Solution by Substitution -

Straight to:

Practice 1, setting $y=y$
Practice 2, Substitute $y$ in the middle of an expression
Practice 3, Rearrange before substituting

Making y=y

A system of equations in two unknowns is solved when the $y$ coordinate on both graphs are equal, for some particular value of $x$ .

Example 1:

Solve the system:

$\begin{cases}y=-x+16\\ y=3x-8 \end{cases}$

The graphing technique helps us to understand the algebra technique. There is one point that lies on both lines. Let’s call that point $(a,b)$ .

Graph showing system of linear equations intersecting at point (a,b)

We are looking for a value of $x$ that gives the same answer to both equations. In other words, for some particular value of $x$ the answer to $-x+16$ is equal to the answer to $3x-8$ .

$-x+16=3x-8$

Use algebra to calculate this particular value of $x$ :

$\begin{align*}-x+16&=3x-8 &\text{add x to both sides}\\[10pt] 16&=4x-8 &\text{add 8 to both sides}\\[10pt] 24&=4x &\text{divide both sides by 4}\\[10pt] 6&=x &\text{swap sides, just because.}\\[10pt] x&=6 &\end{align*}$

We must also calculate the particular value of $y$ . We now know that $x=6$ .

$\begin{cases}y=-x+16\\ y=3x-8 \end{cases}$

Use either equation. Using $y=-x+16$ we have $y=-6+16=10$ .

Solution: $x=6$ , $y=10$ .

Check: Let’s make sure that the point $(6,10)$ lies on the other equation.

$\begin{align*}y=&3x-8\\[10pt]10=&3(6)-8\\[10pt]10=&18-8\checkmark\end{align}$

Graph showing system of linear equations intersecting at point (6,10)

Learn another way:

Check out Khan Academy: Solving by Substitution

Practice 1

Take the second expression for $y$ and substitute it into the first equation. Solve to find the $x$ coordinate that yields the same $y$ coordinate on both lines. Remember that if the lines are parallel, you will not reach a solution. If the lines are cooincident, there are infinitely many solutions- any point on the coincident lines will work.

Substitute into the middle of an expression

In the first example, $y$ is the subject of both equations. That means that both equations start with “ $y=$ “.

Sometimes the value $y$ appears in other places in the equation.

At a point of intersection, the value $y$ is simply a particular number. When we solve a system of equations, we can either

rearrange the equations to make $y$ the subject of both equations,
leave the $y$ where it is and substitute an equivalent calculation.

Example 2:

$\begin{cases}2x+y=21 \\ y=2x-15 \end{cases}$

Show Example 2 Solution

At this point the algebra can become a little trickier especially if you have to contend with negative numbers, brackets and fractions: plenty of practice will give you the opportunity to both make and resolve all the possible errors.

Example 3:

$\begin{cases}x-4y=-42 \\ y=-2x+6 \end{cases}$

Show Example 3 Solution

Example 4:

$\begin{cases}x-2y=9 \\ y=\frac{x-11}{3} \end{cases}$

Show Example 4 Solution

Practice 2:

Substituting when neither variable is the subject of either equation

Rearrange one or the other of the equations to make either $x$ or $y$ the subject. Then substitute into the other equation to find one part of the solution. Complete by finding the second part. Enter the values into the x, y input boxes to check the solution and to see the graphical representation of the solution.