Solve (b+1/5)^2-(b+1/5)(1-b/5)-(1-b/5)^2=frac{b^2}{25}+frac{3}{5}

$$( \frac { b + 1 } { 5 } ) ^ { 2 } - ( \frac { b + 1 } { 5 } ) ( \frac { 1 - b } { 5 } ) - ( \frac { 1 - b } { 5 } ) ^ { 2 } = \frac { b ^ { 2 } } { 25 } + \frac { 3 } { 5 }$$

Solve for b

$$b=4$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 25, the least common multiple of 5,25.

$$25\left(\left(\frac{b+1}{5}\right)^{2}-\frac{b+1}{5}\times \frac{1-b}{5}\right)-25\times \left(\frac{1-b}{5}\right)^{2}=b^{2}+15$$

To raise \frac{b+1}{5} to a power, raise both numerator and denominator to the power and then divide.

$$25\left(\frac{\left(b+1\right)^{2}}{5^{2}}-\frac{b+1}{5}\times \frac{1-b}{5}\right)-25\times \left(\frac{1-b}{5}\right)^{2}=b^{2}+15$$

Multiply \frac{b+1}{5} times \frac{1-b}{5} by multiplying numerator times numerator and denominator times denominator.

$$25\left(\frac{\left(b+1\right)^{2}}{5^{2}}-\frac{\left(b+1\right)\left(1-b\right)}{5\times 5}\right)-25\times \left(\frac{1-b}{5}\right)^{2}=b^{2}+15$$

Multiply 5 and 5 to get 25.

$$25\left(\frac{\left(b+1\right)^{2}}{5^{2}}-\frac{\left(b+1\right)\left(1-b\right)}{25}\right)-25\times \left(\frac{1-b}{5}\right)^{2}=b^{2}+15$$

To add or subtract expressions, expand them to make their denominators the same. Expand 5^{2}.

$$25\left(\frac{\left(b+1\right)^{2}}{25}-\frac{\left(b+1\right)\left(1-b\right)}{25}\right)-25\times \left(\frac{1-b}{5}\right)^{2}=b^{2}+15$$

Since \frac{\left(b+1\right)^{2}}{25} and \frac{\left(b+1\right)\left(1-b\right)}{25} have the same denominator, subtract them by subtracting their numerators.

$$25\times \frac{\left(b+1\right)^{2}-\left(b+1\right)\left(1-b\right)}{25}-25\times \left(\frac{1-b}{5}\right)^{2}=b^{2}+15$$

Do the multiplications in \left(b+1\right)^{2}-\left(b+1\right)\left(1-b\right).

$$25\times \frac{b^{2}+2b+1-b+b^{2}+b-1}{25}-25\times \left(\frac{1-b}{5}\right)^{2}=b^{2}+15$$

Combine like terms in b^{2}+2b+1-b+b^{2}+b-1.

$$25\times \frac{2b^{2}+2b}{25}-25\times \left(\frac{1-b}{5}\right)^{2}=b^{2}+15$$

Express 25\times \frac{2b^{2}+2b}{25} as a single fraction.

$$\frac{25\left(2b^{2}+2b\right)}{25}-25\times \left(\frac{1-b}{5}\right)^{2}=b^{2}+15$$

Cancel out 25 and 25.

$$2b^{2}+2b-25\times \left(\frac{1-b}{5}\right)^{2}=b^{2}+15$$

To raise \frac{1-b}{5} to a power, raise both numerator and denominator to the power and then divide.

$$2b^{2}+2b-25\times \frac{\left(1-b\right)^{2}}{5^{2}}=b^{2}+15$$

Express 25\times \frac{\left(1-b\right)^{2}}{5^{2}} as a single fraction.

$$2b^{2}+2b-\frac{25\left(1-b\right)^{2}}{5^{2}}=b^{2}+15$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-b\right)^{2}.

$$2b^{2}+2b-\frac{25\left(1-2b+b^{2}\right)}{5^{2}}=b^{2}+15$$

Calculate 5 to the power of 2 and get 25.

$$2b^{2}+2b-\frac{25\left(1-2b+b^{2}\right)}{25}=b^{2}+15$$

Cancel out 25 and 25.

$$2b^{2}+2b-\left(1-2b+b^{2}\right)=b^{2}+15$$

To find the opposite of 1-2b+b^{2}, find the opposite of each term.

$$2b^{2}+2b-1+2b-b^{2}=b^{2}+15$$

Combine 2b and 2b to get 4b.

$$2b^{2}+4b-1-b^{2}=b^{2}+15$$

Combine 2b^{2} and -b^{2} to get b^{2}.

$$b^{2}+4b-1=b^{2}+15$$

Subtract b^{2} from both sides.

$$b^{2}+4b-1-b^{2}=15$$

Combine b^{2} and -b^{2} to get 0.

$$4b-1=15$$

Add 1 to both sides.

$$4b=15+1$$

Add 15 and 1 to get 16.

$$4b=16$$

Divide both sides by 4.

$$b=\frac{16}{4}$$

Divide 16 by 4 to get 4.

$$b=4$$