SAT Math · Algebra · free lesson

Systems of Linear Equations — Substitution

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

Two equations at once

A system of linear equations is two equations in the same two letters, such as $y = 2x + 1$ and $3x + y = 11$ . A solution of the system is one pair $(x, y)$ that makes both equations true at the same time.

Each equation by itself has a whole line of solutions. A pair that works in both must lie on both lines, so it is the point where the two lines cross.

Definition.

The solution of a system of two linear equations is the point where their graphs cross. Its two numbers make both equations true.

Substitution

You do not need a graph to find the crossing point. The first equation says that $y$ is the same thing as $2x + 1$ . So in the second equation, write $2x + 1$ in place of $y$ . That leaves one equation with one letter, which you can solve.

\begin{aligned} 3x + 2x + 1 &= 11 \\ x &= 2 \\ y &= 2x + 1 \\ &= 2 \times 2 + 1 \\ &= 5 \end{aligned}

Once $x$ is known, the first equation gives $y$ . The solution is $({2}, {5})$ , the black point of the picture. Check it in the equation you did not use for the last step:

\begin{aligned} L &= 3x + y \\ &= 3 \times 2 + 5 \\ &= 11 \end{aligned}

The check gives $11$ , the right side of the second equation, so the pair works in both.

Rule.

Substitution in three steps. First, get one letter alone in one equation. Second, put what it equals into the other equation and solve. Third, go back for the other letter.

When no letter is alone yet

In the system $x + 2y = 10$ and $3x - y = 9$ , nothing is alone. Pick the letter that is easiest to free: one with no number in front of it. Here that is the $x$ of the first equation. Take $2y$ from both sides to get $x = 10 - 2y$ , then put $10 - 2y$ in place of $x$ in the second equation.

\begin{aligned} 3(10 - 2y) - y &= 9 \\ y &= 3 \\ x &= 10 - 2y \\ &= 10 - 2 \times 3 \\ &= 4 \\ L &= 3x - y \\ &= 3 \times 4 - 3 \\ &= 9 \end{aligned}

The solution is $({4}, {3})$ , and the check gives the $9$ of the second equation. Notice the parentheses around $10 - 2y$ : the 3 multiplies all of it. Leaving them out is the most common slip in this method.

Move a line and watch the solution move

The blue line is $x + y = 7$ and stays where it is. The red line is $y = mx + 1$ . As its slope $m$ changes, the engine solves the system again and marks the crossing point.

A steeper red line meets the blue line sooner, so the solution slides up and to the left. Every slope gives exactly one crossing point, and substitution finds it each time.

Walking to the crossing point

The point $P$ walks along $y = 2x + 1$ . The gray segment shows how far it is, straight up or down, from the line $3x + y = 11$ . The gap closes to nothing at exactly one place, where $x$ is 2.

Note.

On the SAT, read what is asked before you stop. A question may want $x$ , or $y$ , or something built from them such as $x + y$ . The wrong options usually include the value of the letter that was not asked for.

Worked examples

Each example is one of the kinds of question the test asks about a system solved by substitution. The numbers in the working are found by the engine.

Example.

1. One letter is already alone. Solve the system $y = 3x - 2$ and $2x + y = 13$ .

The first equation says what $y$ equals. Write $3x - 2$ in place of $y$ in the second equation and solve for $x$ . Then the first equation gives $y$ .

\begin{aligned} 2x + 3x - 2 &= 13 \\ x &= 3 \\ y &= 3x - 2 \\ &= 3 \times 3 - 2 \\ &= 7 \\ L &= 2x + y \\ &= 2 \times 3 + 7 \\ &= 13 \end{aligned}

The solution is $({3}, {7})$ . The check in the second equation gives $13$ , as it should. The picture shows the two lines crossing there.

Example.

2. Two lines given by their equations. At what point do the lines $y = 2x + 3$ and $y = -x + 9$ cross?

Both right sides equal the same $y$ , so they equal each other. This is substitution too: the $y$ of one equation is replaced by what the other says it is.

\begin{aligned} 2x + 3 &= -x + 9 \\ x &= 2 \\ y &= 2x + 3 \\ &= 2 \times 2 + 3 \\ &= 7 \\ y &= -x + 9 \\ &= -2 + 9 \\ &= 7 \end{aligned}

The lines cross at $({2}, {7})$ . Both equations give the same $y$ , which is the check.

Example.

3. Free a letter first. Solve the system $2x + 3y = 16$ and $x - y = 3$ .

No letter is alone. The $x$ of the second equation has no number in front, so it is the easiest to free: add $y$ to both sides to get $x = y + 3$ . Put $y + 3$ in place of $x$ in the first equation, inside parentheses.

\begin{aligned} 2(y + 3) + 3y &= 16 \\ y &= 2 \\ x &= y + 3 \\ &= 2 + 3 \\ &= 5 \\ L &= 2x + 3y \\ &= 2 \times 5 + 3 \times 2 \\ &= 16 \end{aligned}

The solution is $({5}, {2})$ , and the first equation checks: $16$ .

Example.

4. The question asks for something else. If $x = 2y$ and $3x + 4y = 40$ , what is the value of $x + y$ ?

Put $2y$ in place of $x$ , so that $3x$ becomes 3 times $2y$ . Find $y$ , then $x$ , and only then answer what was asked.

\begin{aligned} 3 \times 2y + 4y &= 40 \\ y &= 4 \\ x &= 2y \\ &= 2 \times 4 \\ &= 8 \\ s &= x + y \\ &= 8 + 4 \\ &= 12 \end{aligned}

The answer is $12$ . On the test, $4$ and $8$ would both be among the options, waiting for a reader who stops early.

Example.

5. Is this point the solution? Is $(4, 2)$ the solution of the system $x + 2y = 8$ and $3x - y = 3$ ?

A solution must make both equations true. Put the point into each left side.

\begin{aligned} L &= x + 2y \\ &= 4 + 2 \times 2 \\ &= 8 \end{aligned}

The first equation is satisfied: its left side comes to $8$ .

\begin{aligned} L &= 3x - y \\ &= 3 \times 4 - 2 \\ &= 10 \end{aligned}

The second is not: its left side comes to $10$ , not 3. So $(4, 2)$ lies on the first line but is not the solution. Substitution finds the real one. From the first equation $x = 8 - 2y$ :

\begin{aligned} 3(8 - 2y) - y &= 3 \\ y &= 3 \\ x &= 8 - 2y \\ &= 8 - 2 \times 3 \\ &= 2 \end{aligned}

The solution is $({2}, {3})$ . In the picture, the gray point sits on the blue line only. The black point is on both.

Example.

6. From a situation. A school play sold adult tickets at 9 dollars and student tickets at 5 dollars. It sold 30 more student tickets than adult tickets and took in 990 dollars. How many adult tickets were sold?

Let $a$ be the number of adult tickets and $s$ the number of student tickets. The story gives $s = a + 30$ and $9a + 5s = 990$ . The first equation already has $s$ alone, so put $a + 30$ in place of $s$ .

\begin{aligned} 9a + 5(a + 30) &= 990 \\ a &= 60 \\ s &= a + 30 \\ &= 60 + 30 \\ &= 90 \\ L &= 9a + 5s \\ &= 9 \times 60 + 5 \times 90 \\ &= 990 \end{aligned}

$60$ adult tickets were sold, with $90$ student tickets. The money comes to $990$ dollars, which matches the story.

Practice

Answer SAT-style questions on systems of linear equations — substitution. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise systems of linear equations — substitution with new numbers