Solve 5x-10/3x+2=3x-7/3x-1

$$\frac { 5 x - 10 } { 3 x + 2 } = \frac { 3 x - 7 } { 3 x - 1 }$$

Solve for x (complex solution)

$$x=\frac{5+\sqrt{11}i}{3}\approx 1.666666667+1.105541597i$$

$$x=\frac{-\sqrt{11}i+5}{3}\approx 1.666666667-1.105541597i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -\frac{2}{3},\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by \left(3x-1\right)\left(3x+2\right), the least common multiple of 3x+2,3x-1.

$$\left(3x-1\right)\left(5x-10\right)=\left(3x+2\right)\left(3x-7\right)$$

Use the distributive property to multiply 3x-1 by 5x-10 and combine like terms.

$$15x^{2}-35x+10=\left(3x+2\right)\left(3x-7\right)$$

Use the distributive property to multiply 3x+2 by 3x-7 and combine like terms.

$$15x^{2}-35x+10=9x^{2}-15x-14$$

Subtract 9x^{2} from both sides.

$$15x^{2}-35x+10-9x^{2}=-15x-14$$

Combine 15x^{2} and -9x^{2} to get 6x^{2}.

$$6x^{2}-35x+10=-15x-14$$

Add 15x to both sides.

$$6x^{2}-35x+10+15x=-14$$

Combine -35x and 15x to get -20x.

$$6x^{2}-20x+10=-14$$

Add 14 to both sides.

$$6x^{2}-20x+10+14=0$$

Add 10 and 14 to get 24.

$$6x^{2}-20x+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -20 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 6\times 24}}{2\times 6}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 6\times 24}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-20\right)±\sqrt{400-24\times 24}}{2\times 6}$$

Multiply -24 times 24.

$$x=\frac{-\left(-20\right)±\sqrt{400-576}}{2\times 6}$$

Add 400 to -576.

$$x=\frac{-\left(-20\right)±\sqrt{-176}}{2\times 6}$$

Take the square root of -176.

$$x=\frac{-\left(-20\right)±4\sqrt{11}i}{2\times 6}$$

The opposite of -20 is 20.

$$x=\frac{20±4\sqrt{11}i}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{20±4\sqrt{11}i}{12}$$

Now solve the equation x=\frac{20±4\sqrt{11}i}{12} when ± is plus. Add 20 to 4i\sqrt{11}.

$$x=\frac{20+4\sqrt{11}i}{12}$$

Divide 20+4i\sqrt{11} by 12.

$$x=\frac{5+\sqrt{11}i}{3}$$

Now solve the equation x=\frac{20±4\sqrt{11}i}{12} when ± is minus. Subtract 4i\sqrt{11} from 20.

$$x=\frac{-4\sqrt{11}i+20}{12}$$

Divide 20-4i\sqrt{11} by 12.

$$x=\frac{-\sqrt{11}i+5}{3}$$

The equation is now solved.

$$x=\frac{5+\sqrt{11}i}{3} x=\frac{-\sqrt{11}i+5}{3}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -\frac{2}{3},\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by \left(3x-1\right)\left(3x+2\right), the least common multiple of 3x+2,3x-1.

$$\left(3x-1\right)\left(5x-10\right)=\left(3x+2\right)\left(3x-7\right)$$

Use the distributive property to multiply 3x-1 by 5x-10 and combine like terms.

$$15x^{2}-35x+10=\left(3x+2\right)\left(3x-7\right)$$

Use the distributive property to multiply 3x+2 by 3x-7 and combine like terms.

$$15x^{2}-35x+10=9x^{2}-15x-14$$

Subtract 9x^{2} from both sides.

$$15x^{2}-35x+10-9x^{2}=-15x-14$$

Combine 15x^{2} and -9x^{2} to get 6x^{2}.

$$6x^{2}-35x+10=-15x-14$$

Add 15x to both sides.

$$6x^{2}-35x+10+15x=-14$$

Combine -35x and 15x to get -20x.

$$6x^{2}-20x+10=-14$$

Subtract 10 from both sides.

$$6x^{2}-20x=-14-10$$

Subtract 10 from -14 to get -24.

$$6x^{2}-20x=-24$$

Divide both sides by 6.

$$\frac{6x^{2}-20x}{6}=-\frac{24}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\left(-\frac{20}{6}\right)x=-\frac{24}{6}$$

Reduce the fraction \frac{-20}{6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{10}{3}x=-\frac{24}{6}$$

Divide -24 by 6.

$$x^{2}-\frac{10}{3}x=-4$$

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

$$x^{2}-\frac{10}{3}x+\left(-\frac{5}{3}\right)^{2}=-4+\left(-\frac{5}{3}\right)^{2}$$

Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=-4+\frac{25}{9}$$

Add -4 to \frac{25}{9}.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=-\frac{11}{9}$$

Factor x^{2}-\frac{10}{3}x+\frac{25}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{3}\right)^{2}=-\frac{11}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{3}\right)^{2}}=\sqrt{-\frac{11}{9}}$$

Simplify.

$$x-\frac{5}{3}=\frac{\sqrt{11}i}{3} x-\frac{5}{3}=-\frac{\sqrt{11}i}{3}$$

Add \frac{5}{3} to both sides of the equation.

$$x=\frac{5+\sqrt{11}i}{3} x=\frac{-\sqrt{11}i+5}{3}$$