x - 3 = 17, so 4x = 20, giving x = 5. Check: f(5) = 4(5) - 3 = 17. The distractor x = 4 comes from forgetting to divide by 4 after adding 3, mistakenly writing 4x = 20 and then x = 20/5 = 4 — an arithmetic slip.

Subtract equation 1 from equation 2: x - 2y = -3. Subtract equation 1 from equation 3: -2y + z = -1. From x - 2y = -3, write x = 2y - 3. Substitute into equation 1: (2y - 3) + y + z = 6, giving 3y + z = 9. Combine with -2y + z = -1 by subtracting: 5y = 10, so y = 2, then z = 3, and x = 1. Verify: 1+2+3=6, 2-2+3=3, 1-2+6=5. All three equations hold.

Find the corner points by solving pairs of boundary equations. The lines x + y = 6 and 2x + y = 8 intersect at x = 2, y = 4. The four vertices of the feasible region are (0, 0), (4, 0), (2, 4), and (0, 6). Evaluate P: P(0,0)=0, P(4,0)=20, P(2,4)=10+12=22, P(0,6)=18. The maximum is 22 at (2, 4). Students who overlook the interior intersection point may incorrectly choose 20.

Evaluate the inner function first: g(4) = 4 - 3 = 1. Then apply f to that result: f(1) = 2(1) + 1 = 3. A frequent error is reversing the order and computing g(f(4)) instead: f(4) = 9, then g(9) = 6 — that is a different composition. Always work from the inside out.

Let a, s, c be the counts of adult, student, and child tickets. The system is: a + s + c = 100, 12a + 8s + 4c = 880, and s = 2c. Substituting s = 2c gives a + 3c = 100, so a = 100 - 3c. Substituting into the revenue equation: 12(100 - 3c) + 8(2c) + 4c = 880, which simplifies to 1200 - 16c = 880, giving c = 20. Then s = 40 and a = 40. Verify: 40 + 40 + 20 = 100 and 480 + 320 + 80 = 880.

Each chair uses 2 labor hours and each table uses 5, so total labor is 2x + 5y, which must not exceed 40: 2x + 5y <= 40. Each chair uses 3 kg of wood and each table uses 4 kg, so 3x + 4y <= 36. Quantities cannot be negative, so x >= 0 and y >= 0. Choice B swaps the coefficients of x and y in both constraints. Choice D uses 4x + 3y for wood instead of 3x + 4y.

Compute the slope: m = (13 - 5)/(6 - 2) = 8/4 = 2. Use point-slope form with (2, 5): y - 5 = 2(x - 2), so f(x) = 2x + 1. Then f(10) = 2(10) + 1 = 21. A common error is forgetting the y-intercept term, computing f(10) = 2(10) = 20, and then selecting a nearby distractor such as 19.

Test (2, 3, 4): 2+3+4=9 (true), 4-3+4=5 (true), 2+6-4=4 (true). All three equations are satisfied. Check (1, 4, 4): 2(1)-4+4=2, not 5 — fails the second equation. Check (3, 2, 4): 2(3)-2+4=8, not 5 — fails. Check (2, 4, 3): 2(2)-4+3=3, not 5 — fails. Only (2, 3, 4) works.

Find the corner points of the feasible region (the region satisfying all constraints). Setting x + y = 4 and 2x + y = 6 simultaneously gives x = 2, y = 2. Setting x = 0 in 2x + y >= 6 gives y = 6, so the vertex is (0, 6). Setting y = 0 in 2x + y >= 6 gives x = 3, so the vertex is (3, 0). Evaluate C at each vertex: C(0,6) = 12, C(2,2) = 10, C(3,0) = 9. The minimum is 9 at (3, 0). The distractor 8 might be chosen by students who incorrectly include (0, 4) as a vertex, but (0, 4) fails the constraint 2x + y >= 6.

Substitute f(x) into g: g(f(x)) = f(x) - 4 = (3x + 2) - 4 = 3x - 2. A common error is computing f(g(x)) instead: f(g(x)) = 3(x - 4) + 2 = 3x - 10. Another distractor is 3x + 6, which results from mistakenly adding 4 rather than subtracting it.

Let x be the first number. The second is x + 4 and the third is 2x. Their sum: x + (x + 4) + 2x = 20, giving 4x + 4 = 20, so x = 4. The three numbers are 4, 8, and 8. The largest is 8. A common error is solving 4x = 20 to get x = 5 by forgetting the constant +4, which leads to the distractor 10.

First derive f-inverse: set y = 2x - 4 and solve for x to get f-inverse(x) = (x + 4)/2. Now evaluate the argument of f: f-inverse(x) + 1 = (x + 4)/2 + 1. Apply f to this: f((x+4)/2 + 1) = 2 * ((x+4)/2 + 1) - 4 = (x + 4) + 2 - 4 = x + 2. Set equal to 10: x + 2 = 10, so x = 8. The distractor 10 tempts students who skip the inverse step and solve 2(x+1) - 4 = 10 directly.

The third row [0, 0, 0 | 0] represents 0 = 0, which is always true, indicating the system is consistent but underdetermined. Since there are three unknowns and only two non-trivial equations, z is a free parameter t. The full solution set is x = 5 - 2t, y = 3 + t, z = t for any real t, giving infinitely many solutions. If the last row were [0, 0, 0 | k] with k not equal to 0, the equation 0 = k would be a contradiction and the system would have no solution.

Evaluate P at each corner: P(0,6) = 24, P(2,3) = 2c + 12, P(5,1) = 5c + 4, P(8,0) = 8c. For the minimum to occur at (2, 3), three inequalities must hold: 2c + 12 <= 24 (giving c <= 6), 2c + 12 <= 5c + 4 (giving c >= 8/3), and 2c + 12 <= 8c (giving c >= 2). The binding range is 8/3 <= c <= 6. Testing c = 4: P values are 24, 20, 24, 32 — minimum is 20 at (2, 3). At c = 1, the minimum shifts to (8, 0); at c = 7, it shifts to (0, 6).

First compute f(3): f(3) = (3 + 2)/(3 - 1) = 5/2. Then compute f(5/2): f(5/2) = (5/2 + 2)/(5/2 - 1) = (9/2)/(3/2) = 3. So f(f(3)) = 3. This function has the interesting property that applying it twice returns the original output. The distractor 5/2 is f(3) — the result after only one application — which students choose if they stop too early.

Let o, m, y denote the oldest, middle, and youngest ages. The system is: o + m + y = 45, 2o - m - y = 9, and m = y + 5. Substitute m = y + 5 into equation 1: o + 2y + 5 = 45, so o + 2y = 40. Substitute m = y + 5 into equation 2: 2o - (y + 5) - y = 9, giving 2o - 2y = 14, so o - y = 7 and o = y + 7. Substitute into o + 2y = 40: (y + 7) + 2y = 40, giving 3y = 33, so y = 11. Then m = 16 and o = 18. Verify: 18 + 16 + 11 = 45 and 2(18) - 11 = 25 = 16 + 9.

Evaluate P at each corner: P(0,8) = 8k, P(4,4) = 24 + 4k, P(6,0) = 36. For the maximum to be at (4, 4): need 24 + 4k >= 8k (giving k <= 6) and 24 + 4k >= 36 (giving k >= 3). The valid range is 3 <= k <= 6. At k = 4: P(0,8) = 32, P(4,4) = 40, P(6,0) = 36 — maximum is 40 at (4, 4). At k = 2, the maximum is at (6, 0); at k = 7, the maximum shifts to (0, 8).

Add equations 2 and 3: 3x + z = 9, so z = 9 - 3x. Add equations 1 and 3: 2x + 3z = 13. Substitute z = 9 - 3x: 2x + 3(9 - 3x) = 13, giving 2x + 27 - 9x = 13, so -7x = -14, x = 2. Then z = 9 - 6 = 3. From equation 1: 2 + y + 3 = 6, so y = 1. The solution is (2, 1, 3). Finally, 2(2) - 1 + 3 = 4 - 1 + 3 = 6. Students who confuse the expression with the first equation and answer 6 via shortcut should verify by actually computing the combination.

Substitute x = 2: f(2) = 4(2) - 3 = 8 - 3 = 5. Choice B (8) is a frequent error where students compute 4(2) = 8 but forget to subtract 3.

Function notation f(3) = 10 means: the input value is 3 and the corresponding output is 10. Choice D reverses the roles of input and output, a common misconception about function notation.

Substitute x = 0: f(0) = 3(0) + 7 = 0 + 7 = 7. For any linear function f(x) = mx + b, the value f(0) equals the y-intercept b. Choice D (10) results from adding 3 and 7 rather than substituting x = 0 first.

The Corner Point Theorem states that for a linear objective function over a convex feasible region, the optimal value always occurs at one of the vertices. Interior points are never optimal for strictly linear objectives, ruling out Choice B.

Substitute x = -1: f(-1) = -5(-1) + 3 = 5 + 3 = 8. The key step is recognizing that a negative coefficient times a negative input is positive. Choice A (-2) results from computing -5 + (-1) + 3 by treating multiplication as addition.

Substitute each triple: (2) + (-1) + (-1) = 0, so Choice B works. For Choice A: 1 + 2 + 3 = 6. For Choice C: 0 + 0 + 1 = 1. For Choice D: -1 + 2 + 0 = 1. Only Choice B sums to zero.

Replace every x with (x + 3): f(x + 3) = 4(x + 3) - 5 = 4x + 12 - 5 = 4x + 7. Choice B (4x - 2) is a frequent error where students shift only the constant term by adding 3 to -5, ignoring that 3 must also be multiplied by the leading coefficient 4.

Substitute x = 3 and y = 4: P = 2(3) + 3(4) = 6 + 12 = 18. Choice A (14) results from using a coefficient of 2 for y as well: 2(3) + 2(4) = 14, incorrectly applying the same coefficient to both variables.

Evaluate the inner function first: f(4) = 2(4) - 1 = 7. Then apply g: g(7) = 3(7) + 2 = 23. Choice A (14) results from stopping after computing g(4) = 14 without composing. Choice D (27) comes from reversing the order and computing f(g(4)) = f(14) = 27.

The total flour consumed is 3x (from cakes) plus 4y (from pies). Since at most 48 cups are available, the constraint is 3x + 4y <= 48. Choice A uses the wrong direction (>= instead of <=). Choice B swaps the per-item flour amounts for cakes and pies.

Subtract equation 2 from equation 1: 2y = 6, so y = 3. Subtract equation 3 from equation 1: 2z = 4, so z = 2. Substitute back into equation 1: x = 9 - 3 - 2 = 4. The solution is (4, 3, 2). Choice A lists (3, 4, 2), which swaps the values of x and y.

Compute f(a + 1) = 5(a + 1) + 2 = 5a + 7. Then f(a + 1) - f(a) = (5a + 7) - (5a + 2) = 5. For any linear function f(x) = mx + b, this difference always equals the slope m. Choice C (5a + 5) results from forgetting to subtract f(a), keeping the expression un-simplified.

Evaluate P at each vertex: P(0,4) = 0 + 20 = 20; P(2,6) = 6 + 30 = 36; P(5,1) = 15 + 5 = 20; P(4,0) = 12 + 0 = 12. The maximum is 36, occurring at (2,6). Choice A (20) appears at two different vertices, which may tempt students who check only some corner points.

The slope is m = (18 - 9)/(5 - 2) = 3. Using point (2, 9): 3(2) + b = 9, so b = 3. The function is f(x) = 3x + 3, and f(0) = 3. Choice C (6) is a common error from using slope 3 but computing the y-intercept as 9 - 2 = 7 or misapplying point-slope form.

Let the youngest age be x. Then the middle age is x + 4 and the oldest is 2x. The equation is x + (x + 4) + 2x = 48, giving 4x + 4 = 48, so 4x = 44 and x = 11. Choice A (10) comes from incorrectly setting up 4x = 40, perhaps from a subtraction error with the constant.

Let p, n, b represent pens, notebooks, and binders. The three equations are p + n + b = 14; p + 3n + 5b = 30; and p = n + b. Substituting the third into the first gives 2(n + b) = 14, so n + b = 7 and p = 7. From the cost equation: 7 + 3n + 5b = 30, giving 3n + 5b = 23. Since n = 7 - b: 3(7 - b) + 5b = 23, so 21 + 2b = 23, and b = 1. Choice B (2) results from an arithmetic error when solving 2b = 2.

Vertices occur at intersections of boundary lines within the feasible region. Setting x + y = 7 and x - y = 3 and solving simultaneously gives x = 5, y = 2. The point (5, 2) satisfies all constraints. Choice A (7, 0): x - y = 7 > 3, violating the fourth constraint. Choice D (6, 1): x - y = 5 > 3, also infeasible.

Find a and b: slope a = (13 - 7)/(5 - 2) = 2; then 2(2) + b = 7 gives b = 3, so f(x) = 2x + 3. Evaluate inside-out: f(1) = 2(1) + 3 = 5, then f(f(1)) = f(5) = 2(5) + 3 = 13. Choice A (9) results from computing f(3) instead of f(f(1)), stopping at an intermediate step.

From equation 3, z = x + y. Substituting into equation 1 gives 2x + 3y = 8, and into equation 2 gives 5x + 2y = 9. Multiply the first by 2 and the second by 3: 4x + 6y = 16 and 15x + 6y = 27. Subtracting yields 11x = 11, so x = 1. Then y = 2 and z = 3. Therefore 3x + 2z = 3 + 6 = 9. Choice D (11) results from computing x + y + z = 1 + 2 + 3 = 6 and then adding an incorrect offset.

Set Q equal at both vertices: Q(7, 0) = 7k and Q(4, 6) = 4k + 24. Solving 7k = 4k + 24 gives 3k = 24, so k = 8. When the objective function has equal values at two adjacent vertices, its level lines are parallel to that edge. Choice B (6) results from incorrectly solving 3k = 18, which would apply if the y-coefficient were 3 rather than 4.

Compute f(1) = 1 and f(3) = 7. Then g(1) = 10 and g(7) = 22. Solving m + n = 10 and 7m + n = 22 gives 6m = 12, so m = 2 and n = 8. Thus g(x) = 2x + 8. Since f(2) = 4, we get g(f(2)) = g(4) = 2(4) + 8 = 16. Choice A (14) results from evaluating g at x = 3 instead of at f(2) = 4, confusing the input to g.

Set P(5, 0) = P(4, 2): 5a = 4a + 2b, which simplifies to a = 2b, so a/b = 2. When two adjacent vertices yield the same objective value, the entire connecting edge is optimal. Choice A (1/2) is the reciprocal of the correct answer, a common error from inverting the ratio when setting up the equation.

Substitute f(x) into h: h(2x + 3) = m(2x + 3) + c = 2mx + (3m + c). Matching coefficients with 6x + 11 gives 2m = 6, so m = 3, and 3(3) + c = 11, so c = 2. Therefore h(x) = 3x + 2, and h(5) = 3(5) + 2 = 17. Choice A (13) results from using h(x) = f(x) = 2x + 3 rather than finding h separately.

Let k be the common equal value after the adjustments. Then a = k - 3, b = k/2, and c = k + 3. Substituting into a + b + c = 30 gives (k - 3) + k/2 + (k + 3) = 30, which simplifies to 5k/2 = 30, so k = 12. The originals are a = 9, b = 6, and c = 15. The largest is c = 15. Choice B (12) is the value of k, not one of the original numbers.

Find all vertices by intersecting boundary lines within the feasible region: (0,0), (5,0), (5,3) from x = 5 and x + y = 8, (2,6) from y = 6 and x + y = 8, and (0,6). Evaluate: P(5,0) = 25; P(5,3) = 25 + 12 = 37; P(2,6) = 10 + 24 = 34; P(0,6) = 24. The maximum is 37 at (5,3). Choice A (34) is the value at (2,6), which students may select if they overlook the vertex (5,3) when sketching the region.

Substitute x = 4: f(4) = 3(4) - 5 = 12 - 5 = 7. A common mistake is computing 3(4) = 12 and stopping, which gives distractor 12. Always apply the full function rule including the constant term.

Substitute x = 0: f(0) = -2(0) + 6 = 0 + 6 = 6. The y-intercept of a linear function f(x) = mx + b is always b, which here is 6. Distractor -2 is the slope, not the function value at x = 0.

A linear function has the form \(f(x) = mx + b\), where the variable \(x\) appears to the first power only. \(f(x) = 5x - 2\) fits this form. \(f(x) = x^2 + 3\) is quadratic, \(f(x) = \frac{3}{x}\) is a rational function, and \(f(x) = 2^x\) is exponential — none of these are linear.

To uniquely determine three unknowns, you need exactly three independent equations — one constraint per unknown. With fewer equations the system is underdetermined (infinitely many solutions), and a fourth redundant equation adds no new information.

The feasible region contains every point that satisfies ALL constraints at the same time. Boundary points satisfy at least one constraint with equality, but the full region includes interior points too. Non-negative coordinates describe only the non-negativity constraints, not the full feasible region.

In slope-intercept form f(x) = mx + b, the coefficient m is the slope and b is the y-intercept. Here m = 7 and b = -4. The slope is 7, not -4 (which is the y-intercept).

Substitute x = -3: f(-3) = -(-3) + 5 = 3 + 5 = 8. A frequent error is forgetting that the negative of a negative is positive: -(-3) = +3, not -3. Distractor 2 results from computing -(3) + 5 incorrectly as 5 - 3.

Evaluate the inner function first: g(4) = 4 - 1 = 3. Then apply f: f(3) = 2(3) + 3 = 9. A common mistake is computing f(4) first and then applying g, which gives g(11) = 10 — composition is not commutative, so order matters.

Add equations 1 and 2: 2x + 2z = 16, so x + z = 8. Add equations 1 and 3: 2x + 2y = 20, so x + y = 10. From equation 1, z = 12 - (x + y) = 12 - 10 = 2. Then x = 6 and y = 4. Verify: 6 + 4 + 2 = 12, 6 - 4 + 2 = 4, 6 + 4 - 2 = 8. All check out.

Find the slope: m = (12 - 3)/(4 - 1) = 9/3 = 3. Using point-slope form: f(x) - 3 = 3(x - 1), so f(x) = 3x. Then f(7) = 3(7) = 21. Distractor 24 results from mistakenly adding 3 to f(4) three times (adding slope once per unit), which is correct, but 12 + 9 = 21, not 24.

Evaluate P at each vertex: P(0,0) = 0, P(5,0) = 20, P(3,4) = 12 + 12 = 24, P(0,6) = 18. The maximum is 24 at (3, 4). By the Corner Point Theorem, the optimal value of a linear objective over a bounded feasible region always occurs at a vertex.

Since f(0) = -3, the y-intercept is b = -3. The slope is m = (5 - (-3))/(2 - 0) = 8/2 = 4. So f(x) = 4x - 3. Then f(5) = 4(5) - 3 = 20 - 3 = 17. Distractor 20 ignores the y-intercept, and 13 may result from using m = 3 by mistake.

First find f(-1) = 5(-1) - 3 = -8. Then f(-1) + 15 = -8 + 15 = 7. Set f(x) = 7: 5x - 3 = 7, so 5x = 10, giving x = 2. Distractor 1 comes from solving 5x = 5, which would mean f(x) = 2 rather than 7.

A contradiction like 0 = 5 (a false statement with no variables) means the system is inconsistent — there are no values of the variables that satisfy all equations simultaneously. This is distinct from 0 = 0 (a true statement), which indicates dependent equations and infinitely many solutions.

Substitute y = 2x and z = x + 3 into the first equation: x + 2x + (x + 3) = 15, which gives 4x + 3 = 15, so 4x = 12 and x = 3. Then y = 6 and z = 6. Verify: 3 + 6 + 6 = 15. Distractor 2 comes from solving 4x = 8 instead of 4x = 12.

The total flour used is 2x (from cakes) plus 3y (from pies). Since only 24 cups are available, the total must be less than or equal to 24: 2x + 3y <= 24. The inequality cannot be >= because the baker cannot use more flour than is available, and equality would mean every last cup must be used.

Replace x with (a - 1): f(a - 1) = -4(a - 1) + 10 = -4a + 4 + 10 = -4a + 14. Distractor -4a + 6 comes from computing -4a - 4 + 10, which incorrectly distributes the negative: -4(a - 1) = -4a + 4, not -4a - 4.

The feasible region has vertices at (1, 1), (6, 1), and (1, 6). Evaluate C at each: C(1,1) = 1 + 2 = 3, C(6,1) = 6 + 2 = 8, C(1,6) = 1 + 12 = 13. The minimum is 3 at (1, 1). Since the objective has positive coefficients, the minimum occurs at the vertex closest to the origin that still satisfies all constraints.

Compute g(f(k)) step by step: f(k) = 2k + 1, then g(f(k)) = 3(2k + 1) - 2 = 6k + 3 - 2 = 6k + 1. Set equal to 19: 6k + 1 = 19, so 6k = 18, giving k = 3. Distractor 4 comes from solving 6k = 24, which would arise if the composition were evaluated incorrectly as g(f(k)) = 6k + 5.

The second equation is exactly twice the first, so those two equations are dependent and carry no new information. The distinct equations are x + y + z = 5 and x - y + z = 3. Subtracting gives 2y = 2, so y = 1, and x + z = 4. Since x and z can take any values summing to 4, there are infinitely many solutions of the form (t, 1, 4 - t) for any real t.

The constraints are 2x + y <= 8, x + 3y <= 9, x >= 0, y >= 0. Finding vertices: (0,0) gives P = 0; (4,0) gives P = 12; (0,3) gives P = 12. For the intersection of 2x + y = 8 and x + 3y = 9: substituting y = 8 - 2x gives x + 3(8 - 2x) = 9, so -5x = -15, x = 3, y = 2, and P = 9 + 8 = 17. Maximum profit is 17.

Compute f(f(x)) = a(ax + b) + b = a^2 x + ab + b. Match coefficients with 4x + 9: a^2 = 4 gives a = 2 (taking the positive value), and ab + b = 9 gives 2b + b = 3b = 9, so b = 3. Verify: f(f(x)) = 2(2x + 3) + 3 = 4x + 6 + 3 = 4x + 9. Distractor a = 4 confuses a^2 = 4 with a = 4.

Let the three numbers be a, b, and c. The conditions give a + b + c = 18, 2a = b + c, and c = b + 3. From the first two equations: a + 2a = 18, so 3a = 18 and a = 6. Then b + c = 12. Using c = b + 3: 2b + 3 = 12, so b = 4.5 and c = 7.5. Verify: 6 + 4.5 + 7.5 = 18. Distractor 9 results from incorrectly solving 2a = 18.

Find the vertices of the feasible region: (0,0), (4,0), (4,2), and (0,6). The intersection of x = 4 and x + y = 6 gives (4, 2). Evaluate P: P(0,0) = 0, P(4,0) = 8, P(4,2) = 8 + 10 = 18, P(0,6) = 30. The maximum is 30 at (0,6). Because the objective heavily weights y (coefficient 5 vs 2), maximizing y while satisfying the constraints gives the best result.

The condition f(g(x)) = x means g is the inverse function of f. To find the inverse: set y = 3x + 6, solve for x to get x = (y - 6)/3, so f^(-1)(x) = (x - 6)/3. Therefore g(x) = (x - 6)/3 and g(2) = (2 - 6)/3 = -4/3. Distractor 4/3 comes from computing (2 + 6)/6 or misapplying the inverse formula.