7/8(-32-48x)+36=2/3(-33x-18)10x

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Solution for 7/8(-32-48x)+36=2/3(-33x-18)10x equation:



7/8(-32-48x)+36=2/3(-33x-18)10x
We move all terms to the left:
7/8(-32-48x)+36-(2/3(-33x-18)10x)=0
Domain of the equation: 8(-32-48x)!=0
x∈R
Domain of the equation: 3(-33x-18)10x)!=0
x∈R
We add all the numbers together, and all the variables
7/8(-48x-32)-(2/3(-33x-18)10x)+36=0
We calculate fractions
(21x(-)/(8(-48x-32)*3(-33x-18)10x))+(-16x0/(8(-48x-32)*3(-33x-18)10x))+36=0
We calculate terms in parentheses: +(21x(-)/(8(-48x-32)*3(-33x-18)10x)), so:
21x(-)/(8(-48x-32)*3(-33x-18)10x)
We add all the numbers together, and all the variables
21x0/(8(-48x-32)*3(-33x-18)10x)
We multiply all the terms by the denominator
21x0
We add all the numbers together, and all the variables
21x
Back to the equation:
+(21x)
We calculate terms in parentheses: +(-16x0/(8(-48x-32)*3(-33x-18)10x)), so:
-16x0/(8(-48x-32)*3(-33x-18)10x)
We multiply all the terms by the denominator
-16x0
We add all the numbers together, and all the variables
-16x
Back to the equation:
+(-16x)
We get rid of parentheses
21x-16x+36=0
We add all the numbers together, and all the variables
5x+36=0
We move all terms containing x to the left, all other terms to the right
5x=-36
x=-36/5
x=-7+1/5

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