Solve 1/a+1/a+b=5/8:A=?

$$\frac { 1 } { a } + \frac { 1 } { a + b } = \frac { 5 } { 8 } : A = ?$$

Solve for A

$$A=\frac{5a\left(a+b\right)}{8\left(2a+b\right)}$$

$$a\neq -b\text{ and }a\neq 0\text{ and }a\neq -\frac{b}{2}$$

Steps for Solving Linear Equation

Variable A cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by Aa\left(a+b\right), the least common multiple of a,a+b,A.

$$A\left(a+b\right)+Aa=a\left(a+b\right)\times \frac{5}{8}$$

Use the distributive property to multiply A by a+b.

$$Aa+Ab+Aa=a\left(a+b\right)\times \frac{5}{8}$$

Combine Aa and Aa to get 2Aa.

$$2Aa+Ab=a\left(a+b\right)\times \frac{5}{8}$$

Use the distributive property to multiply a by a+b.

$$2Aa+Ab=\left(a^{2}+ab\right)\times \frac{5}{8}$$

Use the distributive property to multiply a^{2}+ab by \frac{5}{8}.

$$2Aa+Ab=\frac{5}{8}a^{2}+\frac{5}{8}ab$$

Combine all terms containing A.

$$\left(2a+b\right)A=\frac{5}{8}a^{2}+\frac{5}{8}ab$$

The equation is in standard form.

$$\left(2a+b\right)A=\frac{5a^{2}+5ab}{8}$$

Divide both sides by 2a+b.

$$\frac{\left(2a+b\right)A}{2a+b}=\frac{5a\left(a+b\right)}{8\left(2a+b\right)}$$

Dividing by 2a+b undoes the multiplication by 2a+b.

$$A=\frac{5a\left(a+b\right)}{8\left(2a+b\right)}$$

Variable A cannot be equal to 0.

$$A=\frac{5a\left(a+b\right)}{8\left(2a+b\right)}\text{, }A\neq 0$$

Solve for a

$$\left\{\begin{matrix}a=\frac{\sqrt{256A^{2}+25b^{2}}}{10}+\frac{8A}{5}-\frac{b}{2}\text{, }&\left(b\neq 0\text{ and }A\neq 0\right)\text{ or }A>0\\a=-\frac{\sqrt{256A^{2}+25b^{2}}}{10}+\frac{8A}{5}-\frac{b}{2}\text{, }&\left(b\neq 0\text{ and }A\neq 0\right)\text{ or }A<0\end{matrix}\right.$$