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Solving One-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 sentence

Find the value of $x$ that makes $x-(-2.35)=-0.8$ true, and check it.
1. Hint 1
  Subtracting a negative adds its opposite.
2. Hint 2
  Undo that addition on both sides.
Answer

$x=-3.15$ ; both sides equal $-0.8$ .

Full solution

The left side is $x+2.35$ .
Subtract $2.35$ from both sides to isolate $x$ :
$\begin{aligned}x&=-0.8-2.35\\x&=-3.15\end{aligned}$

In the original, the proposed value gives
$-3.15-(-2.35)=-0.8$
The sides agree, so the value solves the equation.

Answer

$x=-3.15$ ; both sides equal $-0.8$ .

Key idea

Undo the operation attached to the variable after interpreting its signs.
Problem 1: Mark complete
Problem 2 Finding $r$

Find $r$ if $-0.6r=4.2$ , and check your value.
1. Hint 1
  The variable is multiplied by a signed decimal coefficient.
2. Hint 2
  Divide both sides by the whole coefficient, including its sign.
Answer

$r=-7$ ; both sides equal $4.2$ .

Full solution

The variable is multiplied by $-0.6$ , so divide both sides by that nonzero coefficient:
$r=\frac{4.2}{-0.6}$

A positive number divided by a negative number is negative, so
$r=-7$

Check in the original:
$-0.6(-7)=4.2$
Both sides equal $4.2$ , so the value solves the equation.

Answer

$r=-7$ ; both sides equal $4.2$ .

Key idea

Divide by the whole coefficient, sign included, and let the division rules settle the sign.
Problem 2: Mark complete
Problem 3 A quotient of $-8$

Find $t$ if $t\div(-\frac{3}{4})=-8$ , and check your value.
1. Hint 1
  The variable is divided by a nonzero fraction.
2. Hint 2
  Multiply both sides by that whole divisor to undo the division.
Answer

$t=6$ ; both sides equal $-8$ .

Full solution

Multiplying both sides by $-\frac{3}{4}$ undoes the division by that same nonzero number:
$t=(-8)(-\frac{3}{4})$

The product is positive, giving
$t=6$
Check by dividing the proposed value by the original divisor:
$6\div(-\frac{3}{4})=-8$

Answer

$t=6$ ; both sides equal $-8$ .

Key idea

To undo division by a fraction, multiply by that fraction on both sides.
Problem 3: Mark complete
Problem 4 An audio clip

An audio clip is played at $1.25$ times its normal speed. If its normal duration is $d$ seconds, its playback duration is $d\div1.25$ seconds. Playback lasts $36$ seconds. Find the normal duration.
1. Hint 1
  The given playback rule becomes an equation when its output is known.
2. Hint 2
  Undo division by $1.25$ on both sides.
Answer

$45$ seconds.

Full solution

Set the playback expression equal to the measured duration:
$d\div1.25=36$

Multiply both sides by $1.25$ to find the normal duration:
$d=36\cdot1.25$
Thus $d=45$ seconds.

Checking the playback, $45\div1.25=36$ seconds, matching the observation.

Answer

$45$ seconds.

Key idea

A known output lets an operation in a duration rule be undone to recover the input.
Problem 4: Mark complete
Problem 5 Flour in a container

After $\frac{3}{4}$ cup of flour is used, a container holds $\frac{5}{8}$ cup. Find how much flour it held before.
1. Hint 1
  The original amount minus the flour used equals the amount remaining.
2. Hint 2
  Add the used amount to both sides to recover the original amount.
Answer

$\frac{11}{8}$ cups, or $1\frac{3}{8}$ cups.

Full solution

Let $f$ be the original number of cups.
The equation is
$f-\frac{3}{4}=\frac{5}{8}$
Add $\frac{3}{4}$ , which is $\frac{6}{8}$ , to both sides:
$\begin{aligned}f&=\frac{5}{8}+\frac{6}{8}\\f&=\frac{11}{8}\end{aligned}$
which is $1\frac{3}{8}$ cups.

For a check, $\frac{11}{8}-\frac{6}{8}=\frac{5}{8}$ cup remains after $\frac{3}{4}$ cup is used, as the statement says.

Answer

$\frac{11}{8}$ cups, or $1\frac{3}{8}$ cups.

Key idea

Recover an original amount by adding back what was removed.
Problem 5: Mark complete
Problem 6 A sensor reading

A sensor displays the true temperature plus a fixed adjustment $e$ , with all temperatures in degrees Celsius. When the true temperature is $-2.8$ , the display reads $1.6$ . Find $e$ and the display reading when the true temperature is $5.2$ .
1. Hint 1
  The same adjustment is added to each true temperature.
2. Hint 2
  Use the known true temperature and its display to form an equation for $e$ .
Answer

Adjustment: $4.4$ degrees Celsius; display reading: $9.6$ degrees Celsius.

Full solution

The known measurement gives
$-2.8+e=1.6$
Add $2.8$ to both sides to isolate the adjustment:
$e=4.4$

For a true temperature of $5.2$ degrees Celsius, the display is
$5.2+4.4=9.6$

The adjustment checks against the earlier reading because
$-2.8+4.4=1.6.$

Answer

Adjustment: $4.4$ degrees Celsius; display reading: $9.6$ degrees Celsius.

Key idea

A fixed adjustment can be found from one known input and output and then applied to another input.
Problem 6: Mark complete
Problem 7 A diver's descent

A diver's height relative to the water surface changes by the same amount, $r$ meters, in each of $8$ minutes, and by $-14$ meters altogether. Write an equation for $r$ , solve it, and check it.
1. Hint 1
  Eight equal changes make up the total change.
2. Hint 2
  Undo the multiplication by $8$ on both sides.
Answer

$8r=-14$ , or an equivalent equation; $r=-1.75$ (or $-\frac{7}{4}$ ), a change of $-1.75$ meters each minute.

Full solution

Eight equal changes of $r$ meters give a total change of $8r$ meters, so
$8r=-14$

The variable is multiplied by $8$ , so divide both sides by the nonzero number $8$ :
$\begin{aligned}r&=\frac{-14}{8}\\r&=-\frac{7}{4}\\r&=-1.75\end{aligned}$
The height drops $1.75$ meters each minute.

Check in the equation:
$8(-1.75)=-14$
Eight changes of $-1.75$ meters make the stated total change.

Answer

$8r=-14$ , or an equivalent equation; $r=-1.75$ (or $-\frac{7}{4}$ ), a change of $-1.75$ meters each minute.

Key idea

Equal repeated changes give a product, which division undoes.
Problem 7: Mark complete
Problem 8 Omar's second equation

Omar solves $x+(-7)=-2$ by adding $7$ to both sides and gets $x=5$ . For $-7x=-2$ he again adds $7$ to both sides and reports $x=5$ once more. Is his second answer right? Find and check the solution of $-7x=-2$ .
1. Hint 1
  Substitute the reported value into $-7x=-2$ before anything else.
2. Hint 2
  In $-7x=-2$ the number $-7$ multiplies $x$ rather than being added to it, so a different inverse operation is needed.
Answer

No; $x=\frac{2}{7}$ ; both sides equal $-2$ .

Full solution

Substituting the reported value gives
$-7(5)=-35$
This is not $-2$ , so $x=5$ is not a solution of $-7x=-2$ .

Here $-7$ multiplies $x$ , so the inverse is division.
Divide both sides by the nonzero coefficient $-7$ :
$x=\frac{-2}{-7}$
A negative number divided by a negative number is positive, so
$x=\frac{2}{7}.$

Its check is
$-7(\frac{2}{7})=-2.$
Adding $7$ to both sides of $-7x=-2$ is allowed, but it gives $-7x+7=5$ , which does not isolate $x$ , so $x=5$ cannot be read from it.
Adding $7$ worked in the first equation because there $-7$ was added to $x$ , not multiplied by it.

Answer

No; $x=\frac{2}{7}$ ; both sides equal $-2$ .

Key idea

Choose an inverse operation from the relationship in an equation, not just its numbers.
Problem 8: Mark complete
Problem 9 Sam's working page

Sam changes $0.8x=-4$ to $1.6x=-8$ , then reports $x=-5$ . Is every step valid? Explain and check the reported value.
1. Hint 1
  A valid change must preserve the equality for exactly the same values of $x$ .
2. Hint 2
  Compare the factor applied to each side in the first step, then identify the coefficient in the new equation.
Answer

Yes; $x=-5$ ; the original sides both equal $-4$ .

Full solution

The first step doubles both sides.
Multiplication by the same nonzero number preserves the solution, even though it does not yet isolate the variable.

Dividing both sides of the new equation by the nonzero number $1.6$ gives
$x=\frac{-8}{1.6}$
Therefore $x=-5$ .

Check in the original equation:
$0.8(-5)=-4$
The result matches its right side, so the reported value is correct.

Answer

Yes; $x=-5$ ; the original sides both equal $-4$ .

Key idea

A valid balance move need not isolate the variable immediately.
Problem 9: Mark complete
Problem 10 Priya's claim

The equation is $ax=b$ , where $a$ is negative. Priya claims that its solution is positive when $b$ is negative and is zero when $b=0$ . Is she right in both cases? Explain without choosing specific values for $a$ .
1. Hint 1
  The coefficient is negative, so it is also nonzero.
2. Hint 2
  Consider the sign of $b\div a$ and then the case with a zero numerator.
Answer

Yes: positive for negative $b$ , and $x=0$ for $b=0$ .

Full solution

Since $a$ is nonzero, divide both sides by $a$ :
$x=\frac{b}{a}$

When $b$ is negative, this is a negative number divided by a negative number, so the solution is positive.
When $b=0$ , zero divided by nonzero $a$ is zero, so $x=0$ .

The zero case checks because $a\cdot0=0$ .
In the negative case, multiplying the quotient $b\div a$ by its nonzero divisor $a$ returns $b$ .

Answer

Yes: positive for negative $b$ , and $x=0$ for $b=0$ .

Key idea

Writing the solution of $ax=b$ as $b\div a$ , with $a$ nonzero, lets the division rules decide whether it is positive, negative or zero without choosing values.
Problem 10: Mark complete

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