How do you solve $\frac{1}{2} (x - 3) + \frac{3}{2 - x} = 5 x$ $1/2(x-3)+3/(2-x)=5x$ ?

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Answer 1 · 2015-12-01T03:37:15

$x = 0, \frac{15}{9}$ $x=0,15/9$

$\frac{x - 3}{2} + \frac{3}{2 - x} = 5 x$ $(x-3)/2+3/(2-x)=5x$

Achieve a common denominator of $2 (2 - x)$ $2(2-x)$ .

$\frac{x - 3}{2} (\frac{2 - x}{2 - x}) + (\frac{3}{2 - x}) (\frac{2}{2}) = 5 x$ $(x-3)/2((2-x)/(2-x))+(3/(2-x))(2/2)=5x$

$\frac{(- x^{2} + 2 x + 3 x - 6) + 6}{4 - 2 x} = 5 x$ $((-x^2+2x+3x-6)+6)/(4-2x)=5x$

$- x^{2} + 5 x = 5 x (4 - 2 x)$ $-x^2+5x=5x(4-2x)$

$- x^{2} + 5 x = - 10 x^{2} + 20 x$ $-x^2+5x=-10x^2+20x$

$9 x^{2} - 15 x = 0$ $9x^2-15x=0$

$x (9 x - 15) = 0$ $x(9x-15)=0$

${(x=0),(9x-15=0rarrx=15/9):}$

Check to make sure that neither answer will cause a denominator to be $0$ .

How do you solve 1/2(x-3)+3/(2-x)=5x12(x−3)+32−x=5x1/2(x-3)+3/(2-x)=5x?