Understanding systems of equations is essential for solving many algebra problems on the SAT math section.
Systems of equations involve two or more equations with two or more variables. On the SAT you will solve systems with two equations and two variables.
Example: 2x + 3y = 12 and x - 4y = -2. The goal is to find values of x and y that satisfy both equations simultaneously.
There are three primary methods: Substitution, Elimination, and Graphing.
Substitution
Solve one equation for one variable, then substitute that expression into the other. Best when one equation is easy to solve for a variable (coefficient of 1 or -1).
Example: 2y + 3x = 14 and y - 2x = 0. From second: y = 2x. Substitute: 2(2x) + 3x = 14, so 7x = 14, x = 2. Then y = 2(2) = 4.
Example Problem 1
Solve: 4x + y = 22 and 2x - y = 6. From second: y = 2x - 6. Substitute: 4x + (2x-6) = 22, 6x = 28, x = 14/3. Then y = 2(14/3) - 6 = 10/3.
Example Problem 2
Solve: 3x + 4y = 12 and x - 2y = 3. From second: x = 2y + 3. Substitute: 3(2y+3) + 4y = 12, 10y + 9 = 12, y = 3/10. Then x = 2(3/10) + 3 = 18/5.
Elimination
Add or subtract equations to eliminate one variable. Best when equations are in standard form and coefficients can be easily aligned.
Example: 2y + 4x = 20 and y - x = 1. Multiply second by 2: 2y - 2x = 2. Subtract from first: 6x = 18, x = 3. Then 2y + 12 = 20, y = 4.
Example Problem 1
Solve: 6x + 5y = 7 and 3x - 2y = 4. Multiply second by 2: 6x - 4y = 8. Subtract: 9y = -1, y = -1/9. Substitute back: x = 34/27.
Example Problem 2
Solve: 5x - 3y = 2 and 2x + y = 9. Multiply second by 3: 6x + 3y = 27. Add: 11x = 29, x = 29/11. Then y = 41/11.
Graphing
Plot each equation and find their intersection point. Convert to slope-intercept form for easy graphing. Best for visual understanding and verifying solutions from other methods.
Example: 2y + 4x = 20 becomes y = -2x + 10. And y - x = 1 becomes y = x + 1. Intersection: (3, 4).
Graphing shows the nature of solutions: intersecting lines (one solution), parallel lines (no solution), or coincident lines (infinite solutions).
Caution: Graphing is less reliable for precise answers since small errors can lead to incorrect solutions. Use it to verify answers from algebraic methods.
Practice Questions
x = y + 4. Substitute: 3(y+4) + 2y = 14, 5y + 12 = 14, y = 2/5. Then x = 2/5 + 4 = 22/5.
Multiply second by 3: 6x + 3y = 9. Add: 10x = 14, x = 7/5. Then 2(7/5) + y = 3, y = 1/5.
Convert: y = -(1/2)x + 13/2 and y = 3x - 4. Set equal: -(1/2)x + 13/2 = 3x - 4. Solve: x = 3, y = 5.
Substitution, Elimination, and Graphing. Choose based on which is fastest for the given problem.
Use substitution when one equation has a coefficient of 1/-1. Use elimination when both are in standard form and coefficients align easily.
No solution: solving gives a false statement (2=4, parallel lines). Infinite solutions: gives a true statement (0=0, same line). One solution: specific values for both variables.