Solve the equation: $-4(3-x)=3(4x-6)$

Israel Hale
2022-08-01
Answered

Solve the equation: $-4(3-x)=3(4x-6)$

You can still ask an expert for help

Wayne Everett

Answered 2022-08-02
Author has **19** answers

Step 1

Given: $-4(3-x)=3(4x-6)$

Step 2

Now, $-12+4x=12x-18\phantom{\rule{0ex}{0ex}}4x-12x=-18+12\phantom{\rule{0ex}{0ex}}-8x=-6\phantom{\rule{0ex}{0ex}}8x=6\phantom{\rule{0ex}{0ex}}x=\frac{6}{8}\phantom{\rule{0ex}{0ex}}x=\frac{3}{4}$

Given: $-4(3-x)=3(4x-6)$

Step 2

Now, $-12+4x=12x-18\phantom{\rule{0ex}{0ex}}4x-12x=-18+12\phantom{\rule{0ex}{0ex}}-8x=-6\phantom{\rule{0ex}{0ex}}8x=6\phantom{\rule{0ex}{0ex}}x=\frac{6}{8}\phantom{\rule{0ex}{0ex}}x=\frac{3}{4}$

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Does there exist $a,b,c,d$ such that $\frac{a+b+c+d}{4}$ is an integer?

Let $a,b,c,d$ are different positive integers, such that a,b,c, nor d is a multiple of a different variable.

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Does there exist $a,b,c,d$ for

$\phantom{\rule{0ex}{0ex}}\frac{a+b+c+d}{4}=I\phantom{\rule{0ex}{0ex}}$

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Let $a,b,c,d$ are different positive integers, such that a,b,c, nor d is a multiple of a different variable.

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Does there exist $a,b,c,d$ for

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such that I is an integer?

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Solve

$\frac{x-2}{x+2}=\frac{5}{7}$

Select the correct choice below and, if necessary, fill in the answer box to complete your choice

A. The solution is x=?. (Simplify your answer ype an integerora fraction)

There is no solution.

$\frac{x-2}{x+2}=\frac{5}{7}$

Select the correct choice below and, if necessary, fill in the answer box to complete your choice

A. The solution is x=?. (Simplify your answer ype an integerora fraction)

There is no solution.