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3 years ago

-4|2w+6|= -32 how do you solve it

1 answer:

7 0

Isolate the absolute value, confirm the absolute value is not equal to a negative number, split the absolute value into 2 equations (one positive and one negative), then solve both equations.

-4|2w + 6| = -32

|2w + 6| = 8 <em>divided both sides by -4 (confirmed that 8 is positive)</em>

<u>positive:</u> <u>negative:</u>

2w + 6 = 8 -(2w + 6) = 8

2w = 2 <em>(subtracted 6 from both sides)</em> 2w + 6 = -8 (<em>divided both sides by -1)</em>

w = 1 <em>(divided both sides by 2) </em>2w = -14<em> (subtracted 6 from both sides)</em>

w = -7 <em>(divided both sides by 2)</em>

Answer: 1, -7

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The solution of the given equation is:

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<h2>Step-by-step explanation:</h2>

The expression is given by:

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Now on solving for the given equation

$\dfrac{-8}{2(y-4)}=\dfrac{5}{y+4}-\dfrac{7y+8}{(y-4)(y+4)}$

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$a^2-b^2=(a-b)(a+b)\\\\so,\\\\y^2-16=y^2-4^2\\\\i.e.\\\\y^2-16=(y-4)(y+4)$

Hence, we get:

$\dfrac{-4}{y-4}=\dfrac{5\times (y-4)-(7y+8)}{(y+4)(y-4)}\\\\i.e.\\\\\dfrac{-4}{y-4}=\dfrac{5y-20-7y-8}{(y-4)(y+4)}$

$-4\times (y-4)(y+4)=(-2y-28)(y-4)\\\\i.e.\\\\(y-4)(-4\times (y+4))=(-2y-28)(y-4)\\\\i.e.\\\\(y-4)(-4y+16)=(-2y-28)(y-4)\\\\i.e.$

$(y-4)(-4y+16)-(-2y-28)(y-4)=0\\\\i.e.\\\\(y-4)(-4y-16+2y+28)=0\\\\i.e.\\\\(y-4)(-2y+12)=0\\\\i.e.$

$y-4=0\ or\ -2y+12=0\\\\i.e.\\\\y=4\ or\ 2y=12\\\\i.e.\\\\y=4\ or\ y=6$

but y≠ 4

since, the denominator of the term in the left side of the given equality and the second term in the right side of the given equality will be zero and hence, the expression will be not defined.

Hence, the value of y is: 6

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