Solving Linear Systems by Elimination

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

Add.

$\begin{array}{crcrcr}& 3x& +& 4y& -& 7\\ +& 5x& -& 4y& +& 2\end{array}$

Answer

$8x-5$

Add each column of like terms: $3x+5x=8x$, then $4y+(-4y)=0$, then $-7+2=-5$. The $y$-terms cancel out completely.

Question 2

Subtract.

$\begin{array}{crcrcr}& 7x& +& 3y& +& 5\\ -& 2x& +& 3y& -& 4\end{array}$

Answer

$5x+9$

Subtract every term in the second row from the term above it: $7x-2x=5x$, then $3y-3y=0$, then $5-(-4)=9$. This time it is subtraction that makes the $y$-terms vanish.

Question 3

Add equation A and equation B.

$\begin{array}{rlrl}& \text{A}:& 3x+4y& =17\\ & \text{B}:& 5x-4y& =7\end{array}$

What happens? Find $x$ and $y$.

Answer

The solution is $(3,2)$.

Equations add just like expressions do: add the left sides, and add the right sides. The $y$-terms have opposite coefficients, so they are eliminated.

$$\begin{array}{crcrcr}& 3x& +& 4y& =& 17\\ +& 5x& -& 4y& =& 7\\ & 8x& & & =& 24\end{array}$$

So $x=3$. Substitute into A: $3(3)+4y=17$, so $4y=8$ and $y=2$.

Check: $3(3)+4(2)=17$ and $5(3)-4(2)=7$.

Question 4

Solve the system.

$\begin{array}{rlrl}& \text{A}:& 7x+3y& =17\\ & \text{B}:& 2x+3y& =7\end{array}$

Answer

The solution is $(2,1)$.

The $y$-terms match exactly, with the same sign. Adding would give $6y$, so instead subtract B from A to eliminate $y$.

$$\begin{array}{crcrcr}& 7x& +& 3y& =& 17\\ -& 2x& +& 3y& =& 7\\ & 5x& & & =& 10\end{array}$$

So $x=2$. Substitute into B: $2(2)+3y=7$, so $3y=3$ and $y=1$.

Check: $7(2)+3(1)=17$ and $2(2)+3(1)=7$.

Question 5

Solve the system.

$\begin{array}{rlrl}& \text{A}:& 3x+2y& =11\\ & \text{B}:& 7x-4y& =17\end{array}$

Nothing matches yet. Can you make one pair of terms match?

Answer

The solution is $(3,1)$.

Multiply both sides of A by $2$. The new equation is equivalent to A, and its $y$-term, $+4y$, is now the opposite of the $-4y$ in B. Then add.

$$\begin{array}{crcrcrl}& 6x& +& 4y& =& 22& \text{A}\times 2\\ +& 7x& -& 4y& =& 17& \text{B}\\ & 13x& & & =& 39& \end{array}$$

So $x=3$. Substitute into A: $3(3)+2y=11$, so $2y=2$ and $y=1$.

Check: $3(3)+2(1)=11$ and $7(3)-4(1)=17$.

Question 6

Solve the system.

$\begin{array}{rlrl}& \text{A}:& 6x-5y& =-37\\ & \text{B}:& 2x+3y& =11\end{array}$

Answer

The solution is $(-2,5)$.

This time, multiply B by $3$, so that its $x$-term matches the $6x$ in A. Then subtract, and watch the signs: $-5y-9y=-14y$, and $-37-33=-70$.

$$\begin{array}{crcrcrl}& 6x& -& 5y& =& -37& \text{A}\\ -& 6x& +& 9y& =& 33& \text{B}\times 3\\ & & & -14y& =& -70& \end{array}$$

So $y=5$. Substitute into B: $2x+3(5)=11$, so $2x=-4$ and $x=-2$.

Check: $6(-2)-5(5)=-37$ and $2(-2)+3(5)=11$.

Question 7

Solve the system.

$\begin{array}{rlrl}& \text{A}:& 2x+3y& =11\\ & \text{B}:& 3x-5y& =7\end{array}$

Can you still make a pair of terms match?

Answer

The solution is $(4,1)$.

No single multiplier works here, so multiply both equations. Multiplying A by $5$ and B by $3$ turns the $y$-terms into $+15y$ and $-15y$. Then add.

$$\begin{array}{crcrcrl}& 10x& +& 15y& =& 55& \text{A}\times 5\\ +& 9x& -& 15y& =& 21& \text{B}\times 3\\ & 19x& & & =& 76& \end{array}$$

So $x=4$. Substitute into A: $2(4)+3y=11$, so $3y=3$ and $y=1$.

Check: $2(4)+3(1)=11$ and $3(4)-5(1)=7$.

You could instead eliminate $x$, with A $\times 3$ and B $\times 2$, then subtract. You get the same answer either way.

Question 8

Solve the system.

$\begin{array}{rlrl}& \text{A}:& 4x+3y& =17\\ & \text{B}:& 6x-5y& =-22\end{array}$

Answer

The solution is $\left(\frac{1}{2},5\right)$.

Multiply both equations again: A by $5$ and B by $3$ give $+15y$ and $-15y$. Then add.

$$\begin{array}{crcrcrl}& 20x& +& 15y& =& 85& \text{A}\times 5\\ +& 18x& -& 15y& =& -66& \text{B}\times 3\\ & 38x& & & =& 19& \end{array}$$

So $x=\frac{19}{38}=\frac{1}{2}$. A fraction is a perfectly good answer. Substitute into A: $4\left(\frac{1}{2}\right)+3y=17$, so $3y=15$ and $y=5$.

Check: $4\left(\frac{1}{2}\right)+3(5)=17$ and $6\left(\frac{1}{2}\right)-5(5)=-22$.

The next four questions were bonus cards, for groups that finished early.

Question 9

$\begin{array}{rlrl}& \text{A}:& 5x-3y& =22\\ & \text{B}:& 2x-3y& =7\end{array}$

A group subtracted B from A and wrote $3x-6y=15$. What went wrong? Fix it and solve.

Answer

The solution is $(5,1)$.

The group subtracted the coefficients of $y$ wrongly. Subtracting a negative is adding: $-3y-(-3y)=-3y+3y=0$. So the $y$-terms are eliminated, and the result is $3x=15$, not $3x-6y=15$.

So $x=5$. Substitute into B: $2(5)-3y=7$, so $-3y=-3$ and $y=1$.

Check: $5(5)-3(1)=22$ and $2(5)-3(1)=7$.

Question 10

Solve each system. What would each graph look like?

$\text{I}:\begin{array}{rl}4x-6y& =10\\ 6x-9y& =21\end{array}\text{II}:\begin{array}{rl}4x+6y& =10\\ 6x+9y& =15\end{array}$

Answer

I: no solution. Multiply the first equation by $3$ and the second by $2$: they become $12x-18y=30$ and $12x-18y=42$. Subtracting eliminates both variables and leaves $0=-12$, which is never true. The graph would show two parallel lines that never meet.

II: infinitely many solutions. The same multipliers give $12x+18y=30$ and $12x+18y=30$. Subtracting leaves $0=0$, which is always true. The two equations describe the same line, so the graph would show one line drawn twice (coincident lines).

Question 11

Solve the system.

$\begin{array}{rlrl}& \text{A}:& 3x& =5-2y\\ & \text{B}:& 4x-22& =5y\end{array}$

Answer

The solution is $(3,-2)$.

Rearrange each equation first so that the $x$-terms, the $y$-terms and the constants line up:

$$\begin{array}{rl}3x+2y& =5\\ 4x-5y& =22\end{array}$$

Then multiply the first equation by $5$ and the second by $2$ to get $15x+10y=25$ and $8x-10y=44$. Adding gives $23x=69$, so $x=3$. Substitute into $3x+2y=5$: $9+2y=5$, so $y=-2$.

Check in the original equations: $3(3)=9$ and $5-2(-2)=9$; $4(3)-22=-10$ and $5(-2)=-10$.

Question 12

$\begin{array}{rlrl}& \text{A}:& 3x+8y& =31\\ & \text{B}:& 5x-4y& =17\end{array}$

You could eliminate $x$ or $y$. Which takes less work? Solve it that way.

Answer

The solution is $(5,2)$.

Eliminating $y$ takes less work: only B needs multiplying. B $\times 2$ gives $10x-8y=34$, whose $y$-term is the opposite of the $+8y$ in A. (Eliminating $x$ would mean multiplying both equations.)

$$\begin{array}{crcrcrl}& 3x& +& 8y& =& 31& \text{A}\\ +& 10x& -& 8y& =& 34& \text{B}\times 2\\ & 13x& & & =& 65& \end{array}$$

So $x=5$. Substitute into A: $3(5)+8y=31$, so $8y=16$ and $y=2$.

Check: $3(5)+8(2)=31$ and $5(5)-4(2)=17$.

Summary of key ideas

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

Solving Linear Systems by Elimination

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: why some systems are awkward to solve by graphing or substitution, the four steps of elimination, and how to multiply one or both equations to make a pair of coefficients match.

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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