Calculate: (4)/(x-5)+(4)/(x-6)=(x^2-37)/(x^2-11x+30)

Expression: $\frac{ 4 }{ x-5 }+\frac{ 4 }{ x-6 }=\frac{ {x}^{2}-37 }{ {x}^{2}-11x+30 }$

Determine the defined range

$\begin{array} { l }\frac{ 4 }{ x-5 }+\frac{ 4 }{ x-6 }=\frac{ {x}^{2}-37 }{ {x}^{2}-11x+30 },& \begin{array} { l }x≠5,& x≠6\end{array}\end{array}$

Move the expression to the left-hand side and change its sign

$\frac{ 4 }{ x-5 }+\frac{ 4 }{ x-6 }-\frac{ {x}^{2}-37 }{ {x}^{2}-11x+30 }=0$

Write $-11x$ as a difference

$\frac{ 4 }{ x-5 }+\frac{ 4 }{ x-6 }-\frac{ {x}^{2}-37 }{ {x}^{2}-5x-6x+30 }=0$

Factor out $x$ from the expression

$\frac{ 4 }{ x-5 }+\frac{ 4 }{ x-6 }-\frac{ {x}^{2}-37 }{ x \times \left( x-5 \right)-6x+30 }=0$

Factor out $-6$ from the expression

$\frac{ 4 }{ x-5 }+\frac{ 4 }{ x-6 }-\frac{ {x}^{2}-37 }{ x \times \left( x-5 \right)-6\left( x-5 \right) }=0$

Factor out $x-5$ from the expression

$\frac{ 4 }{ x-5 }+\frac{ 4 }{ x-6 }-\frac{ {x}^{2}-37 }{ \left( x-5 \right) \times \left( x-6 \right) }=0$

Write all numerators above the least common denominator $\left( x-5 \right) \times \left( x-6 \right)$

$\frac{ 4\left( x-6 \right)+4\left( x-5 \right)-\left( {x}^{2}-37 \right) }{ \left( x-5 \right) \times \left( x-6 \right) }=0$

Distribute $4$ through the parentheses

$\frac{ 4x-24+4\left( x-5 \right)-\left( {x}^{2}-37 \right) }{ \left( x-5 \right) \times \left( x-6 \right) }=0$

Distribute $4$ through the parentheses

$\frac{ 4x-24+4x-20-\left( {x}^{2}-37 \right) }{ \left( x-5 \right) \times \left( x-6 \right) }=0$

When there is a $-$ in front of an expression in parentheses, change the sign of each term of the expression and remove the parentheses

$\frac{ 4x-24+4x-20-{x}^{2}+37 }{ \left( x-5 \right) \times \left( x-6 \right) }=0$

Collect like terms

$\frac{ 8x-24-20-{x}^{2}+37 }{ \left( x-5 \right) \times \left( x-6 \right) }=0$

Calculate the sum or difference

$\frac{ 8x-7-{x}^{2} }{ \left( x-5 \right) \times \left( x-6 \right) }=0$

When the quotient of expressions equals $0$, the numerator has to be $0$

$8x-7-{x}^{2}=0$

Use the commutative property to reorder the terms

$-{x}^{2}+8x-7=0$

Change the signs on both sides of the equation

${x}^{2}-8x+7=0$

Write $-8x$ as a difference

${x}^{2}-x-7x+7=0$

Factor out $x$ from the expression

$x \times \left( x-1 \right)-7x+7=0$

Factor out $-7$ from the expression

$x \times \left( x-1 \right)-7\left( x-1 \right)=0$

Factor out $x-1$ from the expression

$\left( x-1 \right) \times \left( x-7 \right)=0$

When the product of factors equals $0$, at least one factor is $0$

$\begin{array} { l }x-1=0,\\x-7=0\end{array}$

Solve the equation for $x$

$\begin{array} { l }x=1,\\x-7=0\end{array}$

Solve the equation for $x$

$\begin{array} { l }\begin{array} { l }x=1,\\x=7\end{array},& \begin{array} { l }x≠5,& x≠6\end{array}\end{array}$

Check if the solution is in the defined range

$\begin{array} { l }x=1,\\x=7\end{array}$

The equation has $2$ solutions

$\begin{array} { l }x_1=1,& x_2=7\end{array}$