Solve x=7/5x-3

$$x = \frac { 7 } { 5 x - 3 }$$

Solve for x

$$x = \frac{\sqrt{149} + 3}{10} \approx 1.520655562$$

$$x=\frac{3-\sqrt{149}}{10}\approx -0.920655562$$

Steps Using the Quadratic Formula

Subtract \frac{7}{5x-3} from both sides.

$$x-\frac{7}{5x-3}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{5x-3}{5x-3}.

$$\frac{x\left(5x-3\right)}{5x-3}-\frac{7}{5x-3}=0$$

Since \frac{x\left(5x-3\right)}{5x-3} and \frac{7}{5x-3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(5x-3\right)-7}{5x-3}=0$$

Do the multiplications in x\left(5x-3\right)-7.

$$\frac{5x^{2}-3x-7}{5x-3}=0$$

Variable x cannot be equal to \frac{3}{5} since division by zero is not defined. Multiply both sides of the equation by 5x-3.

$$5x^{2}-3x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -3 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 5\left(-7\right)}}{2\times 5}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 5\left(-7\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-3\right)±\sqrt{9-20\left(-7\right)}}{2\times 5}$$

Multiply -20 times -7.

$$x=\frac{-\left(-3\right)±\sqrt{9+140}}{2\times 5}$$

Add 9 to 140.

$$x=\frac{-\left(-3\right)±\sqrt{149}}{2\times 5}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{149}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{3±\sqrt{149}}{10}$$

Now solve the equation x=\frac{3±\sqrt{149}}{10} when ± is plus. Add 3 to \sqrt{149}.

$$x=\frac{\sqrt{149}+3}{10}$$

Now solve the equation x=\frac{3±\sqrt{149}}{10} when ± is minus. Subtract \sqrt{149} from 3.

$$x=\frac{3-\sqrt{149}}{10}$$

The equation is now solved.

$$x=\frac{\sqrt{149}+3}{10} x=\frac{3-\sqrt{149}}{10}$$

Steps for Completing the Square

Subtract \frac{7}{5x-3} from both sides.

$$x-\frac{7}{5x-3}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{5x-3}{5x-3}.

$$\frac{x\left(5x-3\right)}{5x-3}-\frac{7}{5x-3}=0$$

Since \frac{x\left(5x-3\right)}{5x-3} and \frac{7}{5x-3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(5x-3\right)-7}{5x-3}=0$$

Do the multiplications in x\left(5x-3\right)-7.

$$\frac{5x^{2}-3x-7}{5x-3}=0$$

Variable x cannot be equal to \frac{3}{5} since division by zero is not defined. Multiply both sides of the equation by 5x-3.

$$5x^{2}-3x-7=0$$

Add 7 to both sides. Anything plus zero gives itself.

$$5x^{2}-3x=7$$

Divide both sides by 5.

$$\frac{5x^{2}-3x}{5}=\frac{7}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{3}{5}x=\frac{7}{5}$$

Divide -\frac{3}{5}, the coefficient of the x term, by 2 to get -\frac{3}{10}. Then add the square of -\frac{3}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{5}x+\left(-\frac{3}{10}\right)^{2}=\frac{7}{5}+\left(-\frac{3}{10}\right)^{2}$$

Square -\frac{3}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{5}x+\frac{9}{100}=\frac{7}{5}+\frac{9}{100}$$

Add \frac{7}{5} to \frac{9}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{5}x+\frac{9}{100}=\frac{149}{100}$$

Factor x^{2}-\frac{3}{5}x+\frac{9}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{10}\right)^{2}=\frac{149}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{10}\right)^{2}}=\sqrt{\frac{149}{100}}$$

Simplify.

$$x-\frac{3}{10}=\frac{\sqrt{149}}{10} x-\frac{3}{10}=-\frac{\sqrt{149}}{10}$$

Add \frac{3}{10} to both sides of the equation.

$$x=\frac{\sqrt{149}+3}{10} x=\frac{3-\sqrt{149}}{10}$$