Solve Equations using the Distributive Property 2

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Question 1 of 4

1. Question
Solve

$-7(3-y)=-28$
- $y=$ (-1)
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Inverse Operations

When moving a term to the other side of an equation, the operation is inversed.

Distributive Property

$a$ $(b+c)=$ $a$ $b+$ $a$ $c$

To solve for $y$ , it needs to be alone on one side.

Start by expanding the left side of the equation by using the Distributive Property.

$-7$ $(3-$ $y$ $)$ $=$ $-28$

$-7$ $(3)-($ $-7$ $)($ $y$ $)$ $=$ $-28$

$-21+7$ $y$ $=$ $-28$

Next, move $-21$ to the other side by adding $21$ to both sides of the equation.

$-21+7$ $y$ $=$ $-28$

$-21+7$ $y$ $+21$ $=$ $-28$ $+21$

$7$ $y$ $=$ $-7$ $-21+21$ cancels out

Finally, remove $7$ by dividing both sides of the equation by $7$ .

$7$ $y$ $=$ $-7$

$7$ $y$ $divide7$ $=$ $-7$ $divide7$

$y$ $=$ $-1$ $7divide7$ cancels out

Check our work

To confirm our answer, substitute $y=-1$ to the original equation.

$-7(3-y)$ $=$ $-28$

$-7[3-(-1)]$ $=$ $-28$ Substitute $y=-1$

$-7(3+1)$ $=$ $-28$

$-7(4)$ $=$ $-28$

$-28$ $=$ $-28$

Since the equation is true, the answer is correct.

$y=-1$
Question 2 of 4

2. Question
Solve for $x$

$5-(3-2x)=10$
- $x=$ (4)
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To solve for $x$ , get $x$ by itself

Subtract $5$ from both sides of the equation

$5-(3-2x)$ $-5$ $=$ $10$ $-5$

$5$ $-(3-2x)$ $-5$ $=$ $10$ $-5$ $5-5$ cancels out

$-3+2x$ $=$ $5$ Distribute negative sign

Next, add $3$ to both sides of the equation

$-3+2x$ $+3$ $=$ $5$ $+3$

$-3$ $+2x$ $+3$ $=$ $5$ $+3$ $-3+3$ cancels out

$2x$ $=$ $8$

Finally, divide both sides of the equation by $2$

$2x$ $divide2$ $=$ $8$ $divide2$

$2$ $x$ $divide2$ $=$ $8$ $divide2$ $times2divide2$ cancels out

$x$ $=$ $4$

$x=4$

Question 3 of 4

Solve for

$k$

$-9(-8+k)=252$

$k=$ (-20)

Question 4 of 4

Solve for

$a$

$-12(a+11)=-108$

$a=$ (-2)

$-9$ $(-8+k)$	$=$	$252$
$((-8)xx$ $(-9)$ $)+(kxx$ $(-9)$ $)$	$=$	$252$
$72-9k$	$=$	$252$

$72-9k$ $-72$	$=$	$252$ $-72$	Subtract $72$ from both sides
$72$ $-9k$ $-72$	$=$	$252$ $-72$	$72-72$ cancels out
$-9k$	$=$	$180$
$-9k$ $-:(-9)$	$=$	$180$ $-:(-9)$	Divide both sides by $-9$
$-9$ $k$ $-:(-9)$	$=$	$180$ $-:(-9)$	$times(-9)divide(-9)$ cancels out
$k$	$=$	$-20$

$-12$ $(a+11)$	$=$	$-108$
$(axx$ $(-12)$ $)+(11xx$ $(-12)$ $)$	$=$	$-108$
$-12a-132$	$=$	$-108$

$-12a-132$ $+132$	$=$	$-108$ $+132$	Add $132$ to both sides
$-12a$ $-132$ $+132$	$=$	$-108$ $+132$	$-132+132$ cancels out
$-12a$	$=$	$24$
$-12a$ $divide(-12)$	$=$	$24$ $divide(-12)$	Divide both sides by $-12$
$-12$ $a$ $divide(-12)$	$=$	$24$ $divide(-12)$	$times(-12)divide(-12)$ cancels out
$a$	$=$	$-2$

Solve Equations using the Distributive Property 2

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Quiz summary

Information

1. Question

Hint

Great Work!

Inverse Operations

Distributive Property

2. Question

Hint

Correct!

3. Question

Fantastic!

4. Question

Exceptional!

Quizzes

$-7$ $(3-$ $y$ $)$	$=$	$-28$
$-7$ $(3)-($ $-7$ $)($ $y$ $)$	$=$	$-28$
$-21+7$ $y$	$=$	$-28$

$-21+7$ $y$	$=$	$-28$
$-21+7$ $y$ $+21$	$=$	$-28$ $+21$
$7$ $y$	$=$	$-7$	$-21+21$ cancels out

$7$ $y$	$=$	$-7$
$7$ $y$ $divide7$	$=$	$-7$ $divide7$
$y$	$=$	$-1$	$7divide7$ cancels out

$-7(3-y)$	$=$	$-28$
$-7[3-(-1)]$	$=$	$-28$	Substitute $y=-1$
$-7(3+1)$	$=$	$-28$
$-7(4)$	$=$	$-28$
$-28$	$=$	$-28$

$5-(3-2x)$ $-5$	$=$	$10$ $-5$
$5$ $-(3-2x)$ $-5$	$=$	$10$ $-5$	$5-5$ cancels out
$-3+2x$	$=$	$5$	Distribute negative sign

$-3+2x$ $+3$	$=$	$5$ $+3$
$-3$ $+2x$ $+3$	$=$	$5$ $+3$	$-3+3$ cancels out
$2x$	$=$	$8$

$2x$ $divide2$	$=$	$8$ $divide2$
$2$ $x$ $divide2$	$=$	$8$ $divide2$	$times2divide2$ cancels out
$x$	$=$	$4$