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find the solution set 2x+(3x-(4x-5))=x/4-2 a.28/3 b.-28/3 c.9 d-9Asked by chevlios
Find the solution set
2x + [3x-(4x-5) ]=x/4-2
a.28/3
b.-28/3
c.9
d.-9
2x + [3x-(4x-5) ]=x/4-2
a.28/3
b.-28/3
c.9
d.-9
Answers
Answered by
Steve
OK. Now you show me yours, and I'll show you mine.
Just clear the fraction and get rid of the parens. Then it's clear sailing.
Just clear the fraction and get rid of the parens. Then it's clear sailing.
Answered by
chevlios
2x+3x-4x+5=x/2
x-5=x/2
-5=x/2-x
i don't know what next ,,,
x-5=x/2
-5=x/2-x
i don't know what next ,,,
Answered by
Steve
you must multiply everything by 4 to clear the fraction. Why did you continue on with another fraction? And your arithmetic is wrong, as well.
2x + [3x-(4x-5) ]=x/4-2
8x + 4[3x-(4x-5) ] = x-8
8x + 12x - 16x + 20 = x - 8
4x+20 = x-8
3x = -28
x = -28/3
2x + [3x-(4x-5) ]=x/4-2
8x + 4[3x-(4x-5) ] = x-8
8x + 12x - 16x + 20 = x - 8
4x+20 = x-8
3x = -28
x = -28/3
Answered by
chevlios
ok i get this tnx mr steve
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