Solve (8/5+1/3):x=x*(1+14/15)

$$( \frac { 8 } { 5 } + \frac { 1 } { 3 } ) : x = x \cdot ( 1 + \frac { 14 } { 15 } )$$

Solve for x

$$x=1$$

$$x=-1$$

Steps by Finding Square Root

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 15x, the least common multiple of x,15.

$$15\left(\frac{8}{5}+\frac{1}{3}\right)=x\left(1+\frac{14}{15}\right)\times 15x$$

Least common multiple of 5 and 3 is 15. Convert \frac{8}{5} and \frac{1}{3} to fractions with denominator 15.

$$15\left(\frac{24}{15}+\frac{5}{15}\right)=x\left(1+\frac{14}{15}\right)\times 15x$$

Since \frac{24}{15} and \frac{5}{15} have the same denominator, add them by adding their numerators.

$$15\times \frac{24+5}{15}=x\left(1+\frac{14}{15}\right)\times 15x$$

Add 24 and 5 to get 29.

$$15\times \frac{29}{15}=x\left(1+\frac{14}{15}\right)\times 15x$$

Cancel out 15 and 15.

$$29=x\left(1+\frac{14}{15}\right)\times 15x$$

Multiply x and x to get x^{2}.

$$29=x^{2}\left(1+\frac{14}{15}\right)\times 15$$

Convert 1 to fraction \frac{15}{15}.

$$29=x^{2}\left(\frac{15}{15}+\frac{14}{15}\right)\times 15$$

Since \frac{15}{15} and \frac{14}{15} have the same denominator, add them by adding their numerators.

$$29=x^{2}\times \frac{15+14}{15}\times 15$$

Add 15 and 14 to get 29.

$$29=x^{2}\times \frac{29}{15}\times 15$$

Cancel out 15 and 15.

$$29=x^{2}\times 29$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}\times 29=29$$

Divide both sides by 29.

$$x^{2}=\frac{29}{29}$$

Divide 29 by 29 to get 1.

$$x^{2}=1$$

Take the square root of both sides of the equation.

$$x=1 x=-1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 15x, the least common multiple of x,15.

$$15\left(\frac{8}{5}+\frac{1}{3}\right)=x\left(1+\frac{14}{15}\right)\times 15x$$

Least common multiple of 5 and 3 is 15. Convert \frac{8}{5} and \frac{1}{3} to fractions with denominator 15.

$$15\left(\frac{24}{15}+\frac{5}{15}\right)=x\left(1+\frac{14}{15}\right)\times 15x$$

Since \frac{24}{15} and \frac{5}{15} have the same denominator, add them by adding their numerators.

$$15\times \frac{24+5}{15}=x\left(1+\frac{14}{15}\right)\times 15x$$

Add 24 and 5 to get 29.

$$15\times \frac{29}{15}=x\left(1+\frac{14}{15}\right)\times 15x$$

Cancel out 15 and 15.

$$29=x\left(1+\frac{14}{15}\right)\times 15x$$

Multiply x and x to get x^{2}.

$$29=x^{2}\left(1+\frac{14}{15}\right)\times 15$$

Convert 1 to fraction \frac{15}{15}.

$$29=x^{2}\left(\frac{15}{15}+\frac{14}{15}\right)\times 15$$

Since \frac{15}{15} and \frac{14}{15} have the same denominator, add them by adding their numerators.

$$29=x^{2}\times \frac{15+14}{15}\times 15$$

Add 15 and 14 to get 29.

$$29=x^{2}\times \frac{29}{15}\times 15$$

Cancel out 15 and 15.

$$29=x^{2}\times 29$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}\times 29=29$$

Subtract 29 from both sides.

$$x^{2}\times 29-29=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$29x^{2}-29=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 29 for a, 0 for b, and -29 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 29\left(-29\right)}}{2\times 29}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 29\left(-29\right)}}{2\times 29}$$

Multiply -4 times 29.

$$x=\frac{0±\sqrt{-116\left(-29\right)}}{2\times 29}$$

Multiply -116 times -29.

$$x=\frac{0±\sqrt{3364}}{2\times 29}$$

Take the square root of 3364.

$$x=\frac{0±58}{2\times 29}$$

Multiply 2 times 29.

$$x=\frac{0±58}{58}$$

Now solve the equation x=\frac{0±58}{58} when ± is plus. Divide 58 by 58.

$$x=1$$

Now solve the equation x=\frac{0±58}{58} when ± is minus. Divide -58 by 58.

$$x=-1$$

The equation is now solved.

$$x=1 x=-1$$