What is expand 3/4 (2x+10=2(3-2x)+7) ?

The solution to expand 3/4 (2x+10=2(3-2x)+7) is (3x)/2+15/2 =-4x+45/4

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expand 3/4 (2x+10=2(3-2x)+7)

expand

\frac{3}{4} (2 x + 10 = 2 (3 - 2 x) + 7)

\frac{3 x}{2} + \frac{15}{2} = - 4 x + \frac{45}{4}

Solution steps

\frac{3}{4} (2 x + 10 = 2 (3 - 2 x) + 7)

= \frac{3}{4} (2 x + 10 \cdot 2 = (3 - 2 x) + 7)

Distribute parentheses

\frac{3}{4} \cdot 2 x + \frac{3}{4} \cdot 10 = 2 (3 - 2 x) + \frac{3}{4} \cdot 7

2 \cdot \frac{3}{4} x + 10 \cdot \frac{3}{4} = 2 (3 - 2 x) + 7 \cdot \frac{3}{4}

Simplify

2 \cdot \frac{3}{4} x + 10 \cdot \frac{3}{4} = 2 (3 - 2 x) + 7 \cdot \frac{3}{4} : \frac{3 x}{2} + \frac{15}{2} = - 4 x + \frac{45}{4}

\frac{3 x}{2} + \frac{15}{2} = - 4 x + \frac{45}{4}

\frac{n ( 5 n - 7 )}{( n + 1 ) ^{2}}

\frac{K}{s ^{2} \cdot ( s t + 1 )}

(e^{t c o s (x)} (t) - e^{t s i n (x)} (t))^{2}

2 ((3, - 4, 5, 0, 0) - 5 (4, 8, - 2, - 1, 3) + (- 4, 7, - 5))

\frac{sec ( x )}{2 sin ( x )}

What is expand 3/4 (2x+10=2(3-2x)+7) ?
The solution to expand 3/4 (2x+10=2(3-2x)+7) is (3x)/2+15/2 =-4x+45/4