Solving Simultaneous Equations by Elimination

Solve a pair of linear equations by adding or subtracting them to eliminate one variable, multiplying one or both equations first when the coefficients do not match.

Before you start

Substitution (lesson 2.6) works best when a variable has coefficient $1$ or $-1$ . Elimination works whatever the coefficients are.
Subtracting a negative term is adding: $2y - (-y) = 3y$ . Take care with signs too in $-3y - 5y = -8y$ .
Multiplying every term of an equation by the same number gives an equivalent equation (lesson 2.1): $x + y = 5$ and $2x + 2y = 10$ have the same solutions.

Eliminating by Subtracting or Adding

If two equations are both true, you may subtract one from the other, left side from left side and right side from right side, and the result is true too. You may also add them. If one variable has the same coefficient in both equations, subtracting removes it; if its coefficients are opposite, such as $2y$ and $-2y$ , adding removes it. That variable is eliminated, and an equation in one unknown is left.

Solving by elimination

(1) Write both equations in the form $ax + by = c$ , with the $x$ terms, the $y$ terms and the numbers lined up.

(2) If needed, multiply one or both equations so that one variable has coefficients of the same size in both.

(3) Same signs: subtract the equations. Opposite signs: add them.

(4) Solve for the remaining variable, then substitute into one of the equations to find the other.

(5) Check both values in both original equations.

Example 1: Equal coefficients: subtract

Solve $3x + 2y = 16$ and $3x - y = 10$ .

Solution

$3x + 2y = 16$ $\quad$ (1)

$3x - y = 10$ $\quad$ (2)

Both equations have $3x$ , with the same sign. Subtract (2) from (1):

$(3x + 2y) - (3x - y) = 16 - 10$

$3x - 3x + 2y - (-y) = 6$

$3y = 6$ , so $y = 2$ .

Substitute $y = 2$ into (2): $3x - 2 = 10$ , so $3x = 12$ and $x = 4$ .

Check in (1): $3(4) + 2(2) = 12 + 4 = 16\text{,}$ true. Check in (2): $3(4) - 2 = 10$ , true.

$x = 4$ , $y = 2$

Example 2: Opposite coefficients: add

Solve $4x + y = 14$ and $2x - y = 4$ .

Solution

$4x + y = 14$ $\quad$ (1)

$2x - y = 4$ $\quad$ (2)

The $y$ terms are $+y$ and $-y$ : opposite signs. Add (1) and (2):

$(4x + y) + (2x - y) = 14 + 4$

$6x = 18$ , so $x = 3$ .

Substitute into (1): $12 + y = 14$ , so $y = 2$ .

Check in (1): $12 + 2 = 14$ , true. Check in (2): $6 - 2 = 4$ , true.

$x = 3$ , $y = 2$

Example 3: Subtracting negative terms

Solve $2x - 3y = -4$ and $2x + 5y = 12$ .

Solution

$2x - 3y = -4$ $\quad$ (1)

$2x + 5y = 12$ $\quad$ (2)

Both have $2x$ . Subtract (2) from (1), term by term, in brackets:

$(2x - 3y) - (2x + 5y) = -4 - 12$

$2x - 2x - 3y - 5y = -16$

$-8y = -16$ , so $y = 2$ .

Substitute into (2): $2x + 10 = 12$ , so $2x = 2$ and $x = 1$ .

Check in (1): $2(1) - 3(2) = 2 - 6 = -4\text{,}$ true. Check in (2): $2 + 10 = 12$ , true.

Subtracting (1) from (2) instead gives $8y = 16$ : the same $y$ , with fewer minus signs. Either order is right.

$x = 1$ , $y = 2$

Multiplying First

If no variable has coefficients of the same size, make them the same. Multiply every term of one equation (or of both) by a number, so that the coefficients of the chosen variable become equal to their lowest common multiple.

Example 4: Multiplying one equation

Solve $2x + 3y = 13$ and $x + y = 5$ .

Solution

$2x + 3y = 13$ $\quad$ (1)

$x + y = 5$ $\quad$ (2)

Multiply every term of (2) by $2$ so that both have $2x$ : $2x + 2y = 10$ $\quad$ (3)

Subtract (3) from (1): $(2x + 3y) - (2x + 2y) = 13 - 10\text{,}$ so $y = 3$ .

Substitute into (2): $x + 3 = 5$ , so $x = 2$ .

Check in (1): $4 + 9 = 13$ , true. Check in (2): $2 + 3 = 5$ , true.

$x = 2$ , $y = 3$

Example 5: Multiplying both equations

Solve $3x + 4y = 10$ and $5x - 3y = -22$ .

Solution

$3x + 4y = 10$ $\quad$ (1)

$5x - 3y = -22$ $\quad$ (2)

The $y$ terms have opposite signs, so eliminate $y$ by adding. The lowest common multiple of $4$ and $3$ is $12$ .

(1) $\times 3$ : $9x + 12y = 30$ $\quad$ (3)

(2) $\times 4$ : $20x - 12y = -88$ $\quad$ (4)

Add (3) and (4): $29x = -58$ , so $x = -2$ .

Substitute into (1): $3(-2) + 4y = 10$ , so $-6 + 4y = 10$ , $4y = 16$ and $y = 4$ .

Check in (1): $-6 + 16 = 10$ , true. Check in (2): $5(-2) - 3(4) = -10 - 12 = -22\text{,}$ true.

$x = -2$ , $y = 4$

Which variable to eliminate?

Either variable will do; the answer is the same. Choose the one that needs the smaller numbers: look at the lowest common multiple of its two coefficients.

If one variable already has coefficients of the same size, or one coefficient is a multiple of the other, eliminate that one.

If the signs are opposite, you add; many people find adding safer than subtracting.

Example 6: Choosing the variable

Solve $4x + 3y = 5$ and $6x - 5y = 17$ , choosing which variable to eliminate and saying why.

Solution

$4x + 3y = 5$ $\quad$ (1)

$6x - 5y = 17$ $\quad$ (2)

To eliminate $x$ : the lowest common multiple of $4$ and $6$ is $12$ (multiply by $3$ and $2$ ). To eliminate $y$ : the lowest common multiple of $3$ and $5$ is $15$ (multiply by $5$ and $3$ ). Eliminating $x$ uses smaller numbers.

(1) $\times 3$ : $12x + 9y = 15$ $\quad$ (3)

(2) $\times 2$ : $12x - 10y = 34$ $\quad$ (4)

Same signs, so subtract (4) from (3): $9y - (-10y) = 15 - 34\text{,}$ so $19y = -19$ and $y = -1$ .

Substitute into (1): $4x - 3 = 5$ , so $4x = 8$ and $x = 2$ .

Check in (1): $8 + 3(-1) = 5$ , true. Check in (2): $12 - 5(-1) = 12 + 5 = 17\text{,}$ true.

$x = 2$ , $y = -1$ (eliminating $x$ , because $12$ is smaller than $15$ )

Fractions, and Terms Not Lined Up

Example 7: A fractional answer

Solve $6x + 3y = 5$ and $3x - 3y = 4$ .

Solution

$6x + 3y = 5$ $\quad$ (1)

$3x - 3y = 4$ $\quad$ (2)

The $y$ terms are $+3y$ and $-3y$ . Add: $9x = 9$ , so $x = 1$ .

Substitute into (2): $3 - 3y = 4$ , so $-3y = 1$ and $y = -\frac{1}{3}$ .

Check in (1): $6 + 3(-\frac{1}{3}) = 6 - 1 = 5\text{,}$ true. Check in (2): $3 - 3(-\frac{1}{3}) = 3 + 1 = 4\text{,}$ true.

$x = 1$ , $y = -\frac{1}{3}$

Elimination needs the like terms lined up: $x$ under $x$ , $y$ under $y$ , numbers on the right. If an equation is written another way, rearrange it into the form $ax + by = c$ first.

Example 8: Rearranging before eliminating

Solve $3y = 2x + 1$ and $4x + y = 12$ .

Solution

Subtract $2x$ from both sides of the first equation: $-2x + 3y = 1$ $\quad$ (1)

$4x + y = 12$ $\quad$ (2)

Eliminate $x$ : (1) $\times 2$ gives $-4x + 6y = 2$ $\quad$ (3). The $x$ terms of (2) and (3) are opposite, so add: $7y = 14$ , $y = 2$ .

Substitute into (2): $4x + 2 = 12$ , so $4x = 10$ and $x = 2.5$ .

Check in the original first equation: $3(2) = 6$ and $2(2.5) + 1 = 6$ , true. Check in (2): $10 + 2 = 12$ , true.

$x = 2\frac{1}{2}$ , $y = 2$

Common Mistakes

Subtracting only some terms. In $(3x + 2y) - (3x - y)$ the whole second equation is taken away: $2y - (-y) = 3y$ , not $y$ .
Multiplying one side only. $x + y = 5$ times $2$ is $2x + 2y = 10$ ; every term, including the number on the right, is multiplied.
Adding when you should subtract. Same signs: subtract. Opposite signs: add. Adding $3x + 2y = 16$ and $3x - y = 10$ gives $6x + y = 26$ , which eliminates nothing.
Eliminating with terms not lined up. Rearrange $3y = 2x + 1$ to $-2x + 3y = 1$ first.
Checking only in the equation used to find the second value. Check in both original equations.

Practice Questions

Solve each pair by elimination on paper before you open its solution, and check in both equations. The questions get harder as you go.

Question 1

Solve $x + y = 7$ and $x - y = 3$ .

ShowHide solution

The $y$ terms are opposite. Add: $2x = 10$ , so $x = 5$ . Then $5 + y = 7$ , so $y = 2$ .

Check: $5 + 2 = 7$ ; $5 - 2 = 3$ .

$x = 5$ , $y = 2$

Question 2

Solve $5x + 2y = 17$ and $3x + 2y = 11$ .

ShowHide solution

Both have $2y$ . Subtract the second from the first: $2x = 6$ , so $x = 3$ . Then $9 + 2y = 11$ , so $y = 1$ .

Check: $15 + 2 = 17$ ; $9 + 2 = 11$ .

$x = 3$ , $y = 1$

Question 3

Solve $2x - 5y = -12$ and $2x + 3y = 4$ .

ShowHide solution

Both have $2x$ . Subtract: $(2x - 5y) - (2x + 3y) = -12 - 4\text{,}$ so $-8y = -16$ and $y = 2$ .

Then $2x + 6 = 4$ , so $2x = -2$ and $x = -1$ .

Check: $-2 - 10 = -12$ ; $-2 + 6 = 4$ .

$x = -1$ , $y = 2$

Question 4

Solve $3x + y = 14$ and $5x + 2y = 23$ .

ShowHide solution

Multiply the first by $2$ : $6x + 2y = 28$ . Subtract the second: $x = 5$ .

Then $15 + y = 14$ , so $y = -1$ .

Check: $15 - 1 = 14$ ; $25 - 2 = 23$ .

$x = 5$ , $y = -1$

Question 5

Solve $4x - 3y = 1$ and $x + 2y = 3$ .

ShowHide solution

Multiply the second by $4$ : $4x + 8y = 12$ . Subtract it from the first: $-3y - 8y = 1 - 12$ , so $-11y = -11$ and $y = 1$ .

Then $x + 2 = 3$ , so $x = 1$ .

Check: $4 - 3 = 1$ ; $1 + 2 = 3$ .

$x = 1$ , $y = 1$

Question 6

Solve $5x + 3y = 23$ and $2x + 7y = 15$ .

ShowHide solution

Eliminate $x$ (lowest common multiple $10$ ): first $\times 2$ : $10x + 6y = 46$ ; second $\times 5$ : $10x + 35y = 75$ .

Subtract: $29y = 29$ , so $y = 1$ . Then $5x + 3 = 23$ , so $x = 4$ .

Check: $20 + 3 = 23$ ; $8 + 7 = 15$ .

$x = 4$ , $y = 1$

Question 7

Solve $2x + 3y = -1$ and $5x - 2y = 26$ .

ShowHide solution

The $y$ terms have opposite signs: eliminate $y$ (lowest common multiple $6$ ). First $\times 2$ : $4x + 6y = -2$ ; second $\times 3$ : $15x - 6y = 78$ .

Add: $19x = 76$ , so $x = 4$ . Then $8 + 3y = -1$ , so $3y = -9$ and $y = -3$ .

Check: $8 - 9 = -1$ ; $20 + 6 = 26$ .

$x = 4$ , $y = -3$

Question 8

Solve $4x + 6y = 7$ and $2x - 3y = 0.5$ .

ShowHide solution

Multiply the second by $2$ : $4x - 6y = 1$ . Add it to the first: $8x = 8$ , so $x = 1$ .

Then $4 + 6y = 7$ , so $6y = 3$ and $y = \frac{1}{2}$ .

Check: $4 + 3 = 7$ ; $2 - 1.5 = 0.5$ .

$x = 1$ , $y = \frac{1}{2}$

Question 9

Solve $2y = x + 4$ and $3x + 4y = 28$ .

ShowHide solution

Rearrange the first: $-x + 2y = 4$ . Multiply by $3$ : $-3x + 6y = 12$ .

Add it to $3x + 4y = 28$ : $10y = 40$ , so $y = 4$ . Then $2(4) = x + 4$ , so $x = 4$ .

Check: $2(4) = 8$ and $4 + 4 = 8$ ; $12 + 16 = 28$ .

$x = 4$ , $y = 4$

Question 10

Solve $3x + 2y = 8$ and $5x - 4y = 6$ . Say which variable you chose to eliminate, and why.

ShowHide solution

For $x$ , the lowest common multiple of $3$ and $5$ is $15$ ; for $y$ , of $2$ and $4$ it is $4$ . Eliminate $y$ : only the first equation needs multiplying.

First $\times 2$ : $6x + 4y = 16$ . The second is $5x - 4y = 6$ .

The $y$ terms have opposite signs, so add: $11x = 22$ and $x = 2$ . Then $3(2) + 2y = 8$ , so $2y = 2$ and $y = 1$ .

Check: $6 + 2 = 8$ ; $10 - 4 = 6$ .

$x = 2$ , $y = 1$ ; $y$ was eliminated because $4$ is smaller than $15$ , and only one equation had to be multiplied (eliminating $x$ gives the same answer).

Question 11

The sum of two numbers is $40$ and their difference is $12$ . Find the numbers.

ShowHide solution

Let the numbers be $x$ and $y$ , with $x$ the larger: $x + y = 40$ and $x - y = 12$ .

Add: $2x = 52$ , so $x = 26$ . Then $y = 40 - 26 = 14$ .

Check: $26 + 14 = 40$ ; $26 - 14 = 12$ .

The numbers are $26$ and $14$ .

Question 12

At a supermarket, 3 kg of rice and 2 bottles of cooking oil cost $ $17.50$ , and 5 kg of rice and 2 bottles of the same oil cost $ $21.50$ . Find the cost of 1 kg of rice and of one bottle of oil.

ShowHide solution

Let 1 kg of rice cost $ $x$ and a bottle of oil $ $y$ : $3x + 2y = 17.5$ and $5x + 2y = 21.5$ .

Both have $2y$ . Subtract the first from the second: $2x = 4$ , so $x = 2$ .

Then $6 + 2y = 17.5$ , so $2y = 11.5$ and $y = 5.75$ .

Check: $3(2) + 2(5.75) = 6 + 11.5 = 17.5\text{;}$ $5(2) + 2(5.75) = 10 + 11.5 = 21.5\text{.}$

1 kg of rice costs $ $2$ and a bottle of oil costs $ $5.75$ .

Key Points

Line up the equations in the form $ax + by = c$ and number them.
Make the coefficients of one variable the same size, multiplying one or both equations (every term) if needed.
Same signs: subtract. Opposite signs: add. Put the subtracted equation in brackets.
Choose the variable whose coefficients have the smaller lowest common multiple, or that already match.
Solve for one variable, substitute to find the other, and check in both original equations.
Answers may be negative or fractions.