Solve 2x-5/3x-3=6x-10/4x+1

$$\frac{ 2x-5 }{ 3x-3 } = \frac{ 6x-10 }{ 4x+1 }$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{5}i+3}{2}\approx 1.5-1.118033989i$$

$$x=\frac{3+\sqrt{5}i}{2}\approx 1.5+1.118033989i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -\frac{1}{4},1 since division by zero is not defined. Multiply both sides of the equation by 3\left(x-1\right)\left(4x+1\right), the least common multiple of 3x-3,4x+1.

$$\left(4x+1\right)\left(2x-5\right)=\left(3x-3\right)\left(6x-10\right)$$

Use the distributive property to multiply 4x+1 by 2x-5 and combine like terms.

$$8x^{2}-18x-5=\left(3x-3\right)\left(6x-10\right)$$

Use the distributive property to multiply 3x-3 by 6x-10 and combine like terms.

$$8x^{2}-18x-5=18x^{2}-48x+30$$

Subtract 18x^{2} from both sides.

$$8x^{2}-18x-5-18x^{2}=-48x+30$$

Combine 8x^{2} and -18x^{2} to get -10x^{2}.

$$-10x^{2}-18x-5=-48x+30$$

Add 48x to both sides.

$$-10x^{2}-18x-5+48x=30$$

Combine -18x and 48x to get 30x.

$$-10x^{2}+30x-5=30$$

Subtract 30 from both sides.

$$-10x^{2}+30x-5-30=0$$

Subtract 30 from -5 to get -35.

$$-10x^{2}+30x-35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 30 for b, and -35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-30±\sqrt{30^{2}-4\left(-10\right)\left(-35\right)}}{2\left(-10\right)}$$

Square 30.

$$x=\frac{-30±\sqrt{900-4\left(-10\right)\left(-35\right)}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-30±\sqrt{900+40\left(-35\right)}}{2\left(-10\right)}$$

Multiply 40 times -35.

$$x=\frac{-30±\sqrt{900-1400}}{2\left(-10\right)}$$

Add 900 to -1400.

$$x=\frac{-30±\sqrt{-500}}{2\left(-10\right)}$$

Take the square root of -500.

$$x=\frac{-30±10\sqrt{5}i}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{-30±10\sqrt{5}i}{-20}$$

Now solve the equation x=\frac{-30±10\sqrt{5}i}{-20} when ± is plus. Add -30 to 10i\sqrt{5}.

$$x=\frac{-30+10\sqrt{5}i}{-20}$$

Divide -30+10i\sqrt{5} by -20.

$$x=\frac{-\sqrt{5}i+3}{2}$$

Now solve the equation x=\frac{-30±10\sqrt{5}i}{-20} when ± is minus. Subtract 10i\sqrt{5} from -30.

$$x=\frac{-10\sqrt{5}i-30}{-20}$$

Divide -30-10i\sqrt{5} by -20.

$$x=\frac{3+\sqrt{5}i}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{5}i+3}{2} x=\frac{3+\sqrt{5}i}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -\frac{1}{4},1 since division by zero is not defined. Multiply both sides of the equation by 3\left(x-1\right)\left(4x+1\right), the least common multiple of 3x-3,4x+1.

$$\left(4x+1\right)\left(2x-5\right)=\left(3x-3\right)\left(6x-10\right)$$

Use the distributive property to multiply 4x+1 by 2x-5 and combine like terms.

$$8x^{2}-18x-5=\left(3x-3\right)\left(6x-10\right)$$

Use the distributive property to multiply 3x-3 by 6x-10 and combine like terms.

$$8x^{2}-18x-5=18x^{2}-48x+30$$

Subtract 18x^{2} from both sides.

$$8x^{2}-18x-5-18x^{2}=-48x+30$$

Combine 8x^{2} and -18x^{2} to get -10x^{2}.

$$-10x^{2}-18x-5=-48x+30$$

Add 48x to both sides.

$$-10x^{2}-18x-5+48x=30$$

Combine -18x and 48x to get 30x.

$$-10x^{2}+30x-5=30$$

Add 5 to both sides.

$$-10x^{2}+30x=30+5$$

Add 30 and 5 to get 35.

$$-10x^{2}+30x=35$$

Divide both sides by -10.

$$\frac{-10x^{2}+30x}{-10}=\frac{35}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$x^{2}+\frac{30}{-10}x=\frac{35}{-10}$$

Divide 30 by -10.

$$x^{2}-3x=\frac{35}{-10}$$

Reduce the fraction \frac{35}{-10} to lowest terms by extracting and canceling out 5.

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

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-\frac{7}{2}+\frac{9}{4}$$

Add -\frac{7}{2} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-3x+\frac{9}{4}=-\frac{5}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{5}i}{2} x-\frac{3}{2}=-\frac{\sqrt{5}i}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{3+\sqrt{5}i}{2} x=\frac{-\sqrt{5}i+3}{2}$$