PSAT 8/9 Math: Systems of Equations
A system of equations asks you to find values that satisfy two equations at once, and the PSAT 8/9 tests this through substitution, elimination, word problems, and graph reading. This free, complete lesson covers solving by substitution and elimination, systems with no solution or infinitely many solutions, word problems, and reading systems from graphs. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback and full explanations. Everything here is free, and you can keep practicing afterward with the linked quizzes below.
1. Solving by Substitution
Substitution works well when one equation is already solved for a variable (or can easily be). Substitute that expression into the other equation, solve for the remaining variable, then plug back in to find the first one.
Worked Example: Solve the system: y = 4x + 1 and x = −2.
Substitute x = −2 into the first equation: y = 4(−2) + 1 = −8 + 1 = −7.
Solution: (−2, −7)
1. Solve the system: y = 4x + 1 and x = −2. (see worked example above)
Explanation: Substitute x = −2: y = 4(−2) + 1 = −7. The solution is (−2, −7).
2. Solve the system: y = 2x + 3 and y = 5x − 9. Find x.
Explanation: Since both equal y, set them equal: 2x + 3 = 5x − 9. Add 9: 2x + 12 = 5x. Subtract 2x: 12 = 3x. Divide by 3: x = 4.
3. Solve the system: x = 3y and 2x + y = 21. Find y.
Explanation: Substitute x = 3y into 2x + y = 21: 2(3y) + y = 21 → 6y + y = 21 → 7y = 21 → y = 3.
2. Solving by Elimination
Elimination works well when adding or subtracting the two equations cancels out one variable. If needed, multiply one or both equations first so that one variable's coefficients match (or are opposites).
Worked Example: Solve the system: (3/2)x + 4y = 20 and (1/2)x + 4y = 12. Find the value of x.
Since the 4y terms match, subtract the second equation from the first: (3/2)x − (1/2)x = 20 − 12.
x = 8.
x = 8
1. Solve the system: (3/2)x + 4y = 20 and (1/2)x + 4y = 12. Find the value of x. (see worked example above)
Explanation: Since 4y appears in both equations, subtract: (3/2)x − (1/2)x = 20 − 12 → x = 8.
2. Solve the system: 3x + 2y = 16 and 3x − 2y = 4. Find y.
Explanation: Subtract the equations: (3x + 2y) − (3x − 2y) = 16 − 4 → 4y = 12 → y = 3.
3. Solve the system: 2x + y = 11 and x + y = 7. Find x.
Explanation: Subtract: (2x + y) − (x + y) = 11 − 7 → x = 4.
3. No Solution and Infinitely Many Solutions
Two linear equations represent parallel lines (no solution) if they have the same slope but different y-intercepts. They represent the same line (infinitely many solutions) if one equation is just a multiple of the other.
Worked Example: The system is 5x + 2y = 4 and 15x + 6y = 12. How many solutions does this system have?
Multiply the first equation by 3: 15x + 6y = 12. This matches the second equation exactly.
The two equations represent the same line, so there are infinitely many solutions.
1. The system is 5x + 2y = 4 and 15x + 6y = 12. How many solutions does this system have? (see worked example above)
Explanation: Multiplying the first equation by 3 gives 15x + 6y = 12, identical to the second equation, so both equations describe the same line: infinitely many solutions.
2. A system is 3x + 2y = 5 and 6x + 4y = 12. How many solutions does this system have?
Explanation: Multiplying the first equation by 2 gives 6x + 4y = 10, which has the same left side as the second equation (6x + 4y) but a different right side (10 vs. 12). These are parallel lines with no intersection: no solution.
3. For what value of k does the system y = 3x + 4 and y = kx + 4 have infinitely many solutions?
Explanation: Since both lines already share the same y-intercept (4), they'll be the exact same line (infinitely many solutions) only if their slopes also match: k = 3.
4. Word Problems and Graphs
A word problem with two unknown quantities often becomes a system: one equation for a total count, and another for a total value (like cost or weight). On a graph, the solution to a system is simply the point where the two lines intersect.
Worked Example: Notebooks cost \$3 each and folders cost \$2 each. A student buys a total of 10 items for \$24. How many notebooks did they buy?
Let n = notebooks, f = folders. n + f = 10 and 3n + 2f = 24.
From the first equation, f = 10 − n. Substitute: 3n + 2(10 − n) = 24 → 3n + 20 − 2n = 24 → n + 20 = 24.
n = 4 notebooks
1. Notebooks cost \$3 each and folders cost \$2 each. A student buys a total of 10 items for \$24. How many notebooks did they buy? (see worked example above)
Explanation: n + f = 10 and 3n + 2f = 24. Substituting f = 10 − n gives n + 20 = 24, so n = 4.
2. A 40-inch rope is cut into two pieces, x and y, where x is 5 more than twice y. Find the length of x.
Explanation: x + y = 40 and x = 2y + 5. Substitute: (2y + 5) + y = 40 → 3y + 5 = 40 → 3y = 35 → y = 35/3. Then x = 40 − 35/3 = 120/3 − 35/3 = 85/3 inches.
3. Two lines are graphed: one with equation y = 2x − 1 and one with equation y = −x + 5. At what point do they intersect?
Explanation: Set the equations equal: 2x − 1 = −x + 5 → 3x = 6 → x = 2. Then y = 2(2) − 1 = 3. The intersection point is (2, 3).
Common Mistakes to Avoid
- Substituting into the same equation you started with instead of the other equation, always plug your isolated expression into the equation you haven't used yet.
- Adding instead of subtracting (or vice versa) during elimination, check whether the matching terms have the same sign (subtract) or opposite signs (add) to cancel them.
- Stopping at "no solution" or "infinitely many" without checking both the slope AND the intercept. Same slope + different intercept = no solution; same slope + same intercept = infinitely many.
- Solving for the wrong variable in a word problem, always double check which quantity the question is actually asking for.
- Forgetting to solve for the second variable after finding the first one, a full solution to a system is an ordered pair, not just one number.
Frequently Asked Questions
How do I decide between substitution and elimination?
Use substitution when one equation is already solved for a variable, or easily can be (like y = ... or x = ...). Use elimination when the coefficients of one variable match or are opposites, or can be made to match by multiplying.
What does it mean graphically when a system has no solution?
The two equations represent parallel lines, lines with the same slope that never cross, so there's no point that satisfies both equations at once.
Where can I practice more problems like these?
The Systems of Equations quizzes in the PSAT 8/9 Math Question Bank include additional original problems on this topic, along with quizzes covering every other Heart of Algebra, Advanced Math, Problem-Solving and Data Analysis, and Geometry and Trigonometry skill tested on the PSAT 8/9. You can also browse the full PSAT 8/9 Math Question Bank.