Solve 1-b/frac{b{5}}+b/frac{6{5}-b/5(5b+1)}=1

$$\frac { 1 - b } { \frac { b } { 5 } } + \frac { b } { \frac { 6 } { 5 } - \frac { b } { 5 ( 5 b + 1 ) } } = 1$$

Solve for b

$$b=\frac{19-\sqrt{861}}{50}\approx -0.20685603$$

$$b=6$$

$$b=\frac{\sqrt{861}+19}{50}\approx 0.96685603$$

Solution Steps

Divide 1-b by \frac{b}{5} by multiplying 1-b by the reciprocal of \frac{b}{5}.

$$\frac{\left(1-b\right)\times 5}{b}+\frac{b}{\frac{6}{5}-\frac{b}{5\left(5b+1\right)}}=1$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 5 and 5\left(5b+1\right) is 5\left(5b+1\right). Multiply \frac{6}{5} times \frac{5b+1}{5b+1}.

$$\frac{\left(1-b\right)\times 5}{b}+\frac{b}{\frac{6\left(5b+1\right)}{5\left(5b+1\right)}-\frac{b}{5\left(5b+1\right)}}=1$$

Since \frac{6\left(5b+1\right)}{5\left(5b+1\right)} and \frac{b}{5\left(5b+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(1-b\right)\times 5}{b}+\frac{b}{\frac{6\left(5b+1\right)-b}{5\left(5b+1\right)}}=1$$

Do the multiplications in 6\left(5b+1\right)-b.

$$\frac{\left(1-b\right)\times 5}{b}+\frac{b}{\frac{30b+6-b}{5\left(5b+1\right)}}=1$$

Combine like terms in 30b+6-b.

$$\frac{\left(1-b\right)\times 5}{b}+\frac{b}{\frac{29b+6}{5\left(5b+1\right)}}=1$$

Variable b cannot be equal to -\frac{1}{5} since division by zero is not defined. Divide b by \frac{29b+6}{5\left(5b+1\right)} by multiplying b by the reciprocal of \frac{29b+6}{5\left(5b+1\right)}.

$$\frac{\left(1-b\right)\times 5}{b}+\frac{b\times 5\left(5b+1\right)}{29b+6}=1$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of b and 29b+6 is b\left(29b+6\right). Multiply \frac{\left(1-b\right)\times 5}{b} times \frac{29b+6}{29b+6}. Multiply \frac{b\times 5\left(5b+1\right)}{29b+6} times \frac{b}{b}.

$$\frac{\left(1-b\right)\times 5\left(29b+6\right)}{b\left(29b+6\right)}+\frac{b\times 5\left(5b+1\right)b}{b\left(29b+6\right)}=1$$

Since \frac{\left(1-b\right)\times 5\left(29b+6\right)}{b\left(29b+6\right)} and \frac{b\times 5\left(5b+1\right)b}{b\left(29b+6\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(1-b\right)\times 5\left(29b+6\right)+b\times 5\left(5b+1\right)b}{b\left(29b+6\right)}=1$$

Do the multiplications in \left(1-b\right)\times 5\left(29b+6\right)+b\times 5\left(5b+1\right)b.

$$\frac{145b+30-145b^{2}-30b+25b^{3}+5b^{2}}{b\left(29b+6\right)}=1$$

Combine like terms in 145b+30-145b^{2}-30b+25b^{3}+5b^{2}.

$$\frac{115b+30-140b^{2}+25b^{3}}{b\left(29b+6\right)}=1$$

Use the distributive property to multiply b by 29b+6.

$$\frac{115b+30-140b^{2}+25b^{3}}{29b^{2}+6b}=1$$

Variable b cannot be equal to any of the values -\frac{6}{29},0 since division by zero is not defined. Multiply both sides of the equation by b\left(29b+6\right).

$$115b+30-140b^{2}+25b^{3}=b\left(29b+6\right)$$

Use the distributive property to multiply b by 29b+6.

$$115b+30-140b^{2}+25b^{3}=29b^{2}+6b$$

Subtract 29b^{2} from both sides.

$$115b+30-140b^{2}+25b^{3}-29b^{2}=6b$$

Combine -140b^{2} and -29b^{2} to get -169b^{2}.

$$115b+30-169b^{2}+25b^{3}=6b$$

Subtract 6b from both sides.

$$115b+30-169b^{2}+25b^{3}-6b=0$$

Combine 115b and -6b to get 109b.

$$109b+30-169b^{2}+25b^{3}=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$25b^{3}-169b^{2}+109b+30=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 30 and q divides the leading coefficient 25. List all candidates \frac{p}{q}.

$$±\frac{6}{5},±6,±30,±\frac{3}{5},±3,±15,±\frac{2}{5},±2,±10,±\frac{6}{25},±\frac{1}{5},±1,±5,±\frac{3}{25},±\frac{2}{25},±\frac{1}{25}$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$b=6$$

By Factor theorem, b-k is a factor of the polynomial for each root k. Divide 25b^{3}-169b^{2}+109b+30 by b-6 to get 25b^{2}-19b-5. Solve the equation where the result equals to 0.

$$25b^{2}-19b-5=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 25 for a, -19 for b, and -5 for c in the quadratic formula.

$$b=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 25\left(-5\right)}}{2\times 25}$$

Do the calculations.

$$b=\frac{19±\sqrt{861}}{50}$$

Solve the equation 25b^{2}-19b-5=0 when ± is plus and when ± is minus.

$$b=\frac{19-\sqrt{861}}{50} b=\frac{\sqrt{861}+19}{50}$$

List all found solutions.

$$b=6 b=\frac{19-\sqrt{861}}{50} b=\frac{\sqrt{861}+19}{50}$$