Preparing for the SAT requires a solid understanding of algebra, as it’s a key part of the math section. In this post, we’ve compiled 25 essential SAT-style algebra questions, covering topics such as linear equations, inequalities, systems of equations, quadratic functions, and more. Challenge yourself with these questions and test your skills before exam day!
Part 1: Linear Equations and Inequalities
1. Solve for x:
3x + 5 = 2x + 10
2. If 4x – 7 = 2x + 9 , what is the value of x?
3. Solve the inequality:
5x – 3 >= 2x + 1
4. Solve for y:
2(3y – 4) = 5y + 8
5. What is the slope of the line that passes through the points (2, 3) and (6, 7)?
Part 2: Systems of Equations
6. Solve the system of equations:
3x + 2y = 12
2x – y = 1
7. If x + y = 5 and 2x – 3y = 4, what is the value of x and y?
8. Solve the system of equations by substitution:
y = 2x + 1
3x + 2y = 14
9. For the system of equations 4x + 3y = 9 and 2x – y = 5, find the values of x and y.
10. Solve for x and y:
5x – 2y = 8
3x + 4y = -7
Part 3: Quadratic Functions
11. Solve the quadratic equation:
x² – 5x + 6 = 0
12. Solve for x:
2x²+ 3x – 2 = 0
13. The function f(x) = x² – 4x + 3 is graphed in the xy-plane. What are the coordinates of the vertex?
14. What are the solutions to the quadratic equation x²- 4x = 12 ?
15. If the sum of two numbers is 12 and their product is 32, what are the two numbers?
Part 4: Exponents and Polynomials
16. Simplify:
(3x^2)(2x^3)
17. Simplify the expression:
\frac{4x^3y^2}{2xy}
18. Factor the expression completely:
x^2 – 9x + 14
19. Simplify the expression:
(2x^3 – 5x^2 + 3x) – (x^3 – 2x + 1)
20. Solve for x:
4x^3 = 64
Part 5: Word Problems
21. The length of a rectangle is 3 times its width. If the perimeter of the rectangle is 48 units, what are the dimensions of the rectangle?
22. John is 5 years older than twice his sister’s age. If John is 23 years old, how old is his sister?
23. A car rental company charges $25 per day plus $0.20 per mile driven. If Sarah rented a car and was charged $43, how many miles did she drive?
24. The sum of three consecutive integers is 51. What are the integers?
25. A train travels 300 miles at a constant speed. If it had gone 10 miles per hour faster, the trip would have taken 1 hour less. What is the speed of the train?
Next Steps
Algebra is a crucial component of the SAT math section. Be sure to practice questions like these to get a feel for the types of problems you’ll encounter on test day.
If you’re looking for solutions to these questions or more practice problems, find the solved solutions below:
SAT Algebra Practice Questions – Step-by-Step Solutions
Now, let’s walk through the step-by-step solutions for each of those questions. Follow along and see how each problem is tackled.
Part 1: Linear Equations and Inequalities
1. Solve for x:
3x + 5 = 2x + 10
• Subtract 2x from both sides:
x + 5 = 10
• Subtract 5 from both sides:
x = 5
2. Solve for x:
4x – 7 = 2x + 9
• Subtract 2x from both sides:
2x – 7 = 9
• Add 7 to both sides:
2x = 16
• Divide by 2:
x = 8
3. Solve the inequality:
5x – 3 ≥ 2x + 9
• Subtract 2x from both sides:
3x – 3 ≥ 9
• Add 3 to both sides:
3x ≥ 12
• Divide by 3:
x ≥ 4
4. Solve for y:
2(3y – 4) = 5y + 8
• Distribute the 2:
6y – 8 = 5y + 8
• Subtract 5y from both sides:
y – 8 = 8
• Add 8 to both sides:
y = 16
5. Find the slope:
The slope of a line through points (x₁, y₁) and (x₂, y₂) is given by:
slope = (y₂ – y₁) / (x₂ – x₁)
• slope = (7 – 3) / (6 – 2) = 4 / 4 = 1
Part 2: Systems of Equations6.
6. Solve the system:
3x + 2y = 12
2x – y = 1
• From the second equation, solve for y:
y = 2x – 1
• Substitute into the first equation:
3x + 2(2x – 1) = 12
3x + 4x – 2 = 12
7x = 14
x = 2
• Substitute x = 2 into y = 2x – 1:
y = 2(2) – 1 = 3
• Solution: x = 2, y = 3
7. Solve for x and y:
x + y = 5
2x – 3y = 4
• Solve the first equation for x:
x = 5 – y
• Substitute into the second equation:
2(5 – y) – 3y = 4
10 – 2y – 3y = 4
10 – 5y = 4
-5y = -6
y = 6/5
• Substitute y = 6/5 into x = 5 – y:
x = 5 – 6/5 = 19/5
8. Solve by substitution:
y = 2x + 1
3x + 2y = 14
• Substitute y = 2x + 1 into the second equation:
3x + 2(2x + 1) = 14
3x + 4x + 2 = 14
7x = 12
x = 12/7
• Substitute x = 12/7 into y = 2x + 1:
y = 2(12/7) + 1 = 24/7 + 1 = 31/7
- Solve the system:
4x + 3y = 9
2x – y = 5
• Solve the second equation for y:
y = 2x – 5
• Substitute into the first equation:
4x + 3(2x – 5) = 9
4x + 6x – 15 = 9
10x = 24
x = 2.4
• Substitute x = 2.4 into y = 2x – 5:
y = 2(2.4) – 5 = 4.8 – 5 = -0.2
10. Solve the system:
5x – 2y = 8
3x + 4y = -7
• Multiply the first equation by 2 and the second by 5 to eliminate y:
10x – 4y = 16
15x + 20y = -35
• Add the two equations:
25x = -19
x = -19/25
• Substitute into 5x – 2y = 8 to solve for y.
Part 3: Quadratic Functions
11. Solve the quadratic equation:
x² – 5x + 6 = 0• Factor:
(x – 2)(x – 3) = 0
• Solutions:
x = 2, x = 3
12. Solve for x:
2x² + 3x – 2 = 0• Factor:
(2x – 1)(x + 2) = 0
• Solutions:
x = 1/2, x = -2
13. Find the vertex of f(x) = x² - 4x + 3: • Use the formula x = -b/2a:
x = 4/2 = 2
• Plug into the function:
f(2) = (2)² – 4(2) + 3 = -1
• Vertex: (2, -1)
14. Solve x² - 4x = 12: • Rearrange:
x² – 4x – 12 = 0
• Factor or use the quadratic formula:
x = [4 ± √(16 + 48)] / 2 = [4 ± √64] / 2
• Solutions:
x = 6, x = -2
15. Find the two numbers: • Let the numbers be x and 12 - x. Their product is 32:
x(12 – x) = 32
• Solve:
x² – 12x + 32 = 0
• Factor:
(x – 8)(x – 4) = 0
• Solutions: x = 8 and x = 4
Part 4:
Exponents and Polynomials
16. Simplify: (3x²)(2x³)
• Multiply the coefficients: 3 × 2 = 6
• Add the exponents: x² × x³ = x⁵
• Solution: 6x⁵
17. Simplify the expression: (4x³y²) / (2xy)
• Divide the coefficients: 4 / 2 = 2 • Subtract the exponents of x: x³ / x = x²
• Subtract the exponents of y: y² / y = y
• Solution: 2x²y
18. Factor the expression completely: x² - 9x + 14 • Find two numbers that multiply to 14 and add to -9: (-7) and (-2)
• Factor the expression: (x - 7)(x - 2)
• Solution: (x - 7)(x - 2)
19. Simplify the expression: (2x³ - 5x² + 3x) - (x³ - 2x + 1) • Distribute the negative sign: 2x³ - 5x² + 3x - x³ + 2x - 1 • Combine like terms: (2x³ - x³) + (-5x²) + (3x + 2x) - 1 • Simplified expression: x³ - 5x² + 5x - 1 • Solution: x³ - 5x² + 5x - 1
20. Solve for x:
4x³ = 64
• Divide both sides by 4: x³ = 16
• Take the cube root of both sides: x = 2 • Solution: x = 2
Part 5: Word Problems
21. The length of a rectangle is 3 times its width. If the perimeter of the rectangle is 48 units, what are the dimensions of the rectangle?
• Let the width be w and the length be 3w.
• The perimeter formula is P = 2(length + width).
• Substitute the values: 48 = 2(3w + w)
• Simplify: 48 = 2(4w) → 48 = 8w
• Solve for w: w = 6
• Length: 3w = 18
• Solution: Width = 6 units, Length = 18 units
22. John is 5 years older than twice his sister’s age. If John is 23 years old, how old is his sister? • Let John’s sister’s age be x.
• John’s age is 5 years more than twice his sister’s age: 23 = 2x + 5 • Subtract 5 from both sides: 18 = 2x
• Divide by 2: x = 9 • Solution: John’s sister is 9 years old
23. A car rental company charges $25 per day plus $0.20 per mile driven. If Sarah rented a car and was charged $43, how many miles did she drive?
• Let the number of miles driven be m.
• The total cost is given by: 25 + 0.20m = 43
• Subtract 25 from both sides: 0.20m = 18
• Divide by 0.20: m = 18 / 0.20 = 90 • Solution: Sarah drove 90 miles
24. The sum of three consecutive integers is 51. What are the integers?
• Let the three integers be x, x + 1, and x + 2.
• The sum is: x + (x + 1) + (x + 2) = 51
• Simplify: 3x + 3 = 51
• Subtract 3 from both sides: 3x = 48
• Divide by 3: x = 16
• The three integers are: 16, 17, and 18
• Solution: The integers are 16, 17, and 18
25. A train travels 300 miles at a constant speed. If it had gone 10 miles per hour faster, the trip would have taken 1 hour less. What is the speed of the train?
• Let the speed of the train be x mph.
• The time taken at speed x is 300 / x hours.
• The time taken at speed (x + 10) is 300 / (x + 10) hours.
• The difference in time is 1 hour: 300 / x - 300 / (x + 10) = 1
• Multiply through by x(x + 10) to eliminate the denominators: 300(x + 10) - 300x = x(x + 10)
• Simplify: 3000 = x² + 10x • Rearrange: x² + 10x - 3000 = 0
• Factor the quadratic: (x - 50)(x + 60) = 0
• Solution: x = 50 mph (since speed can’t be negative)
Conclusion
By working through these solutions, you should now have a clearer understanding of how to approach a variety of algebra problems on the SAT. Whether it’s solving systems of equations, factoring quadratics, or handling word problems, practice is key to improving your SAT math performance.
Stay tuned for more practice problems and tips for acing the SAT!
Good luck with your SAT preparation!
