How do you solve 49−[3+10y−4(y+2)]=−5(3y−6)−[5(y−1)−9y+21]?

2 Answers
G_Ozdilek
Jul 16, 2018

Arrange the equation

Explanation:

49−3−10y+4y+8=−15y+30−5y+5+9y−21

after getting rid of parantheses:

49−3+8−30−5+21=−15y−5y+9y+10y−4y

Now combine numbers on one side and y terms on the other:

40=−24y+19y

40=−5y

y=−405=−8

Your answer is y=−8

Tony B
Jul 16, 2018

y=−8

Explanation:

Given: 49−[3+10y−4(y+2)]=−5(3y−6)−[5(y−1)−9y+21]
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider −[3+10y−4(y+2)]

Multiply inside the square brackets by −1

−3−10y+4(y+2)

−3−10y+4y+8

−3+8−10y+4y

dd+5dddd−6y

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider −5(3y−6)

Multiply everything inside the brackets by −5

−15y+30
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider −[5(y−1)−9y+21]

Multiply inside the square brackets by −1

−5(y−1)+9y−21

−5y+5+9y−21

+5−21−5y+9y

d−16dddd+4y
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Putting it all back together

49+5−6y=−15y+30−16+4y

54−6y=−11y+14

11y−6y=14−54

5y=−40

y=−8

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