RASMUSSEN MAT 121 COLLEGE ALGEBRA FINAL EXAM 2026/2027 | 100
VERIFIED QUESTIONS & ANSWERS | INSTANT DOWNLOAD PDF
SECTION 1: LINEAR EQUATIONS, INEQUALITIES & LINES (Questions 1–10)
1. Solve for x: 3x − 7 = 2x + 5
A. x = −12
B. x = 2
C. x = 12
D. x = −2
Answer: C
Rationale: Subtract 2x from both sides: x − 7 = 5. Add 7 to both sides: x = 12. Checking: 3(12) − 7 = 29
and 2(12) + 5 = 29, so the solution satisfies the equation.
2. Solve for x: 5(x − 2) = 3x + 4
A. x = 3
B. x = 7
C. x = −7
D. x = 1
Answer: B
Rationale: Distribute: 5x − 10 = 3x + 4. Subtract 3x: 2x − 10 = 4. Add 10: 2x = 14. Divide by 2: x = 7.
Check: 5(7−2) = 25 and 3(7)+4 = 25.
3. Solve for y in terms of x: 2y + 4x = 10
A. y = 5 + 2x
B. y = 5 − 2x
C. y = 10 − 4x
D. y = 2x − 5
Answer: B
Rationale: Isolate y: 2y = 10 − 4x, then divide both sides by 2: y = 5 − 2x. Dividing 10 by 2 gives 5 and
−4x by 2 gives −2x.
4. Solve the inequality: −2x + 5 > 11
,A. x > −3
B. x < −3
C. x > 3
D. x < 3
Answer: B
Rationale: Subtract 5: −2x > 6. Divide by −2 and reverse the inequality sign (dividing by a negative
reverses the inequality): x < −3.
5. Solve: |x − 3| = 7
A. x = 10 only
B. x = −4 only
C. x = 10 or x = −4
D. x = 7 or x = −7
Answer: C
Rationale: An absolute value equation splits into two cases: x − 3 = 7 gives x = 10; x − 3 = −7 gives x =
−4. Both values satisfy |x − 3| = 7.
6. Solve the inequality: |2x + 1| < 9
A. x > 4 or x < −5
B. −5 < x < 4
C. x < 4
D. −4 < x < 5
Answer: B
Rationale: |2x + 1| < 9 means −9 < 2x + 1 < 9. Subtract 1: −10 < 2x < 8. Divide by 2: −5 < x < 4.
7. Find the slope of the line passing through (2, 3) and (6, 11).
A. 2
B. 1/2
C. 4
D. −2
Answer: A
Rationale: Slope m = (y₂ − y₁)/(x₂ − x₁) = (11 − 3)/(6 − 2) = 8/4 = 2.
8. Find the equation of the line with slope 3 passing through the point (1, 2).
, A. y = 3x + 2
B. y = 3x − 1
C. y = 3x + 1
D. y = −3x + 5
Answer: B
Rationale: Point-slope form: y − 2 = 3(x − 1). Distribute: y − 2 = 3x − 3. Add 2: y = 3x − 1.
9. Find the x-intercept of the line 2x + 3y = 12.
A. (0, 4)
B. (6, 0)
C. (4, 0)
D. (0, 6)
Answer: B
Rationale: The x-intercept occurs where y = 0: 2x + 0 = 12, so x = 6. The x-intercept is (6, 0). The point
(0, 4) is the y-intercept.
10. Which equation represents the line perpendicular to y = 4x − 2 passing through (0, 5)?
A. y = 4x + 5
B. y = −4x + 5
C. y = −(1/4)x + 5
D. y = (1/4)x + 5
Answer: C
Rationale: Perpendicular slopes are negative reciprocals. The given slope is 4, so the perpendicular
slope is −1/4. With y-intercept 5, the equation is y = −(1/4)x + 5.
SECTION 2: FUNCTIONS AND GRAPHS (Questions 11–20)
11. If f(x) = 2x² − 3x + 1, find f(−2).
A. 3
B. 15
C. −1
D. 11
Answer: B
Rationale: f(−2) = 2(−2)² − 3(−2) + 1 = 2(4) + 6 + 1 = 8 + 6 + 1 = 15.
VERIFIED QUESTIONS & ANSWERS | INSTANT DOWNLOAD PDF
SECTION 1: LINEAR EQUATIONS, INEQUALITIES & LINES (Questions 1–10)
1. Solve for x: 3x − 7 = 2x + 5
A. x = −12
B. x = 2
C. x = 12
D. x = −2
Answer: C
Rationale: Subtract 2x from both sides: x − 7 = 5. Add 7 to both sides: x = 12. Checking: 3(12) − 7 = 29
and 2(12) + 5 = 29, so the solution satisfies the equation.
2. Solve for x: 5(x − 2) = 3x + 4
A. x = 3
B. x = 7
C. x = −7
D. x = 1
Answer: B
Rationale: Distribute: 5x − 10 = 3x + 4. Subtract 3x: 2x − 10 = 4. Add 10: 2x = 14. Divide by 2: x = 7.
Check: 5(7−2) = 25 and 3(7)+4 = 25.
3. Solve for y in terms of x: 2y + 4x = 10
A. y = 5 + 2x
B. y = 5 − 2x
C. y = 10 − 4x
D. y = 2x − 5
Answer: B
Rationale: Isolate y: 2y = 10 − 4x, then divide both sides by 2: y = 5 − 2x. Dividing 10 by 2 gives 5 and
−4x by 2 gives −2x.
4. Solve the inequality: −2x + 5 > 11
,A. x > −3
B. x < −3
C. x > 3
D. x < 3
Answer: B
Rationale: Subtract 5: −2x > 6. Divide by −2 and reverse the inequality sign (dividing by a negative
reverses the inequality): x < −3.
5. Solve: |x − 3| = 7
A. x = 10 only
B. x = −4 only
C. x = 10 or x = −4
D. x = 7 or x = −7
Answer: C
Rationale: An absolute value equation splits into two cases: x − 3 = 7 gives x = 10; x − 3 = −7 gives x =
−4. Both values satisfy |x − 3| = 7.
6. Solve the inequality: |2x + 1| < 9
A. x > 4 or x < −5
B. −5 < x < 4
C. x < 4
D. −4 < x < 5
Answer: B
Rationale: |2x + 1| < 9 means −9 < 2x + 1 < 9. Subtract 1: −10 < 2x < 8. Divide by 2: −5 < x < 4.
7. Find the slope of the line passing through (2, 3) and (6, 11).
A. 2
B. 1/2
C. 4
D. −2
Answer: A
Rationale: Slope m = (y₂ − y₁)/(x₂ − x₁) = (11 − 3)/(6 − 2) = 8/4 = 2.
8. Find the equation of the line with slope 3 passing through the point (1, 2).
, A. y = 3x + 2
B. y = 3x − 1
C. y = 3x + 1
D. y = −3x + 5
Answer: B
Rationale: Point-slope form: y − 2 = 3(x − 1). Distribute: y − 2 = 3x − 3. Add 2: y = 3x − 1.
9. Find the x-intercept of the line 2x + 3y = 12.
A. (0, 4)
B. (6, 0)
C. (4, 0)
D. (0, 6)
Answer: B
Rationale: The x-intercept occurs where y = 0: 2x + 0 = 12, so x = 6. The x-intercept is (6, 0). The point
(0, 4) is the y-intercept.
10. Which equation represents the line perpendicular to y = 4x − 2 passing through (0, 5)?
A. y = 4x + 5
B. y = −4x + 5
C. y = −(1/4)x + 5
D. y = (1/4)x + 5
Answer: C
Rationale: Perpendicular slopes are negative reciprocals. The given slope is 4, so the perpendicular
slope is −1/4. With y-intercept 5, the equation is y = −(1/4)x + 5.
SECTION 2: FUNCTIONS AND GRAPHS (Questions 11–20)
11. If f(x) = 2x² − 3x + 1, find f(−2).
A. 3
B. 15
C. −1
D. 11
Answer: B
Rationale: f(−2) = 2(−2)² − 3(−2) + 1 = 2(4) + 6 + 1 = 8 + 6 + 1 = 15.
