What is expand 12((X-7)/4+(25(X-2))/3 =(5X+35)/4+5/2 (X-7)) ?

The solution to expand 12((X-7)/4+(25(X-2))/3 =(5X+35)/4+5/2 (X-7)) is 103X-221=(125X)/4+35/4-210

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expand 12((X-7)/4+(25(X-2))/3 =(5X+35)/4+5/2 (X-7))

expand

12 (\frac{X - 7}{4} + \frac{25 ( X - 2 )}{3} = \frac{5 X + 35}{4} + \frac{5}{2} (X - 7))

103 X - 221 = \frac{125 X}{4} + \frac{35}{4} - 210

Solution steps

12 (\frac{X - 7}{4} + \frac{25 ( X - 2 )}{3} = \frac{5 X + 35}{4} + \frac{5}{2} (X - 7))

= 12 (\frac{X - 7}{4} + = \frac{25 ( X - 2 )}{3} \cdot \frac{5 X + 35}{4} + \frac{5}{2} (X - 7))

Distribute parentheses

12 \cdot \frac{X - 7}{4} + 12 \cdot \frac{25 ( X - 2 )}{3} = \frac{5 X + 35}{4} + 12 \cdot \frac{5}{2} (X - 7)

Simplify

12 \cdot \frac{X - 7}{4} + 12 \cdot \frac{25 ( X - 2 )}{3} = \frac{5 X + 35}{4} + 12 \cdot \frac{5}{2} (X - 7) : 103 X - 221 = \frac{125 X}{4} + \frac{35}{4} - 210

103 X - 221 = \frac{125 X}{4} + \frac{35}{4} - 210

f (x) (\frac{2 ( x - 1 )}{3} + \frac{3 ( 2 x + 1 )}{9} = 9)

- \frac{1}{x ( x + 1 )}

(3 x - y^{2} + 1) d x

V (- 4, - 3) \cdot P (- 7, - 6)

\frac{t ^{2} - 1}{( t ^{2} + 1 ) ^{2}}

What is expand 12((X-7)/4+(25(X-2))/3 =(5X+35)/4+5/2 (X-7)) ?
The solution to expand 12((X-7)/4+(25(X-2))/3 =(5X+35)/4+5/2 (X-7)) is 103X-221=(125X)/4+35/4-210