SAT Math: Systems of Equations Practice Questions
Test yourself on Systems of Equations with 12 original SAT practice questions. Pick an answer to see instant feedback and a full explanation.
Free original practice questions for study purposes. Open Exam Prep is an independent study resource and is not affiliated with, endorsed by, or sponsored by the makers of SAT.
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Answer the questions below — you get instant feedback and a full explanation for each.
1. What is the solution (x, y) to the system: 2x + y = 7 and x − y = 2?
Explanation. Add the equations: (2x + y) + (x − y) = 7 + 2 gives 3x = 9, so x = 3. Substitute into x − y = 2: 3 − y = 2, so y = 1. The solution is (3, 1).
2. If 3x + 2y = 12 and y = x, what is the value of x?
Explanation. Substitute y = x: 3x + 2x = 12, so 5x = 12 and x = 12/5 = 2.4. Note 12/5 and 2.4 are equal; the answer is 12/5.
3. For what value of c does the system 4x − 6y = 10 and 2x − 3y = c have infinitely many solutions?
Explanation. For infinitely many solutions, the equations must be equivalent. Multiply 2x − 3y = c by 2: 4x − 6y = 2c. Matching 4x − 6y = 10 requires 2c = 10, so c = 5.
4. The system 6x + 2y = 8 and 3x + ky = 9 has no solution. What is the value of k?
Explanation. No solution means the lines are parallel (same slope, different intercept). Divide the first by 2: 3x + y = 4. For parallel lines the x and y coefficients must be proportional: comparing 3x + y and 3x + ky, we need k = 1 (the constants 4 and 9 differ, ensuring no solution).
5. Solve: x + 2y = 9 and 3x − 2y = 7. What is x + y?
Explanation. Add the equations: 4x = 16, so x = 4. Substitute into x + 2y = 9: 4 + 2y = 9, so y = 2.5. Therefore x + y = 4 + 2.5 = 6.5.
6. A store sells pens at $2 each and notebooks at $5 each. If a customer buys 12 items total for $39, how many notebooks did they buy?
Explanation. Let p = pens, n = notebooks. Then p + n = 12 and 2p + 5n = 39. From the first, p = 12 − n. Substitute: 2(12 − n) + 5n = 39, so 24 + 3n = 39, giving 3n = 15 and n = 5. They bought 5 notebooks.
7. The graphs of y = 2x + 1 and y = −x + 7 intersect at what point?
Explanation. Set equal: 2x + 1 = −x + 7, so 3x = 6 and x = 2. Then y = 2(2) + 1 = 5. The intersection is (2, 5).
8. If 5x − 3y = 1 and 5x − 3y = k represent the same line for all points, and the system also includes 10x − 6y = 2, how many solutions does the system 5x − 3y = 1 and 10x − 6y = 2 have?
Explanation. Multiply 5x − 3y = 1 by 2 to get 10x − 6y = 2, which is identical to the second equation. The two equations represent the same line, so there are infinitely many solutions.
9. Solve the system: y = x² and y = x + 6. What are the x-values of the solutions?
Explanation. Set x² = x + 6, so x² − x − 6 = 0. Factor: (x − 3)(x + 2) = 0, giving x = 3 or x = −2.
10. The system kx + 4y = 8 and 2x + y = 5 has exactly one solution for all values of k except one. What value of k gives no unique solution?
Explanation. A unique solution fails when the coefficient ratios are equal: k/2 = 4/1, so k = 8. At k = 8 the lines are parallel (or coincident), so there is not a unique solution.
11. At a concert, adult tickets cost $20 and child tickets cost $12. If 150 tickets were sold for a total of $2,520, how many adult tickets were sold?
Explanation. Let a = adults, c = children. Then a + c = 150 and 20a + 12c = 2520. Substitute c = 150 − a: 20a + 12(150 − a) = 2520, so 20a + 1800 − 12a = 2520, giving 8a = 720 and a = 90.
12. If 3x + 4y = 10 and 6x + 8y = 20, which statement is true?
Explanation. Multiply the first equation by 2: 6x + 8y = 20, which is exactly the second equation. The two equations are equivalent, so the system has infinitely many solutions (every point on the line satisfies both).
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FAQ
What are the main methods for solving systems of equations on the SAT?
Substitution (solve one equation for a variable and plug into the other), elimination (add or subtract equations to cancel a variable), and graphing (intersection point). For linear systems, also know that the number of solutions depends on slopes and intercepts.
How do I determine the number of solutions without solving?
Compare the equations in ax + by = c form. If the coefficient ratios a₁/a₂ and b₁/b₂ are equal AND match the constant ratio c₁/c₂, there are infinitely many solutions. If the coefficient ratios are equal but the constant ratio differs, there are no solutions (parallel lines). Otherwise, exactly one solution.
Are nonlinear systems tested on the SAT?
Yes—commonly a line and a parabola (e.g., y = x² and y = x + 6). Solve by substitution to get a quadratic, then factor or use the quadratic formula. Setting the discriminant to zero is a frequent way the SAT asks for the value that makes the system have exactly one solution.