8th Grade · 8.EE.C.7

Multi-step equations in 8th grade

Multi-step equations are equations that take several moves to solve, such as $3(x + 2) + 2x = 4x + 11$ . In 8th grade, solving linear equations goes well past one or two steps. Your child learns to clear parentheses, combine terms, solve equations with variables on both sides and spot equations that have no solution or infinitely many.

What the standard says

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

8.EE.C.7 Solve linear equations in one variable.
8.EE.C.7.a Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = a, a = a, or a = b results (where a and b are different numbers).
8.EE.C.7.b Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.

What it means, in plain words

A linear equation in one variable has one unknown, usually $x$ , and the $x$ is never squared. Its solution is the number that makes both sides equal. Solving it means doing the same thing to both sides until $x$ stands alone.

Multi-step equations in 8th grade add three kinds of moves. The distributive property clears parentheses by multiplying each term inside, so $3(x + 2)$ becomes $3x + 6$ . Combining like terms joins pieces of the same kind, so $3x + 2x$ becomes $5x$ . In equations with variables on both sides, your child subtracts to gather all the $x$ terms on one side. Fractions and decimals appear too. For $\frac{1}{2}x + 3 = 7$ , multiplying both sides by 2 gives $x + 6 = 14$ , so $x = 8$ .

Not every equation ends with one answer. If the $x$ terms cancel and what is left is true, like $5 = 5$ , every number works, so the equation has infinitely many solutions. If what is left is false, like $5 = -2$ , no number works, so the equation has no solution.

Your child should be able to:

Use the distributive property to clear parentheses, including with negative numbers
Combine like terms and solve equations with variables on both sides
Solve equations that contain fractions or decimals
Tell whether an equation has one solution, no solution or infinitely many

Example

Solve $3(x + 2) + 2x = 4x + 11$ .

Clear the parentheses: $3x + 6 + 2x = 4x + 11$ .
Combine like terms: $5x + 6 = 4x + 11$ .
Subtract $4x$ from both sides: $x + 6 = 11$ .
Subtract 6 from both sides: $x = 5$ .
Check: the left side is $3(7) + 10 = 31$ and the right side is $20 + 11 = 31$ . The solution is $x = 5$ .

Example

How many solutions does $2(x + 4) + x = 3(x + 1) + 6$ have?

Left side: $2x + 8 + x = 3x + 8$ .
Right side: $3x + 3 + 6 = 3x + 9$ .
Subtract $3x$ from both sides: $8 = 9$ .
That is false for every value of $x$ , so the equation has no solution.

Where children usually get stuck

Your child turns $3(x + 2)$ into $3x + 2$ .
The 3 sits right next to the $x$ , so it looks as if it belongs only to the $x$ .
What helps. Pick a number for $x$ , such as 1, and work out both. $3(1 + 2) = 9$ , but $3 \times 1 + 2 = 5$ , so the shortcut changed the value.
Your child turns $-3(x + 2)$ into $-3x + 6$ .
The minus sign gets used on the first term and forgotten on the second. Negative 3 times positive 2 is negative 6.
What helps. Have your child say each product out loud: "negative 3 times $x$ , negative 3 times 2". Saying both makes the second sign hard to skip.
Your child reaches $5 = 5$ and writes $x = 0$ , or crosses out the work and says it went wrong.
Your child expects every equation to end with $x$ equal to a number, so a line with no $x$ in it looks like a mistake.
What helps. Ask whether $5 = 5$ is true. Then put $x = 1$ and $x = 10$ into the original equation and see that both values work.

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
Look at the equation below. Which solution type does it have?
$\displaystyle x = 5$
A
one solution
B
no solution
C
infinitely many solutions
Show the answer and how to solve it
The answer is one solution because x = 5 tells you x equals exactly one number.
1. 1Look at the equation.
2. 2The variable x is already by itself on the left side. It equals one specific number, 5.
3. 3Because x is equal to exactly one value, the equation has one solution.
Common mistake. A common mistake is choosing no solution because the equation looks too simple. x = 5 is already solved — that means x has exactly one answer.
The idea underneath. When x is missing, the equation is just a statement about numbers: a true statement gives infinitely many solutions, and a false statement gives no solution.
2On level
Select all equations that have exactly one solution.
A
$\displaystyle 5x + {}$
$\displaystyle 3 = {}$
$\displaystyle 2x + {}$
$\displaystyle 3$
B
$\displaystyle 4x + {}$
$\displaystyle 7 = {}$
$\displaystyle 4x + {}$
$\displaystyle 7$
C
$\displaystyle 3x + {}$
$\displaystyle 2 = {}$
$\displaystyle 3x + {}$
$\displaystyle 9$
D
$\displaystyle 6x + {}$
$\displaystyle 4 = {}$
$\displaystyle 6x + {}$
$\displaystyle 1$
E
$\displaystyle 2x + {}$
$\displaystyle 5 = {}$
$\displaystyle 7x + {}$
$\displaystyle 3$
Show the answer and how to solve it
The correct choices are 5x + 3 = 2x + 3 and 2x + 5 = 7x + 3 because those are the only equations with different x-coefficients on the two sides.
1. 1Look at the x-coefficients on both sides of each equation. An equation in the form $ax + b = cx + d$ has exactly one solution when $a \neq c$ .
2. 2The equations with different x-coefficients are the ones with exactly one solution.
3. 3If the x-coefficients match but the constants differ, the x-terms cancel and leave a false statement — no solution.
4. 4If both sides are identical, every value of x works, giving infinitely many solutions.
Common mistake. A common mistake is focusing on the constants. For example, 3x + 2 = 3x + 9 has different constants but no solution because the x-terms cancel, while 5x + 3 = 2x + 3 has matching constants but exactly one solution because the x-coefficients differ.
The idea underneath. The number of solutions is determined by what happens to the x-terms after you subtract one side from the other: different x-coefficients leave one x-term, but equal x-coefficients make the x-terms disappear.
3On level
Complete the equation so it has exactly one solution. Drag one coefficient tile into each blank.
Tap or click a tile to drop it into the highlighted blank, or tap an empty blank first to choose a different one. Tap a filled blank to send its tile back.
x + 5 =
x - 3
2
2
4
6
Tap an item to place it in the next empty slot, or drag it to a specific slot. With a keyboard: select an item with Enter or Space, then select a slot to place it. Press Delete or Backspace on a slot to remove an item.
Show the answer and how to solve it
One correct completion is 4x + 5 = 2x - 3. Any left-right pair with different coefficients is correct: (2,4), (2,6), (4,2), (4,6), (6,2), or (6,4).
1. 1Look at the two coefficient blanks. For an equation of the form ax + b = cx + d to have exactly one solution, the coefficients of x must be different.
2. 2The trap is using both 2 tiles. The x terms cancel and the equation becomes 5 = -3, which is never true, so the equation has no solution.
3. 3One valid choice is 4 on the left and 2 on the right. The coefficients are different, so the x terms do not cancel and the equation solves to exactly one value of x.
4. 4Any other left-right pair with different coefficients also produces exactly one solution.
Common mistake. A common mistake is using both 2 tiles since they are the duplicate in the bank. That produces 2x + 5 = 2x - 3, and the x terms cancel to 5 = -3, so the equation has no solution instead of one solution.
The idea underneath. In ax + b = cx + d, equal x-coefficients make the variable terms disappear, but different x-coefficients leave an x-term to solve, giving exactly one solution.
4Stretch
Simplify each side of the equation. Enter A, B, C, and D so the simplified equation fits Ax + B = Cx + D, then choose the solution type.
$\displaystyle 3(x + 2) + 2x - 1 = 4(x - 3) + x + 10$
A B C D Solution type
Show the answer and how to solve it
A = 5, B = 5, C = 5, and D = -2. Because the x-coefficients are equal but the constants differ, the equation has no solution.
1. 1Start by simplifying each side separately.
2. 2On the left side, distribute 3 through (x + 2), then combine the three x-terms and the two constants: $3x + 2x = 5x$ and $6 - 1 = 5$ .
3. 3On the right side, distribute 4 through (x - 3), then combine: $4x + x = 5x$ and $-12 + 10 = -2$ .
4. 4So A = 5, B = 5, C = 5, and D = -2. Subtract $5x$ on both sides to get $5 = -2$ . This is false for every value of $x$ , so the equation has no solution.
Common mistake. A common mistake is selecting 'infinitely many solutions' because A and C are both 5. You must also compare the constants: 5 is not equal to -2, so the equation has no solution.
The idea underneath. After simplifying each side, collect the x-terms on one side. If an x-term remains, division gives exactly one value of $x$ . If the x-terms cancel, a false statement means no solution. A true statement means every value of $x$ works.