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Solving Two-Step Equations: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A decimal coefficient

Find $y$ if $-2.4y-3.8=3.4$ , and check your value in the original equation.
1. Hint 1
  The variable is multiplied by a negative decimal first, and a constant is subtracted after that.
2. Hint 2
  Add $3.8$ to both sides, then divide both sides by the full coefficient, sign included.
Answer

$y=-3$ ; both sides equal $3.4$ .

Full solution

The subtraction of $3.8$ is the outer operation, so undo it first by adding $3.8$ to both sides:
$-2.4y=7.2$

Divide both sides by the full coefficient $-2.4$ .
Since $2.4(3)=7.2$ and a positive number divided by a negative number is negative,
$y=-3$

Check in the original: $-2.4(-3)=7.2$ , and
$7.2-3.8=3.4$
The sides agree.

Answer

$y=-3$ ; both sides equal $3.4$ .

Key idea

Clear the constant first, then divide by the coefficient with its sign.
Problem 1: Mark complete
Problem 2 Fractions on a card

Find $x$ if $\frac{5}{6}-\frac{2}{3}x=\frac{1}{4}$ . Give the answer as a fraction in lowest terms and check it.
1. Hint 1
  The coefficient of $x$ includes the negative sign.
2. Hint 2
  Subtract $\frac{5}{6}$ from both sides, then divide by that signed coefficient.
Answer

$x=\frac{7}{8}$ ; both sides equal $\frac{1}{4}$ .

Full solution

Subtract the constant from both sides, using twelfths on the right:
$-\frac{2}{3}x=-\frac{7}{12}$

Divide by the nonzero coefficient $-\frac{2}{3}$ , or multiply by its reciprocal:
$x=(-\frac{7}{12})(-\frac{3}{2})$
The resulting fraction $\frac{21}{24}$ reduces to $\frac{7}{8}$ .

For the check, $\frac{2}{3}\cdot\frac{7}{8}=\frac{7}{12}$ , so
$\frac{5}{6}-\frac{7}{12}=\frac{1}{4}$

Answer

$x=\frac{7}{8}$ ; both sides equal $\frac{1}{4}$ .

Key idea

An exact fractional solution is checked with the same substitution used for an integer solution.
Problem 2: Mark complete
Problem 3 A decimal outside parentheses

Find $x$ if $1.5(4x-2)+8=20$ , and check it in the equation as written.
1. Hint 1
  Identify the terms on the left after the parentheses have been removed.
2. Hint 2
  Multiply $1.5$ by both terms in the parentheses, then combine the two constant terms before undoing the operations.
Answer

$x=2.5$ , or equivalently $\frac{5}{2}$ ; both sides equal $20$ .

Full solution

Distribute $1.5$ to both terms in the parentheses: $1.5(4x)=6x$ and
$1.5(-2)=-3.$
The left side becomes $6x-3+8$ , which simplifies to $6x+5$ .
The equation is
$6x+5=20$

Subtract $5$ from both sides and divide both sides by $6$ :
$\begin{aligned}6x&=15\\x&=2.5\end{aligned}$
Dividing $15$ by $6$ gives $\frac{15}{6}$ , which reduces to $\frac{5}{2}$ , or $2.5$ as a decimal.

Check in the original at $x=2.5$ : the parentheses give $4(2.5)-2=8$ and the product is $1.5(8)=12$ .
Adding $8$ gives the left side
$12+8=20$
The original sides agree.

Answer

$x=2.5$ , or equivalently $\frac{5}{2}$ ; both sides equal $20$ .

Key idea

Rewrite a crowded side in a simpler form before undoing its operations.
Problem 3: Mark complete
Problem 4 A fractional line

Find $x$ if $\frac{3}{4}x-\frac{1}{2}x+5=3$ , and check the original equation.
1. Hint 1
  The two variable terms have matching variable parts.
2. Hint 2
  Combine their signed coefficients before clearing the constant.
Answer

$x=-8$ ; both sides equal $3$ .

Full solution

The coefficient difference is $\frac{3}{4}-\frac{1}{2}=\frac{1}{4}$ , so the equation becomes
$\frac{1}{4}x+5=3$

Subtract $5$ from both sides, then divide by $\frac{1}{4}$ :
$\begin{aligned}\frac{1}{4}x&=-2\\x&=-8\end{aligned}$

The check uses both original variable terms:
$\begin{aligned}\frac{3}{4}(-8)-\frac{1}{2}(-8)+5&=-6+4+5\\-6+4+5&=3\end{aligned}$

Answer

$x=-8$ ; both sides equal $3$ .

Key idea

Fractional coefficients combine before the usual inverse operations are applied.
Problem 4: Mark complete
Problem 5 Bundles of cardstock

Two bundles contain the same number of cardstock sheets. Three sheets from each bundle are set aside, leaving $50$ sheets in the two bundles together. How many sheets were in each bundle originally? Check your answer.
1. Hint 1
  Each bundle loses its own three sheets.
2. Hint 2
  Let one letter represent the original count in one bundle; double its remaining count to describe the total.
Answer

$28$ sheets in each bundle.

Full solution

Let $c$ be the original count in one bundle.
Each has $c-3$ sheets left, so
$2(c-3)=50$

Distribute and solve:
$\begin{aligned}2c-6&=50\\2c&=56\\c&=28\end{aligned}$
The moves are adding $6$ to both sides, then dividing both sides by $2$ .

Checking the remaining count, each bundle has $28-3=25$ sheets and the two together have $50$ .

Answer

$28$ sheets in each bundle.

Key idea

A repeated change within each group belongs inside the repeated quantity.
Problem 5: Mark complete
Problem 6 A shared bill

Eight friends split a restaurant bill equally. Each friend applies a $3.50 coupon to their own share and then pays the remaining $5. How much was the whole bill before the coupons? Check your answer.
1. Hint 1
  Each friend's share is the whole bill divided by eight.
2. Hint 2
  Write what one friend pays after the coupon, set it equal to five dollars, and undo the subtraction before the division.
Answer

$68 before the coupons.

Full solution

Let $b$ be the whole bill in dollars.
Each share is $\frac{b}{8}$ , and after the coupon each friend pays
$\frac{b}{8}-3.5=5$

Add $3.5$ to both sides, then multiply both sides by $8$ :
$\begin{aligned}\frac{b}{8}&=8.5\\b&=68\end{aligned}$
The whole bill before the coupons was $68.

Check: each share is $68\div8=8.5$ dollars, and $8.5-3.5=5$ dollars, as stated.

Answer

$68 before the coupons.

Key idea

Recover an original amount by undoing, in reverse order, the operations that produced the known result.
Problem 6: Mark complete
Problem 7 A short record

A solution record says, "Divide both sides by $4$ , then add $6$ to both sides," and ends at $x=-2$ . The original left side was $4$ times a difference containing $x$ once. Reconstruct the original equation with that difference in parentheses and a single number on the right.
1. Hint 1
  The recorded moves can be undone to recover the earlier equation.
2. Hint 2
  Undo the last recorded move before undoing the first.
Answer

$4(x-6)=-32$ .

Full solution

Begin with the last line, $x=-2$ .
Undo the addition of $6$ by subtracting $6$ from both sides:
$x-6=-8$

Undo the earlier division by $4$ by multiplying both sides by $4$ :
$4(x-6)=-32$

This has the required form.
Checking the recorded solution in it gives
$4((-2)-6)=-32$
Dividing this equation by $4$ and then adding $6$ recovers $x=-2$ , as the record requires.

Answer

$4(x-6)=-32$ .

Key idea

Reversing valid balance moves can reconstruct an equation from its solution record.
Problem 7: Mark complete
Problem 8 Jo's notebook

Jo starts with $-2(3x-4)+x=13$ , writes $-6x-8+x=13$ , and reports $x=-\frac{21}{5}$ . Is Jo's work correct? If not, identify the first invalid step and give the correct solution.
1. Hint 1
  Check the rewritten equation before checking the final arithmetic.
2. Hint 2
  The outside factor multiplies both terms inside its group, with their signs.
Answer

No; the first invalid step is rewriting $-2(3x-4)$ as $-6x-8$ (it should be $-6x+8$ ); the correct solution is $x=-1$ .

Full solution

The first rewrite is invalid: the product of $-2$ and $-4$ is $+8$ , not $-8$ .
Removing the parentheses correctly gives
$-6x+8+x=13$

Combine the variable terms, subtract $8$ from both sides, and divide by $-5$ :
$\begin{aligned}-5x+8&=13\\-5x&=5\\x&=-1\end{aligned}$

Check the original at $x=-1$ : the parentheses give $-7$ , the product is $14$ , and adding $-1$ gives $13$ .

Answer

No; the first invalid step is rewriting $-2(3x-4)$ as $-6x-8$ (it should be $-6x+8$ ); the correct solution is $x=-1$ .

Key idea

Check a distributed constant with its full sign before continuing to solve.
Problem 8: Mark complete
Problem 9 Ines's estimate

Ines claims that the solution of $3.2-0.9x=1.7$ lies strictly between $1$ and $2$ . Is Ines right? Give the exact solution and explain your verdict.
1. Hint 1
  The claim concerns a range, but the equation still has one exact solution.
2. Hint 2
  Clear the constant, divide by the signed coefficient, and write the quotient as a fraction in lowest terms; then compare it with one and two.
Answer

Yes; $x=\frac{5}{3}$ .

Full solution

Subtract $3.2$ from both sides, then divide both sides by the full coefficient $-0.9$ :
$\begin{aligned}-0.9x&=-1.5\\x&=\frac{-1.5}{-0.9}\\x&=\frac{15}{9}\\x&=\frac{5}{3}\end{aligned}$
A negative divided by a negative is positive, and multiplying the top and bottom of $\frac{1.5}{0.9}$ by $10$ gives $\frac{15}{9}$ , which reduces to $\frac{5}{3}$ .

Since $\frac{3}{3}<\frac{5}{3}<\frac{6}{3}$ , the value $\frac{5}{3}$ is larger than $1$ and smaller than $2$ , so Ines is right.
Its decimal form $1.666\ldots$ never ends, which is why the fraction is the exact answer.

The exact solution checks in the original: $0.9\div3=0.3$ , so $0.9(\frac{5}{3})=0.3(5)$ , which is $1.5$ , and
$3.2-1.5=1.7$

Answer

Yes; $x=\frac{5}{3}$ .

Key idea

An estimate can locate a solution, while exact arithmetic identifies the value that makes the equation true.
Problem 9: Mark complete
Problem 10 Two printed cards

One card reads $\frac{x-6}{7}=1$ and another reads $\frac{x}{7}-6=1$ . For each card, name the operation applied last to $x$ , then solve and check both cards.
1. Hint 1
  Consider which amount the fraction bar divides on each card.
2. Hint 2
  Undo the operation applied last on each left side.
Answer

First card: the division by $7$ was applied last, $x=13$ ; other card: the subtraction of $6$ was applied last, $x=49$ ; on each card both sides equal $1$ .

Full solution

On the first card, the whole difference $x-6$ is divided by $7$ , so the division is applied last and is undone first.
Multiply both sides by $7$ , then add $6$ :
$\begin{aligned}x-6&=7\\x&=13\end{aligned}$
Its check is
$\frac{13-6}{7}=1.$

On the other card, $x$ is divided by $7$ before $6$ is subtracted, so the subtraction is applied last and is undone first.
Add $6$ to both sides, then multiply by $7$ :
$\begin{aligned}\frac{x}{7}&=7\\x&=49\end{aligned}$
Its check is
$\frac{49}{7}-6=1.$

The two cards use the same numbers but have different solutions, because the fraction bars group different amounts.

Answer

First card: the division by $7$ was applied last, $x=13$ ; other card: the subtraction of $6$ was applied last, $x=49$ ; on each card both sides equal $1$ .

Key idea

Grouping decides which operation is applied last, and so which layer comes off first.
Problem 10: Mark complete

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