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Systems of Linear Equations on the SAT

SAT Math · Algebra · Updated October 2026

Systems come up several times on every SAT, both as plain equations and as word problems about tickets, mixtures and prices. Desmos is a big help with the plain ones: graph both lines and click the point where they cross. Word problems still need you to write the two equations yourself.

A system of two linear equations asks for the one point that makes both of them true. Use substitution when one equation already says $y = \ldots$ or $x = \ldots$. Use elimination when adding or subtracting the equations will make one variable disappear.

Before you start solving, read what the question wants. It often asks for $x + y$ or $x - y$, and sometimes you can get that straight from adding or subtracting the equations without finding $x$ and $y$ separately.

Three worked examples

Example 1: elimination after multiplying

Solve $5x - 3y = 1$ and $3x + 2y = 12$.

  1. Multiply the first equation by $2$ and the second by $3$: $10x - 6y = 2$ and $9x + 6y = 36$.
  2. Add: $19x = 38$, so $x = 2$.
  3. Substitute: $3(2) + 2y = 12$, so $y = 3$. The solution is $(2, 3)$.

Example 2: a word problem

Adult tickets cost \$12 and child tickets cost \$7. A group bought $50$ tickets for \$475. How many adult tickets did they buy?

  1. Count: $a + c = 50$. Cost: $12a + 7c = 475$.
  2. Substitute $c = 50 - a$: $12a + 350 - 7a = 475$, so $5a = 125$.
  3. $a = 25$ adult tickets.

Example 3: a system with no solution

For what value of $k$ does the system $kx + 2y = 5$ and $3x + y = 1$ have no solution?

  1. Multiply the second equation by $2$: $6x + 2y = 2$.
  2. With $k = 6$, the left sides match but the right sides are $5$ and $2$. That can never happen, so there is no solution.
  3. $k = 6$.

Substitution, elimination and the special cases

You seeDo this
One equation solved for a variableSubstitute it into the other equation.
Matching or opposite coefficientsAdd or subtract the equations to eliminate a variable.
Coefficients that don't matchMultiply one or both equations first, then eliminate.
DecimalsMultiply by $10$ or $100$ to clear them before you start.
"No solution"Same slope, different intercepts. Set the ratios of the $x$ and $y$ coefficients equal.
"Infinitely many solutions"One equation is a multiple of the other, constants included.
Two things, two totals (a count and a cost)One equation for the count, one for the cost.

Mistakes I see often

Multiplying only one side of an equation. When you multiply to line up coefficients, every term on both sides gets multiplied.

Subtracting when you meant to add. Subtracting a negative changes its sign. Write the subtraction out term by term.

Stopping at $x$ when the question asks for $y$. Or for $x + y$. Reread the last line.

Mixing up "no solution" and "infinitely many". When the left sides match but the constants don't, there is no solution. When everything matches, including the constants, there are infinitely many.

Putting the price in the count equation. In word problems, one equation counts things and the other adds up their values.

Try these

For the word problems, write both equations before you solve anything.

Easy
If $x + y = 10$ and $x - y = 4$, what is the value of $x$?
  • A$3$
  • B$7$
  • C$10$
  • D$14$
Show solution
Answer: B. Add the equations: $2x = 14$, so $x = 7$. Choice A is the value of $y$.
EasyStudent-produced response
If $2x + y = 11$ and $y = x + 2$, what is the value of $y$?
Enter your answer.
Show solution
Answer: $5$. Substitute: $2x + x + 2 = 11$, so $3x = 9$ and $x = 3$. Then $y = 3 + 2 = 5$.
MediumStudent-produced response
If $3x + 2y = 16$ and $x - 2y = 0$, what is the value of $x + y$?
Enter your answer.
Show solution
Answer: $6$. Add the equations: $4x = 16$, so $x = 4$. Then $4 - 2y = 0$ gives $y = 2$, and $x + y = 6$.
Medium
Adult tickets to a show cost \$12 each and child tickets cost \$7 each. A group bought $50$ tickets for a total of \$475. How many adult tickets did the group buy?
  • A$20$
  • B$30$
  • C$35$
  • D$25$
Show solution
Answer: D. $a + c = 50$ and $12a + 7c = 475$. Substituting $c = 50 - a$ gives $5a + 350 = 475$, so $a = 25$. Check: $25 \cdot 12 + 25 \cdot 7 = 300 + 175 = 475$.
Medium
How many solutions does the system $2x + 3y = 6$ and $4x + 6y = 12$ have?
  • AZero
  • BExactly one
  • CExactly two
  • DInfinitely many
Show solution
Answer: D. The second equation is the first one multiplied by $2$, so both describe the same line. Every point on that line is a solution.
MediumStudent-produced response
In the system $kx + 2y = 5$ and $3x + y = 1$, $k$ is a constant. For what value of $k$ does the system have no solution?
Enter your answer.
Show solution
Answer: $6$. Doubling the second equation gives $6x + 2y = 2$. With $k = 6$ the left sides match but the constants $5$ and $2$ do not, so the lines are parallel and there is no solution.
HardStudent-produced response
If $0.2x + 0.5y = 4.1$ and $x + y = 13$, what is the value of $x - y$?
Enter your answer.
Show solution
Answer: $3$. Multiply the first equation by $5$: $x + 2.5y = 20.5$. Subtract $x + y = 13$: $1.5y = 7.5$, so $y = 5$ and $x = 8$. $x - y = 3$.
Hard
If $5x - 3y = 1$ and $3x + 2y = 12$, what is the value of $x + y$?
  • A$5$
  • B$3$
  • C$2$
  • D$6$
Show solution
Answer: A. Multiply to eliminate $y$: $10x - 6y = 2$ and $9x + 6y = 36$. Adding gives $19x = 38$, so $x = 2$, and then $y = 3$. $x + y = 5$.
HardStudent-produced response
In the system $3x - ky = 6$ and $ax + 4y = -12$, $a$ and $k$ are constants. If the system has infinitely many solutions, what is the value of $a + k$?
Enter your answer.
Show solution
Answer: $-4$. The second equation must be a multiple of the first. Since $-12 = -2 \cdot 6$, the second is the first times $-2$: $-6x + 2ky = -12$. So $a = -6$ and $2k = 4$, $k = 2$. $a + k = -4$.
Hard
A chemist mixes a $30\%$ acid solution with a $60\%$ acid solution to make $12$ liters of a $50\%$ acid solution. How many liters of the $30\%$ solution does she use?
  • A$8$
  • B$4$
  • C$6$
  • D$3$
Show solution
Answer: B. $x + y = 12$ and $0.3x + 0.6y = 0.5 \cdot 12 = 6$. Substituting $y = 12 - x$: $0.3x + 7.2 - 0.6x = 6$, so $0.3x = 1.2$ and $x = 4$. Choice A is the amount of the $60\%$ solution.

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