Mixed practice: Algebra

B, 7 Section 1

Expand to get 6x − 3 = 4x + 11. Subtract 4x and add 3: 2x = 14, so x = 7. Substitution gives 39 on both sides. Choice A stops at a wrong sign; choice C reports 2x rather than x.

−3 Section 2

At the y-intercept, x = 0. Then −5y = 15, so y = −3. The point is (0, −3). Setting y = 0 instead would find the x-intercept, not the requested intercept.

C, 42 Section 3

The rate is (21 − 12)/(3 − 0) = 3, consistent with (33 − 21)/(7 − 3) = 3. Because f(0) = 12, the rule is f(x) = 3x + 12. Thus f (10) = 30 + 12 = 42. The table does not have equal input gaps, so compare rates rather than raw differences.

10 Section 4

Add the equations to eliminate y: 3x = 21, so x = 7. Then y = 17 − 7 = 10. The second equation checks: 2(7) − 10 = 4. Reporting 7 would answer the wrong variable.

A, x ≤ −4 Section 5

Subtract 8 to get −3x ≥ 12. Division by −3 reverses the relation, giving x ≤ −4. The endpoint is included because the original comparison includes equality. At x = −5, the left side is 23, confirming the allowed side.

200 Section 1

Let n be pages. The equation is 14 + 0.08n = 30, so 0.08n = 16 and n = 16/0.08 = 200. Check: the variable charge is $16, and the fixed charge raises the total to $30.

A, y = $\frac{\text{3}}{\text{2}}$ x − 2 Section 2

The line passes through (0, −2) and (4, 4). Its slope is (4 − (−2))/(4 − 0) = 6/4 = 3/2, and its y-intercept is −2. Therefore y = (3/2)x − 2. Both a correct slope and a correct intercept are required.

270 Section 3

Substitute zero into the entire input expression: W(0) = 250 − 5(0 − 4) = 250 + 20 = 270. The value 250 is anchored at t = 4, not at t = 0. Expanding to W(t) = 270 − 5t makes the initial value visible.

C, 70 Section 4

Let g + r = 120 and 18g + 10r = 1760. Subtract ten times the count equation from the revenue equation: 8g = 1760 − 1200 = 560, so g = 70. The other 50 tickets bring in $500; the general tickets bring in $1,260.

20 Section 5

The count satisfies 22 + 3.75n ≤ 100, so n ≤ 78/3.75 = 20.8. The greatest integer allowed is 20. Twenty posters cost $97; 21 cost $100.75, which exceeds the budget.

D, 4 Section 1

Substitute the known solution: a(7) = 15 + 13 = 28. Hence a = 4. This is a parameter question, but a supplied solution turns it into a simple equation for the parameter. There is no need to derive a general formula for x.

6 Section 2

The perpendicular slope is 1/2 because (−2)(1/2) = −1. Write y = (1/2)x + b and use the point: 7 = 1 + b, so b = 6. Check both features: the point lies on the line and the slopes multiply to −1.

B, 8 Section 3

The slope is (17 − (−1))/(8 − 2) = 18/6 = 3. Moving from input 2 to input 5 adds 3, so the output adds 3(3) = 9: f(5) = −1 + 9 = 8. Equivalently, input 5 is the midpoint of 2 and 8, so its linear output is the midpoint of −1 and 17.

6 Section 4

Add the equations: 5x + 5y = 30. Factor and divide by 5 to get x + y = 6. The individual values are x = 13/3 and y = 5/3, but computing them is unnecessary and introduces fractions that the requested sum avoids.

D, y ≤ 2x − 1 Section 5

The line passes through (0, −1) and (2, 3), giving slope (3 − (−1))/2 = 2. Its equation is y = 2x − 1. The shaded side is below the line and the boundary is solid, so equality is allowed: y ≤ 2x − 1. For example, (2, 0) is shaded and gives 0 ≤ 3.

4 Section 1

Expand to kx − 3k = 4x + 9. Matching the variable coefficients requires k = 4. Then the constants are −12 and 9, which disagree, so the equation has no solution. If k ≠ 4, the nonzero coefficient k − 4 instead gives one solution.

B, A decrease of 6 kilograms Section 2

The total is 3x + 8y = 240. Changes therefore satisfy 3Δx + 8Δy = 0. With Δx = 16, 48 + 8Δy = 0, so Δy = −6. The negative sign means a decrease of 6 kilograms. The extra $48 spent on one flour is offset by $48 less of the other.

1.2 Section 3

One hour is 60 minutes, so h = t/60. Substitute to get C(t) = 72(t/60) + 25 = 1.2t + 25. Therefore m = 1.2 dollars per minute. The fixed $25 stays unchanged because it is not measured per hour.

C, 23 Section 4

The y coefficients show that the first equation must be twice the second. Hence a − 2 = 6 and 30 = 2b, giving a = 8 and b = 15. Their sum is 23. Matching the whole equation verifies coincidence, not merely parallelism.

16 Section 5

Let c be cartons. Then 53 + 24c ≥ 428, so 24c ≥ 375 and c ≥ 15.625. The least integer is 16. Fifteen cartons produce only 53 + 360 = 413 trays; sixteen produce 437.

B, 27 Section 1

Expand to 2ax − 3a + b = 10x + 7. Matching x coefficients gives 2a = 10, so a = 5. Matching constants gives −15 + b = 7, so b = 22. Thus a + b = 27. The constant on the left is −3a + b, not merely b.

−2 Section 2

The old coordinates are x+2 and y+5 in terms of the new ones. The translated equation is 3(x+2)+2(y+5) = 12, or 3x + 2y = −4. Setting x = 0 gives y = −2. As a check, the original point (0, 6) moves to (−2, 1), which satisfies the new equation.

A, −3 Section 3

The output change is (b + 8) − (b − 12) = 20 for an input change of 5, so m = 4. The first condition then gives 4a + b = b − 12, so 4a = −12 and a = −3. The intercept cancels and need not be known.

12 Section 4

The system describes the single line 3x − 2y = 9. Four times its equation gives 12x − 8y = 36, so k = 12 works. For uniqueness, compare the allowed points (3, 0) and (0, −9/2): the expression gives 3k and 36, respectively. Equality forces k = 12.

C, 2 Section 5

An allowed y requires 3x − 2 < −x + 10, so 4x < 12 and x < 3. The greatest permitted integer is 2. It is attainable, for example with y = 5, since 5 > 4 and 5 ≤ 8. At x = 3, the system would require y > 7 and y ≤ 7, impossible.

4 Section 1

Let u = 5x − 2, so the target is 2u. Multiply the equation (4u + 7)/3 = 2u + 1 by 3 to get 4u + 7 = 6u + 3. Thus 4 = 2u, already the desired value. There is no need to solve for x.

D, 10 Section 2

The intercept point (0, 5) gives 5b = 30, so b = 6. In standard form the slope is −a/b, so −a/6 = −2/3 and a = 4. Therefore a + b = 10. Check: 4x + 6y = 30 rearranges to y = −(2/3)x + 5.

3 Section 3

The slope of f is (14−5)/(4−1) = 3, and its intercept is 2, so f(x) = 3x+2. Thus g(3) = 3(6+c)+2−1 = 19+3c. Setting this equal to 28 gives 3c = 9, so c = 3. Check directly: g(3) = f(9) − 1 = 29 − 1 = 28.

C, $\frac{\text{34}}{\text{9}}$ Section 4

Use y = 2x in the equation without k: 2x + 3(2x) = 18, so x = 9/4 and y = 9/2. The other equation gives (9/4)k − 9/2 = 4, so (9/4)k = 17/2. Therefore k = (17/2)(4/9) = 34/9. Retaining exact fractions avoids rounding a parameter that must satisfy the condition precisely.

26 Section 5

Add the constraints to obtain 4x + 2y ≥ 26. To show that 26 can be reached, solve the boundary equations x + y = 9 and 3x + y = 17. They give x = 4 and y = 5, both nonnegative. This pair satisfies the original inequalities and makes the target 16 + 10 = 26.