Solving Linear Systems by Substitution

In class, we worked through this sequence of questions in random groups at the whiteboards.

If you were away

Work through the questions below in order, on paper, before looking at any answers. Each question builds on the one before it. Surprised by what you discover in a given question? That is where the learning happens.

Once you have an answer, expand the answer box to check your work. If a question didn’t go the way you expected, try to explain why before moving on.

When you reach the end, read the summary of key ideas that we built together in class.

TIP

A solution to a linear system is a point that satisfies both equations. A quick way to be sure of your answer is to substitute the point into each equation and confirm that the left side equals the right side.

Question 1

Solve the system.

$\begin{array}{rlrl}& \text{A}:& y& =3\\ & \text{B}:& 2x+y& =11\end{array}$

Answer

The solution is $(4,3)$.

Equation A tells us that $y$ is worth $3$. So, replace $y$ with $3$ in equation B:

$\begin{array}{rl}2x+3& =11\\ 2x& =8\\ x& =4\end{array}$

Check: in A, $y=3$; in B, $2(4)+3=11$.

Question 2

Solve the system.

$\begin{array}{rlrl}& \text{A}:& y& =2x\\ & \text{B}:& x+y& =12\end{array}$

Answer

The solution is $(4,8)$.

This time, what $y$ is worth is an expression, $2x$. Replace $y$ with $2x$ in equation B, then solve for $x$:

$\begin{array}{rl}x+2x& =12\\ 3x& =12\\ x& =4\end{array}$

Then use equation A to find $y$: $y=2(4)=8$.

Check: in A, $2(4)=8$; in B, $4+8=12$.

Question 3

Solve the system.

$\begin{array}{rlrl}& \text{A}:& y& =2x-3\\ & \text{B}:& 3x+2y& =15\end{array}$

Answer

The solution is $(3,3)$.

Replace $y$ in equation B with the whole expression $2x-3$. Put it in brackets, so that the $2$ multiplies both terms:

$\begin{array}{rl}3x+2(2x-3)& =15\\ 3x+4x-6& =15\\ 7x& =21\\ x& =3\end{array}$

Then, from equation A, $y=2(3)-3=3$.

Check: in A, $2(3)-3=3$; in B, $3(3)+2(3)=15$.

Question 4

Solve the system.

$\begin{array}{rlrl}& \text{A}:& y& =-2x+5\\ & \text{B}:& 4x-3y& =5\end{array}$

Answer

The solution is $(2,1)$.

Substitute $-2x+5$ for $y$ in equation B. Take care when distributing the $-3$: it changes the sign of both terms in the brackets.

$\begin{array}{rl}4x-3(-2x+5)& =5\\ 4x+6x-15& =5\\ 10x& =20\\ x& =2\end{array}$

Then, from equation A, $y=-2(2)+5=1$.

Check: in A, $-2(2)+5=1$; in B, $4(2)-3(1)=5$.

Question 5

Solve the system.

$\begin{array}{rlrl}& \text{A}:& x+y& =5\\ & \text{B}:& 3x+2y& =12\end{array}$

Answer

The solution is $(2,3)$.

Neither variable is isolated yet, so rearrange first. In equation A, both variables have a coefficient of $1$, so it is easy to isolate either one: $y=5-x$. Then substitute into equation B:

$\begin{array}{rl}3x+2(5-x)& =12\\ 3x+10-2x& =12\\ x& =2\end{array}$

Then $y=5-2=3$.

Check: in A, $2+3=5$; in B, $3(2)+2(3)=12$.

Question 6

$\begin{array}{rlrl}& \text{A}:& x-y& =7\\ & \text{B}:& 2x+y& =-10\end{array}$

One group says the solution is $(1,-6)$. Another group says it is $(-1,-8)$.

Who is right? How do you know?

Answer

The second group is right: the solution is $(-1,-8)$.

Check each point in both equations.

For $(1,-6)$: in A, $1-(-6)=7$, which works. But in B, $2(1)+(-6)=-4$, not $-10$. The point is on line A but not on line B, so it is not a solution.

For $(-1,-8)$: in A, $(-1)-(-8)=7$; in B, $2(-1)+(-8)=-10$. Left side equals right side in both equations.

A point that works in only one equation is not a solution. Always check both.

Question 7

Solve the system.

$\begin{array}{rlrl}& \text{A}:& 2x+3y& =7\\ & \text{B}:& 4x-y& =7\end{array}$

Answer

The solution is $(2,1)$.

Which variable, in which equation, is easiest to isolate? The $y$ in equation B has a coefficient of $-1$. Isolating it gives $y=4x-7$. Substitute into equation A:

$\begin{array}{rl}2x+3(4x-7)& =7\\ 2x+12x-21& =7\\ 14x& =28\\ x& =2\end{array}$

Then $y=4(2)-7=1$.

Check: in A, $2(2)+3(1)=7$; in B, $4(2)-1=7$.

Question 8

Solve the system.

$\begin{array}{rlrl}& \text{A}:& 3x+6y& =4\\ & \text{B}:& x-2y& =1\end{array}$

Answer

The solution is $\left(\frac{7}{6},\frac{1}{12}\right)$.

In equation B, $x$ has a coefficient of $1$, so isolate it: $x=1+2y$. Substitute into equation A:

$\begin{array}{rl}3(1+2y)+6y& =4\\ 3+12y& =4\\ 12y& =1\\ y& =\frac{1}{12}\end{array}$

Then $x=1+2\left(\frac{1}{12}\right)=1+\frac{1}{6}=\frac{7}{6}$.

Answers do not have to be whole numbers. Leave them as fractions in lowest terms, written as improper fractions rather than mixed numbers.

Check: in A, $3\left(\frac{7}{6}\right)+6\left(\frac{1}{12}\right)=\frac{7}{2}+\frac{1}{2}=4$; in B, $\frac{7}{6}-2\left(\frac{1}{12}\right)=\frac{7}{6}-\frac{1}{6}=1$.

Question 9

Solve the system.

$\begin{array}{rlrl}& \text{A}:& 2x+y-7& =0\\ & \text{B}:& 3x& =2y\end{array}$

Answer

The solution is $(2,3)$.

The equations are not in the usual form, but the method is the same. From equation A, $y=7-2x$. Substitute into equation B:

$\begin{array}{rl}3x& =2(7-2x)\\ 3x& =14-4x\\ 7x& =14\\ x& =2\end{array}$

Then $y=7-2(2)=3$.

Check: in A, $2(2)+3-7=0$; in B, $3(2)=6$ and $2(3)=6$.

Question 10

Solve the system. What would the graph look like?

$\begin{array}{rlrl}& \text{A}:& y& =2x+1\\ & \text{B}:& 4x-2y& =6\end{array}$

Answer

There is no solution.

Substitute $2x+1$ for $y$ in equation B:

$\begin{array}{rl}4x-2(2x+1)& =6\\ 4x-4x-2& =6\\ -2& =6\end{array}$

The $x$ terms cancel and leave a false statement. No value of $x$ can make $-2$ equal $6$, so the system has no solution.

On a graph, the lines are parallel. Rearranged, equation B is $y=2x-3$: the same slope as line A, $2$, but a different $y$-intercept.

Question 11

Solve the system. What would the graph look like?

$\begin{array}{rlrl}& \text{A}:& x& =2y-3\\ & \text{B}:& 3x-6y& =-9\end{array}$

Answer

There are infinitely many solutions.

Substitute $2y-3$ for $x$ in equation B:

$\begin{array}{rl}3(2y-3)-6y& =-9\\ 6y-9-6y& =-9\\ -9& =-9\end{array}$

The variables cancel and leave a statement that is always true, whatever the value of $y$. Every point on line A is also on line B.

On a graph, the lines are coincident: the same line, written twice. Both equations rearrange to $y=\frac{1}{2}x+\frac{3}{2}$.

Question 12

Solve the system.

$\begin{array}{rlrl}& \text{A}:& 2x+3y& =12\\ & \text{B}:& 5x-2y& =11\end{array}$

Answer

The solution is $(3,2)$.

No variable has a coefficient of $1$, so substitution still works, but it is messy. For example, from equation A, $x=\frac{12-3y}{2}$. Substitute into equation B, then multiply both sides by $2$ to clear the fraction:

$\begin{array}{rl}5\left(\frac{12-3y}{2}\right)-2y& =11\\ 5(12-3y)-4y& =22\\ 60-15y-4y& =22\\ -19y& =-38\\ y& =2\end{array}$

Then $x=\frac{12-3(2)}{2}=3$.

Check: in A, $2(3)+3(2)=12$; in B, $5(3)-2(2)=11$.

Summary of key ideas

Here is the summary of the concepts we consolidated together in class, after the whiteboard work:

Solving Linear Systems by Substitution

NOTE

If the whiteboard conversation didn’t quite click, or you were away, this is the document to read for a concrete summary of the key ideas: how to isolate a variable and substitute it into the other equation, what to do when the answer is a fraction, and how to check a solution with a left-side / right-side check.

Curriculum connection

B1.1

solve systems of two linear equations involving two variables, using the algebraic method of substitution or elimination (Sample problem: Solve $y=\frac{1}{2}x-5$, $3x+2y=-2$ for $x$ and $y$ algebraically, and verify algebraically and graphically);

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Mathematical processes

Reasoning and Proving

develop and apply reasoning skills (e.g., recognition of relationships, generalization through inductive reasoning, use of counter-examples) to make mathematical conjectures, assess conjectures, and justify conclusions, and plan and construct organized mathematical arguments;

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