Solve 2x-3/x-3=x-3/x-5

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

Solve for x

$$x=1$$

$$x=6$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 3,5 since division by zero is not defined. Multiply both sides of the equation by \left(x-5\right)\left(x-3\right), the least common multiple of x-3,x-5.

$$\left(x-5\right)\left(2x-3\right)=\left(x-3\right)\left(x-3\right)$$

Multiply x-3 and x-3 to get \left(x-3\right)^{2}.

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

Use the distributive property to multiply x-5 by 2x-3 and combine like terms.

$$2x^{2}-13x+15=\left(x-3\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$2x^{2}-13x+15=x^{2}-6x+9$$

Subtract x^{2} from both sides.

$$2x^{2}-13x+15-x^{2}=-6x+9$$

Combine 2x^{2} and -x^{2} to get x^{2}.

$$x^{2}-13x+15=-6x+9$$

Add 6x to both sides.

$$x^{2}-13x+15+6x=9$$

Combine -13x and 6x to get -7x.

$$x^{2}-7x+15=9$$

Subtract 9 from both sides.

$$x^{2}-7x+15-9=0$$

Subtract 9 from 15 to get 6.

$$x^{2}-7x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -7 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 6}}{2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 6}}{2}$$

Multiply -4 times 6.

$$x=\frac{-\left(-7\right)±\sqrt{49-24}}{2}$$

Add 49 to -24.

$$x=\frac{-\left(-7\right)±\sqrt{25}}{2}$$

Take the square root of 25.

$$x=\frac{-\left(-7\right)±5}{2}$$

The opposite of -7 is 7.

$$x=\frac{7±5}{2}$$

Now solve the equation x=\frac{7±5}{2} when ± is plus. Add 7 to 5.

$$x=\frac{12}{2}$$

Divide 12 by 2.

$$x=6$$

Now solve the equation x=\frac{7±5}{2} when ± is minus. Subtract 5 from 7.

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

Divide 2 by 2.

$$x=1$$

The equation is now solved.

$$x=6 x=1$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 3,5 since division by zero is not defined. Multiply both sides of the equation by \left(x-5\right)\left(x-3\right), the least common multiple of x-3,x-5.

$$\left(x-5\right)\left(2x-3\right)=\left(x-3\right)\left(x-3\right)$$

Multiply x-3 and x-3 to get \left(x-3\right)^{2}.

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

Use the distributive property to multiply x-5 by 2x-3 and combine like terms.

$$2x^{2}-13x+15=\left(x-3\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$2x^{2}-13x+15=x^{2}-6x+9$$

Subtract x^{2} from both sides.

$$2x^{2}-13x+15-x^{2}=-6x+9$$

Combine 2x^{2} and -x^{2} to get x^{2}.

$$x^{2}-13x+15=-6x+9$$

Add 6x to both sides.

$$x^{2}-13x+15+6x=9$$

Combine -13x and 6x to get -7x.

$$x^{2}-7x+15=9$$

Subtract 15 from both sides.

$$x^{2}-7x=9-15$$

Subtract 15 from 9 to get -6.

$$x^{2}-7x=-6$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-6+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-6+\frac{49}{4}$$

Add -6 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{25}{4}$$

Factor x^{2}-7x+\frac{49}{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}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

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

Add \frac{7}{2} to both sides of the equation.

$$x=6 x=1$$