Calculate: (-7a)/(a^2-100)+(5)/(a-10)=(1)/(a+10)

Expression: $\frac{ -7a }{ {a}^{2}-100 }+\frac{ 5 }{ a-10 }=\frac{ 1 }{ a+10 }$

Determine the defined range

$\begin{array} { l }\frac{ -7a }{ {a}^{2}-100 }+\frac{ 5 }{ a-10 }=\frac{ 1 }{ a+10 },& \begin{array} { l }a≠-10,& a≠10\end{array}\end{array}$

Use $\frac{ -a }{ b }=\frac{ a }{ -b }=-\frac{ a }{ b }$ to rewrite the fraction

$-\frac{ 7a }{ {a}^{2}-100 }+\frac{ 5 }{ a-10 }=\frac{ 1 }{ a+10 }$

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

$-\frac{ 7a }{ {a}^{2}-100 }+\frac{ 5 }{ a-10 }-\frac{ 1 }{ a+10 }=0$

Use ${a}^{2}-{b}^{2}=\left( a-b \right)\left( a+b \right)$ to factor the expression

$-\frac{ 7a }{ \left( a-10 \right) \times \left( a+10 \right) }+\frac{ 5 }{ a-10 }-\frac{ 1 }{ a+10 }=0$

Write all numerators above the least common denominator $\left( a-10 \right) \times \left( a+10 \right)$

$\frac{ -7a+5\left( a+10 \right)-\left( a-10 \right) }{ \left( a-10 \right) \times \left( a+10 \right) }=0$

Distribute $5$ through the parentheses

$\frac{ -7a+5a+50-\left( a-10 \right) }{ \left( a-10 \right) \times \left( a+10 \right) }=0$

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

$\frac{ -7a+5a+50-a+10 }{ \left( a-10 \right) \times \left( a+10 \right) }=0$

Collect like terms

$\frac{ -3a+50+10 }{ \left( a-10 \right) \times \left( a+10 \right) }=0$

Add the numbers

$\frac{ -3a+60 }{ \left( a-10 \right) \times \left( a+10 \right) }=0$

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

$-3a+60=0$

Move the constant to the right-hand side and change its sign

$-3a=-60$

Divide both sides of the equation by $-3$

$\begin{array} { l }a=20,& \begin{array} { l }a≠-10,& a≠10\end{array}\end{array}$

Check if the solution is in the defined range

$a=20$