Solving Systems by Elimination | Grade 8 Math | LevelKite

You already know how to solve a system of equations by substitution. A system of equations is a set of equations that use the same variables. A solution of the system is a pair of values $(x, y)$ that makes every equation true. Substitution gets messy when neither equation is easy to solve for one variable, so this lesson gives you a second method. It rests on one simple fact: when you add equal amounts to equal amounts, the totals are equal.

Try it with numbers. Both $3 + 4 = 7$ and $5 - 4 = 1$ are true. Add the left sides together and add the right sides together: $(3 + 4) + (5 - 4) = 7 + 1$ . Each side equals $8$ , so the new statement is true too. Notice that $+4$ and $-4$ add to $0$ . The $4$ drops out and leaves $3 + 5 = 8$ .

The same idea works when the equations contain variables. Take the system $2x + y = 11$ and $3x - y = 4$ . In every system in this lesson, $x$ and $y$ stand for real numbers.

Suppose a pair $(x, y)$ makes both equations true. Then $3x - y$ and $4$ are the same number, because the second equation says so. So you can add $3x - y$ to the left side of the first equation and $4$ to its right side. You are adding the same amount to both sides, so the new equation is still true for that pair.

Now look at the $y$ -terms. A coefficient is the number that multiplies a variable. Since $y$ means $1y$ and $-y$ means $-1y$ , the coefficients of $y$ are $1$ and $-1$ . These are opposites: two numbers that add to $0$ . So the $y$ -terms add to $0y$ , which equals $0$ , and $y$ drops out. Stack the equations so matching terms line up, then add each column.

\begin{array}{rrcr} & 2x + y & = & 11 \\ + & 3x - y & = & 4 \\ \hline & 5x + 0y & = & 15 \end{array}

Only $x$ is left, in the equation $5x = 15$ . Divide both sides by $5$ , which is allowed because $5$ is not $0$ , to get $x = 3$ . To find $y$ , substitute $3$ for $x$ in either original equation. Using the first one: $2(3) + y = 11$ , so $6 + y = 11$ . Subtract $6$ from both sides to get $y = 5$ .

Now check the pair $(3, 5)$ in both original equations: $2(3) + 5 = 6 + 5 = 11$ and $3(3) - 5 = 9 - 5 = 4$ . Both are true, so the solution is $(3, 5)$ . Use the original equations for the check, because an equation you created could carry a mistake.

This method is called elimination. You combine the equations to eliminate, or remove, one variable. Here that left an equation with just one variable, which you already know how to solve.

Key concept

Elimination: combine equations so one variable drops out

If a pair $(x, y)$ solves a system, the two sides of each equation are equal amounts for that pair. Adding or subtracting equal amounts on both sides keeps an equation true, so the pair also makes the sum and the difference of the two equations true.

One variable has opposite coefficients: add the equations.
One variable has equal coefficients: subtract one equation from the other.
Neither: first multiply both sides of one or both equations by a nonzero number, as you will see later in this lesson.

Every system in this lesson has exactly one solution, so one variable is left. Solve for it, substitute to find the other variable, and check the pair in both original equations.

\text{If } A = B \text{ and } C = D, \text{ then } A + C = B + D \text{ and } A - C = B - D

Worked example

Equal coefficients: subtract the equations

Solve the system $2x + 3y = 16$ and $2x - y = 8$ by elimination.

Compare the coefficients. Both x-coefficients are 2. They are equal, not opposites, so adding would give 4x and nothing would drop out. Subtracting works instead, because 2x minus 2x is 0.
$2x - 2x = 0$
Subtract the second equation from the first. For a pair that solves the system, the two sides of the second equation are the same number, so you are subtracting equal amounts from both sides. Put each left side in parentheses so the subtraction reaches every term.
$(2x + 3y) - (2x - y) = 16 - 8$
Simplify both sides. Subtracting the second set of parentheses changes the sign of each term inside it: subtracting 2x gives minus 2x, and subtracting negative y gives plus y. The x-terms add to 0, the y-terms add to 4y, and 16 minus 8 is 8.
$2x + 3y - 2x + y = 8 \text{, so } 4y = 8$
Divide both sides by 4. This is a valid step because 4 is not 0.
$\frac{4y}{4} = \frac{8}{4} \text{, so } y = 2$
Substitute 2 for y in the first original equation to find x. Multiply 3 by 2, subtract 6 from both sides, then divide both sides by 2, which is not 0.
$2x + 3(2) = 16 \text{, so } 2x + 6 = 16 \text{, so } 2x = 10 \text{, so } x = 5$
Check the pair (5, 2) in both original equations. Both statements are true.
$2(5) + 3(2) = 10 + 6 = 16 \quad\text{and}\quad 2(5) - 2 = 10 - 2 = 8$

Answer

The solution of the system is $(5, 2)$ , so $x = 5$ and $y = 2$ .

Many systems have no variable with opposite or equal coefficients. In that case, you can create matching coefficients yourself. Multiplying both sides of an equation by the same nonzero number (any number except $0$ ) gives an equivalent equation, which is an equation with exactly the same solutions. Equal amounts multiplied by the same number stay equal. And because the number is not $0$ , you could divide by it to undo the step, so no solutions are gained or lost. (Multiplying both sides by $0$ would give $0 = 0$ for every pair, which tells you nothing.)

For example, multiply both sides of $x + 2y = 7$ by $3$ . The left side becomes $3(x + 2y)$ . The distributive property says to multiply each term inside the parentheses by $3$ , so this equals $3x + 6y$ for all real numbers $x$ and $y$ . The right side becomes $3 \cdot 7 = 21$ . The new equation is $3x + 6y = 21$ .

The table tests four pairs in both equations. The value of $3x + 6y$ is always $3$ times the value of $x + 2y$ , so each equation is true exactly when the other is.

Testing four pairs in x + 2y = 7 and in 3x + 6y = 21

Pair (x, y)	Value of x + 2y	Is x + 2y = 7 true?	Value of 3x + 6y	Is 3x + 6y = 21 true?
$(1, 3)$	$1 + 2(3) = 7$	Yes	$3(1) + 6(3) = 21$	Yes
$(5, 1)$	$5 + 2(1) = 7$	Yes	$3(5) + 6(1) = 21$	Yes
$(-1, 4)$	$-1 + 2(4) = 7$	Yes	$3(-1) + 6(4) = 21$	Yes
$(2, 2)$	$2 + 2(2) = 6$	No	$3(2) + 6(2) = 18$	No

Worked example

Multiply one equation, then add

Solve the system $3x + 2y = 10$ and $5x - 4y = 24$ by elimination.

Compare the coefficients. No variable has opposite or equal coefficients yet. The y-coefficients are 2 and -4. Doubling the first equation turns 2y into 4y, which is the opposite of -4y, so only one equation needs to change.
$2 \cdot 2y = 4y \quad\text{and}\quad 4y + (-4y) = 0$
Multiply both sides of the first equation by 2. Because 2 is not 0, the new equation has the same solutions as the first one. Multiply every term, including the 10 on the right side.
$2(3x + 2y) = 2(10) \text{, so } 6x + 4y = 20$
Add the new equation and the second original equation. For a pair that solves the system, this adds equal amounts to both sides. The y-terms are opposites, so they add to 0. Then 6x plus 5x is 11x, and 20 plus 24 is 44.
$(6x + 4y) + (5x - 4y) = 20 + 24 \text{, so } 11x = 44$
Divide both sides by 11, which is not 0.
$\frac{11x}{11} = \frac{44}{11} \text{, so } x = 4$
Substitute 4 for x in the first original equation. Multiply 3 by 4, subtract 12 from both sides, then divide both sides by 2, which is not 0.
$3(4) + 2y = 10 \text{, so } 12 + 2y = 10 \text{, so } 2y = -2 \text{, so } y = -1$
Check the pair (4, -1) in both original equations. Both statements are true.
$3(4) + 2(-1) = 12 - 2 = 10 \quad\text{and}\quad 5(4) - 4(-1) = 20 + 4 = 24$

Answer

The solution of the system is $(4, -1)$ , so $x = 4$ and $y = -1$ .

Worked example

Multiply both equations, then subtract

Solve the system $2x + 3y = 5$ and $3x + 5y = 9$ by elimination.

Choose a variable to eliminate. Either one works. Choose x, because its coefficients, 2 and 3, are smaller than the y-coefficients, 3 and 5. Neither 2 nor 3 is a multiple of the other, so multiply both equations. The smallest positive number that is a multiple of both 2 and 3 is 6, so aim for 6x in each equation.
$3 \cdot 2x = 6x \quad\text{and}\quad 2 \cdot 3x = 6x$
Multiply both sides of the first equation by 3 and both sides of the second equation by 2. Neither multiplier is 0, so each new equation has the same solutions as the equation it came from.
$3(2x + 3y) = 3(5) \text{ gives } 6x + 9y = 15, \quad 2(3x + 5y) = 2(9) \text{ gives } 6x + 10y = 18$
The x-coefficients are now equal, so subtract. Take the first new equation away from the second new equation, subtracting every term. Then 6x minus 6x is 0, 10y minus 9y is y, and 18 minus 15 is 3.
$(6x + 10y) - (6x + 9y) = 18 - 15 \text{, so } y = 3$
Substitute 3 for y in the first original equation. Multiply 3 by 3, subtract 9 from both sides, then divide both sides by 2, which is not 0.
$2x + 3(3) = 5 \text{, so } 2x + 9 = 5 \text{, so } 2x = -4 \text{, so } x = -2$
Check the pair (-2, 3) in both original equations. Both statements are true.
$2(-2) + 3(3) = -4 + 9 = 5 \quad\text{and}\quad 3(-2) + 5(3) = -6 + 15 = 9$

Answer

The solution of the system is $(-2, 3)$ , so $x = -2$ and $y = 3$ .

Takeaway: Elimination works because doing the same valid operation to both sides of an equation keeps its solutions. You add or subtract equal amounts, or you multiply or divide by the same nonzero number. When you need opposite or equal coefficients, multiply first. Then add or subtract the equations so one variable drops out, solve for the variable that is left, and substitute to find the other. Always check your pair in both original equations.

Next, you will choose between substitution and elimination for a system, find how many solutions a system has, and find and correct errors in a solution.