How do you solve $\frac{4}{x - 3} = \frac{x + 5}{5}]$ $\frac { 4} { x - 3} = \frac { x + 5} { 5} ]$ ?

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Answer 1 · 2018-01-11T17:57:57

$x_{1} = 5$ $x_1=5$ and $x_{2} = - 7$ $x_2=-7$

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

Perform this
$\frac{4}{x - 3}$ $4/(x-3)$ χ $\frac{x + 5}{5}$ $(x+5)/5$

$\Leftrightarrow$ $<=>$ $4 \cdot 5 = (x - 3) (x + 5)$ $4*5=(x-3)(x+5)$ $\Leftrightarrow$ $<=>$

$20 = x^{2} + 5 x - 3 x - 15 \Leftrightarrow$ $20=x^2+5x-3x-15 <=>$

$x^{2} + 2 x - 35 = 0$ $x^2+2x-35=0$

Discriminant will be $Δ=b^2-4ac=4-4*1*(-35)=140+4=144$

$x_(1,2)=(-b+-sqrt(Δ))/(2α)$ so

$x_1=(-2+12)/2=5$

$x_2=(-2-12)/2=-14/2=-7$

As a result the roots are $x_1=5$ and $x_2=-7$

How do you solve \frac { 4} { x - 3} = \frac { x + 5} { 5} ]4x−3=x+55]\frac { 4} { x - 3} = \frac { x + 5} { 5} ]?