8th Grade · 8.EE.C.8

Systems of equations in 8th grade

Systems of equations are two equations about the same two unknowns, solved together. In 8th grade, your child works with systems of linear equations, where each equation makes a straight line on a graph. Your child learns what the solution means on that graph, how to find it with algebra and how to turn a word problem into two equations.

What the standard says

The official Common Core wording, quoted exactly. The plain version follows.

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations.
8.EE.C.8.a Understand that solutions to a system of two linear equations in two variables correspond to points of intersection of their graphs, because points of intersection satisfy both equations simultaneously.
8.EE.C.8.b Solve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations. Solve simple cases by inspection.
8.EE.C.8.c Solve real-world and mathematical problems leading to two linear equations in two variables.

What it means, in plain words

A system of equations is a pair of equations with the same two unknowns, $x$ and $y$ . Its solution is a pair of numbers that makes both equations true at once. Each equation is a line on a graph, and every point on the line fits that equation. The point where the two lines cross fits both, so the crossing point is the solution.

Solving systems of equations in 8th grade uses a few methods. Graphing both lines gives an estimate of the crossing point. Substitution uses one equation to say what $y$ equals, then puts that expression into the other equation. Elimination adds or subtracts the two equations so that one unknown cancels out. Some systems can be solved by inspection, which means by looking closely. For example, $3x + 2y = 5$ and $3x + 2y = 6$ can never both be true, so that system has no solution.

Two lines usually cross once, so most systems have one solution. Parallel lines never meet, so their system has no solution. Two equations for the same line share every point, so their system has infinitely many solutions. In a word problem, your child names the two unknowns, writes one equation for each fact and then solves.

Your child should be able to:

Explain why the crossing point of two lines is the solution of the system
Estimate a solution by graphing both equations
Solve a system by substitution or by elimination
Write and solve two equations for a word problem

Example

Solve the system $y = 2x + 1$ and $3x + 2y = 16$ .

The first equation says what $y$ equals, so substitute $2x + 1$ for $y$ in the second: $3x + 2(2x + 1) = 16$ .
Clear the parentheses: $3x + 4x + 2 = 16$ , so $7x + 2 = 16$ .
Subtract 2 from both sides to get $7x = 14$ , then divide by 7: $x = 2$ .
Then $y = 2(2) + 1 = 5$ . Check: $3(2) + 2(5) = 6 + 10 = 16$ .
The solution is $x = 2$ and $y = 5$ , the point $(2, 5)$ where the two lines cross.

Example

A school sold 10 tickets for 62 dollars in all. Adult tickets cost 8 dollars and child tickets cost 5 dollars. How many of each kind did it sell?

Let $a$ be the number of adult tickets and $c$ the number of child tickets. Then $a + c = 10$ and $8a + 5c = 62$ .
Multiply the first equation by 5: $5a + 5c = 50$ .
Subtract it from the second equation. The $c$ terms cancel: $3a = 12$ , so $a = 4$ .
Then $c = 10 - 4 = 6$ . Check: $8 \times 4 + 5 \times 6 = 32 + 30 = 62$ dollars.
The school sold 4 adult tickets and 6 child tickets.

Where children usually get stuck

Your child finds $x = 2$ , writes it as the answer and stops.
An equation with one unknown is finished when $x$ is found, and that habit carries over. A system has two unknowns, so the answer needs both numbers.
What helps. Ask where the two lines cross. A point needs two numbers, so the question sends your child back to find $y$ .
When putting $2x + 1$ in place of $y$ in $3x + 2y = 16$ , your child writes $3x + 2 \times 2x + 1 = 16$ and gets $x = \frac{15}{7}$ .
Without parentheses, the 2 multiplies only the $2x$ and misses the 1.
What helps. Agree on one rule: when a letter is replaced by an expression, the expression goes in parentheses. An odd fraction where a clean answer was expected is also a good reason to look back.
From a hand-drawn graph, your child reads the crossing point as $(2, 3)$ when the lines actually cross at $(1.5, 2.5)$ .
A sketch gives an estimate. When the crossing point falls between grid lines, the eye rounds it to the nearest corner.
What helps. Put the point into both equations. If either one comes out false, the point is not the solution, and substitution or elimination will find the exact one.

Try four practice questions

Real 8th grade questions from our question bank, easiest first. Answer them right here, check each one, and see a worked solution for any question.

1Warm-up
Consider this system of equations.
$\displaystyle \left\{\begin{array}{l}y=2x+1\\3x+2y=7\end{array}\right.$
Which equation results when $2x+1$ replaces $y$ in the second equation?
A
$3x+2(2x+1)=7$
B
$3x+2x+1=7$
C
$3(2x+1)+2y=7$
Show the answer and how to solve it
$3x+2(2x+1)=7$
1. 1The first equation says that $y$ and $2x+1$ have the same value. So $2y$ is twice that whole expression.
2. 2In the second equation, replace $2y$ with $2(2x+1)$ . The equation now contains only $x$ .
Common mistake. Replacing $y$ does not remove the factor $2$ in $2y$ .
The idea underneath. You can replace a variable with an equal expression. Any factor multiplying the variable still multiplies the whole expression.
2On level
Two students want to remove $y$ from this system.
$\displaystyle \left\{\begin{array}{l}4x-y=9\\5x+3y=7\end{array}\right.$
Maya rewrites the first equation as $y=4x-9$ , then substitutes this expression for $y$ in the second equation.
Leo multiplies every term of the first equation by $3$ , then adds the two equations.
Whose method gives an equation with only $x$ ?
A
Both methods give an equation with only $x$ .
B
Only Maya's method gives an equation with only $x$ .
C
Only Leo's method gives an equation with only $x$ .
D
Neither method gives an equation with only $x$ .
Show the answer and how to solve it
Both Maya and Leo obtain $17x=34$ , an equation with only $x$ .
1. 1Maya first isolates $y$ . Subtracting $4x$ from both sides gives $-y=9-4x$ . Multiplying both sides by $-1$ gives $y=4x-9$ .
2. 2In the second equation, $3y$ becomes $3(4x-9)$ . Thus $5x+3(4x-9)=7$ , or $17x-27=7$ . Adding $27$ to both sides gives $17x=34$ . Only $x$ remains.
3. 3Leo's multiplication gives $12x-3y=27$ . Adding the second equation gives $(12x+5x)+(-3y+3y)=27+7$ , so $17x=34$ . The $y$ -terms cancel. Both methods work.
Common mistake. When multiplying an equation by $3$ , multiply every term on both sides: $4x-y=9$ becomes $12x-3y=27$ .
The idea underneath. Substitution and elimination can both remove the same variable. Check the algebra in each method.
3On level
Consider this system of equations.
$\displaystyle \begin{aligned}\text{First:}\quad -2x+3y=4\\\text{Second:}\quad -6x+3y=0\end{aligned}$
Change one equation, then combine the equations to eliminate $x$ .
every term in the
equation by
.
Then combine the updated equations:
.
Show the answer and how to solve it
Multiply the first equation by $3$ , or divide the second equation by $3$ . Then subtract the equations in either order.
1. 1The $x$ coefficients are $-2$ and $-6$ . One way to make them equal is to multiply every term in the first equation by $3$ .
2. 2Subtract the second equation from the updated first equation. The $x$ terms cancel: $-6x-(-6x)=0$ . Subtract each term in the second equation.
3. 3You can also divide every term in the second equation by $3$ , then subtract it from the first. The $x$ coefficients are then both $-2$ , so they cancel by subtraction.
4. 4Reversing either subtraction still eliminates $x$ . It changes the signs of both sides, not whether the $x$ terms cancel.
Common mistake. Choose the equation and action together. Dividing the first equation by $3$ or multiplying the second by $3$ does not make the $x$ coefficients equal.
The idea underneath. Equal coefficients cancel by subtraction. You can make them equal by multiplying one equation or dividing the other.
4Stretch
Consider this system of equations.
$\displaystyle \begin{cases}-x+2y=-2 \\ 4y=2x-4\end{cases}$
The thick solid trace shows the first equation. The dashed trace shows the second equation.
A student says, "The lines overlap, so the system has exactly one solution."
Which statement correctly checks the claim?
A
The claim is correct. Two overlapping lines share only one point.
B
The claim is incorrect. Equal slopes always mean there is no solution.
C
The claim is incorrect. Every point on the shared line satisfies both equations.
D
The claim is correct. Only the shared y-intercept satisfies both equations.
Show the answer and how to solve it
The claim is incorrect. Every point on the shared line satisfies both equations.
1. 1In the first equation, add $x$ to both sides to get $2y=x-2$ . Divide by $2$ : $y=\frac12x-1$ .
2. 2Divide both sides of the second equation by $4$ : $y=\frac{2x-4}{4}=\frac12x-1$ .
3. 3Both equations describe the same line. This explains why the solid and dashed traces overlap on the graph. Every point on that line satisfies both original equations.
4. 4For example, $(0, -1)$ and $(2, 0)$ both lie on the shared line. There are infinitely many points on it, so there are infinitely many solutions. The claim is incorrect.
Common mistake. Count shared points, not visible lines. One shared line contains infinitely many solution points.
The idea underneath. A solution must satisfy both equations. On a graph, it must be a point on both lines.