SAT Systems of Equations: Substitution or Elimination?

Use substitution when one equation already isolates a variable, or can do so with little work. Use elimination when adding or subtracting the equations can cancel a variable, possibly after multiplying an entire equation. Both methods find values that satisfy both equations at the same time.

College Board’s SAT Algebra guide includes systems of two linear equations in two variables and solving with multiple techniques. The examples here are original Kibo practice, not official SAT questions.

Choose the method from the equations

These are choices that can reduce arithmetic. Either method can solve the same system, so use the one whose steps you can track clearly. If solving the resulting single equation is difficult, first review our guide to linear equations with variables on both sides.

Substitution: replace a variable with an equal expression

Solve this system:

$$\begin{cases}y=3x-8\\2x+y=17\end{cases}$$

The first equation already gives $y$. Replace $y$ in the second equation with the whole expression $3x-8$:

$$2x+(3x-8)=17$$

Combine like terms to get $5x-8=17$. Adding $8$ gives $5x=25$, so $x=5$. Substitute into the first equation to get $y=3(5)-8=7$. The solution is $(x,y)=(5,7)$.

Check both original equations: $7=3(5)-8$ and $2(5)+7=17$. A pair that works in only one equation is not a solution of the system.

OpenStax’s substitution lesson explains this replacement method and the check in both original equations. Keep parentheses around a substituted expression, especially when the other equation multiplies or subtracts it.

Elimination: make one variable cancel

Solve this different system:

$$\begin{cases}3x+2y=23\\5x-4y=9\end{cases}$$

Multiply the entire first equation by $2$. This makes its $y$-coefficient the opposite of the second equation’s coefficient:

$$\begin{aligned}6x+4y&=46\\5x-4y&=9\end{aligned}$$

Adding the left sides and adding the right sides gives $11x=55$, so $x=5$. In the original first equation, $3(5)+2y=23$, giving $2y=8$ and $y=4$.

The solution is $(5,4)$. Check: $3(5)+2(4)=23$ and $5(5)-4(4)=9$.

OpenStax’s elimination lesson explains why adding equal quantities preserves equality. When scaling an equation, multiply every term and the constant on the other side. Multiplying just the term you want to cancel changes the equation.

In this example, the scaled first equation is $6x+4y=46$. Keeping $23$ on the right would no longer describe the same line.

What if both variables disappear?

For a pair of linear equations that each describe a line, elimination can produce one solution, a contradiction, or a statement that is always true. Read the remaining statement before deciding what the result means.

A contradiction means no solution

$$\begin{cases}2x-5y=6\\4x-10y=15\end{cases}$$

Doubling the first equation gives $4x-10y=12$. Subtract it from the second equation and you get $0=3$. This cannot be true. The same expression cannot equal both $12$ and $15$, so no ordered pair satisfies both equations.

The equations describe distinct parallel lines. Such a system is called inconsistent.

A repeated equation means infinitely many solutions

Now change the second equation’s constant to $12$:

$$\begin{cases}2x-5y=6\\4x-10y=12\end{cases}$$

The second equation is exactly twice the first. Subtracting twice the first from the second gives $0=0$. These equations describe the same line, and every point on that line satisfies both. For example, $(3,0)$ works because $2(3)-5(0)=6$ and $4(3)-10(0)=12$.

There are infinitely many solutions, but not every possible pair is a solution: the pair must still satisfy $2x-5y=6$. The equation $0=0$ does not tell you that $x=0$ or $y=0$.

This is a dependent system. The OpenStax lesson also covers these two outcomes. The same-line interpretation explains why you cannot determine a unique pair.

Four original practice questions

Choose a method before calculating. Read each question’s requested answer carefully; one asks for a sum rather than an ordered pair.

1. Solve $y=2x+3$ and $3x+y=28$. Give $(x,y)$.
2. If $4x+3y=26$ and $4x-y=10$, what is $x+y$?
3. How many solutions satisfy both $3x+2y=14$ and $9x+6y=45$?
4. How many solutions satisfy both $5x-y=9$ and $10x-2y=18$?

Answers and checks

1. Substitution gives $3x+(2x+3)=28$, so $5x=25$ and $x=5$. Then $y=13$. The pair is $(5,13)$: $13=2(5)+3$ and $3(5)+13=28$.
2. Subtract the second equation from the first. The left side becomes $(4x+3y)-(4x-y)=4y$, so $4y=16$ and $y=4$. Then $4x-4=10$ gives $x=\frac{7}{2}$. Both checks hold: $4\left(\frac{7}{2}\right)+3(4)=26$ and $4\left(\frac{7}{2}\right)-4=10$. The requested sum is $x+y=\frac{15}{2}$.
3. No solution. Tripling the first equation gives $9x+6y=42$. Subtracting it from the second gives the contradiction $0=3$.
4. Infinitely many solutions. The second equation is twice the first, including its constant. Every pair on $y=5x-9$ works. For example, $(2,1)$ gives $5(2)-1=9$ and $10(2)-2(1)=18$.

Review the step where the error happened

If your answer fails a check, inspect the substituted parentheses, the terms you multiplied, or the signs in a subtraction. Write down that specific step in our practice error log, then solve a fresh problem without looking at its answer. When both variables cancel, classify the remaining statement instead of forcing a numerical pair.

Try a free SAT practice question with Kibo to continue.

Prepared with AI assistance. Original problems were verified with separate exact calculations; no human review is claimed. Sources checked October 13, 2026. SAT is a registered trademark of the College Board, which is not affiliated with or endorsing this guide.