Solve Equations with Variables on Both Sides using the Distributive Property 3

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Question 1 of 4

Solve

$-4(2y-3)=1/2(12-4y)$

$y=$ (1)

Question 2 of 4

Solve

$2/3(n-5)=3/4(2n+10)$

$n=$ (-13)

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Inverse Operations

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

Distributive Property

$a$

$(b+c)=$ $a$

$b+$ $a$

$c$

Get $n$ alone to the left side and all constants to the right.

First, remove the fractions by finding the Least Common Denominator

$(LCD)$ of the denominators

$4$ and

$3$ .

Multiples of

$3$ :

$3 6 9$ $12$

$15$

Multiples of

$4$ :

$4 8$ $12$

$16 20$

The

$LCD$ of

$3$ and

$4$ is $12$

Multiply the

$LCD$ to both sides of the equation to remove the fractions. Use the Distributive Property.

$\frac{2}{3}(\color{#00880A}{n}-5)\color{#D800AD}{\times12}$	$=$	$\frac{3}{4}(2\color{#00880A}{n}+10)\color{#D800AD}{\times12}$

$\frac{24}{3}(\color{#00880A}{n}-5)$	$=$	$\frac{36}{4}(2\color{#00880A}{n}+10)$

$8$ $($ $n$ $-5)$	$=$	$9$ $(2$ $n$ $+10)$
$8$ $n$ $+$ $8$ $(-5)$	$=$	$9$ $(2$ $n$ $)+$ $9$ $(10)$
$8$ $n$ $-40$	$=$	$18$ $n$ $+90$	Simplify

Next, move

$8n$ to the other side by subtracting

$8n$ from both sides of the equation.

$8$ $n$ $-40$	$=$	$18$ $n$ $+90$
$8$ $n$ $-40$ $-8n$	$=$	$18$ $n$ $+90$ $-8n$
$-40$	$=$	$10$ $n$ $+90$	$8n-8n$ cancels out

Now, move

$90$ to the other side by subtracting

$90$ from both sides of the equation.

$-40$	$=$	$10$ $n$ $+90$
$-40$ $-90$	$=$	$10$ $n$ $+90$ $-90$
$-130$	$=$	$10$ $n$	$90-90$ cancels out

Finally, remove

$10$ by dividing both sides of the equation by

$10$ .

$-130$	$=$	$10$ $n$
$-130$ $divide10$	$=$	$10$ $n$ $divide10$
$-13$	$=$	$n$	$10divide10$ cancels out
$n$	$=$	$-13$

Check our work

To confirm our answer, substitute

$n=-13$ to the original equation.

$2/3(n-5)$	$=$	$3/4(2n+10)$

$2/3(-13-5)$	$=$	$3/4(2(-13)+10)$	Substitute $n=-13$

$2/3(-18)$	$=$	$3/4(-26+10)$

$(-36)/3$	$=$	$3/4(-16)$

$-12$	$=$	$(-48)/4$

$-12$	$=$	$-12$

Since the equation is true, the answer is correct.

$n=-13$

Question 3 of 4

Solve

$(3(p-3))/4=(2(2p-7))/5$

$p=$ (11)

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Inverse Operations

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

Distributive Property

$a$

$(b+c)=$ $a$

$b+$ $a$

$c$

Get $p$ alone to the left side and all constants to the right.

First, remove the fractions by finding the Least Common Denominator

$(LCD)$ of the denominators

$4$ and

$5$ .

Multiples of

$4$ :

$4 8 12 16$ $20$

Multiples of

$5$ :

$5 10 15$ $20$

$25$

The

$LCD$ of

$4$ and

$5$ is $20$

Multiply the

$LCD$ to both sides of the equation to remove the fractions. Use the Distributive Property.

$\left[\frac{3(\color{#00880A}{p}-3)}{4}\right]\color{#D800AD}{\times20}$	$=$	$\left[\frac{2(2\color{#00880A}{p}-7)}{5}\right]\color{#D800AD}{\times20}$

$\frac{20\times3}{4}(\color{#00880A}{p}-3)$	$=$	$\frac{20\times2}{5}(2\color{#00880A}{p}-7)$

$\frac{60}{4}(\color{#00880A}{p}-3)$	$=$	$\frac{40}{5}(2\color{#00880A}{p}-7)$

$15$ $($ $p$ $-3)$	$=$	$8$ $(2$ $p$ $-7)$
$15$ $p$ $+$ $15$ $(-3)$	$=$	$8$ $(2$ $p$ $)+$ $8$ $(-7)$
$15$ $p$ $-45$	$=$	$16$ $p$ $-56$	Simplify

Next, move

$15p$ to the other side by subtracting

$15p$ from both sides of the equation.

$15$ $p$ $-45$	$=$	$16$ $p$ $-56$
$15$ $p$ $-45$ $-15p$	$=$	$16$ $p$ $-56$ $-15p$
$-45$	$=$	$p$ $-56$	$15p-15p$ cancels out

Finally, move

$-56$ to the other side by adding

$56$ to both sides of the equation.

$-45$	$=$	$p$ $-56$
$-45$ $+56$	$=$	$p$ $-56$ $+56$
$11$	$=$	$p$	$-56+56$ cancels out
$p$	$=$	$11$

Check our work

To confirm our answer, substitute

$p=11$ to the original equation.

$(3(p-3))/4$	$=$	$(2(2p-7))/5$

$(3(11-3))/4$	$=$	$(2(2(11)-7))/5$	Substitute $p=11$

$(3(8))/4$	$=$	$(2(22-7))/5$

$24/4$	$=$	$(2(15))/5$

$24/4$	$=$	$30/5$

$6$	$=$	$6$

Since the equation is true, the answer is correct.

$p=11$

Question 4 of 4

Solve for

$x$

$(2x+5)/4 = (x-1)/3$

$x=$ (-19/2)

$-4$ $(2$ $y$ $-3)$	$=$	$1/2$ $(12-4$ $y$ $)$

$-4$ $(2$ $y$ $)$ $-4$ $(-3)$	$=$	$1/2$ $(12)+$ $1/2$ $(-4$ $y$ $)$

$-8$ $y$ $+12$	$=$	$6-2$ $y$

$-8$ $y$ $+12$	$=$	$6-2$ $y$
$-8$ $y$ $+12$ $+2y$	$=$	$6-2$ $y$ $+2y$
$-6$ $y$ $+12$	$=$	$6$	$-2y+2y$ cancels out

$-6$ $y$ $+12$	$=$	$6$
$-6$ $y$ $+12$ $-12$	$=$	$6$ $-12$
$-6$ $y$	$=$	$-6$	$12-12$ cancels out

$-6$ $y$	$=$	$-6$
$-6$ $y$ $divide-6$	$=$	$-6$ $divide-6$
$y$	$=$	$1$	$6divide6$ cancels out

$-4(2y-3)$	$=$	$1/2(12-4y)$

$-4[2(1)-3]$	$=$	$1/2[12-4(1)]$	Substitute $y=1$

$-4(2-3)$	$=$	$1/2(12-4)$

$-4(-1)$	$=$	$1/2(8)$

$4$	$=$	$4$

Solve Equations with Variables on Both Sides using the Distributive Property 3

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Quiz summary

Information

1. Question

Hint

Fantastic!

Inverse Operations

Distributive Property

2. Question

Hint

Correct!

Inverse Operations

Distributive Property

3. Question

Hint

Great Work!

Inverse Operations

Distributive Property

4. Question

Hint

Well Done!

Quizzes

$frac{2x+5}{4}$ $times4$	$=$	$frac{x-1}{3}$ $times4$

$frac{4(2x+5)}{4}$	$=$	$frac{x-1}{3}$ $times4$

$2x+5$	$=$	$frac{4(x-1)}{3}$	The coefficient $frac{4}{4}$ cancels out

$2x+5$	$=$	$frac{(x xx 4)-(1xx4)}{3}$	Distribute the value inside the parenthesis

$2x+5$	$=$	$frac{4x-4}{3}$

$(2x+5)$ $times3$	$=$	$frac{4x-4}{3}$ $times3$

$(2x+5)$ $times3$	$=$	$frac{3(4x-4)}{3}$

$3(2x+5)$	$=$	$4x-4$	The coefficient $frac{3}{3}$ cancels out
$(2x xx 3)+(5xx3)$	$=$	$4x-4$	Distribute the value inside the parenthesis
$6x+15$	$=$	$4x-4$

$6x+15$ $-4x$	$=$	$4x-4$ $-4x$
$6x+15$ $-4x$	$=$	$4x$ $-4$ $-4x$	$4x-4x$ cancels out
$2x+15$	$=$	$-4$

$2x+15$ $-15$	$=$	$-4$ $-15$
$2x$ $+15$ $-15$	$=$	$-4$ $-15$	$15-15$ cancels out
$2x$	$=$	$-19$

$2x$ $divide2$	$=$	$-19$ $divide2$
$2$ $x$ $divide2$	$=$	$-19$ $divide2$	$times2divide2$ cancels out

$x$	$=$	$-frac{19}{2}$