Solve x-7/x-3=x-5/x-4-1

$$\frac { x - 7 } { x - 3 } = \frac { x - 5 } { x - 4 } - 1$$

Solve for x

$$x=5$$

Steps Using the Quadratic Formula

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

$$\left(x-4\right)\left(x-7\right)=\left(x-3\right)\left(x-5\right)+\left(x-4\right)\left(x-3\right)\left(-1\right)$$

Use the distributive property to multiply x-4 by x-7 and combine like terms.

$$x^{2}-11x+28=\left(x-3\right)\left(x-5\right)+\left(x-4\right)\left(x-3\right)\left(-1\right)$$

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

$$x^{2}-11x+28=x^{2}-8x+15+\left(x-4\right)\left(x-3\right)\left(-1\right)$$

Use the distributive property to multiply x-4 by x-3 and combine like terms.

$$x^{2}-11x+28=x^{2}-8x+15+\left(x^{2}-7x+12\right)\left(-1\right)$$

Use the distributive property to multiply x^{2}-7x+12 by -1.

$$x^{2}-11x+28=x^{2}-8x+15-x^{2}+7x-12$$

Combine x^{2} and -x^{2} to get 0.

$$x^{2}-11x+28=-8x+15+7x-12$$

Combine -8x and 7x to get -x.

$$x^{2}-11x+28=-x+15-12$$

Subtract 12 from 15 to get 3.

$$x^{2}-11x+28=-x+3$$

Add x to both sides.

$$x^{2}-11x+28+x=3$$

Combine -11x and x to get -10x.

$$x^{2}-10x+28=3$$

Subtract 3 from both sides.

$$x^{2}-10x+28-3=0$$

Subtract 3 from 28 to get 25.

$$x^{2}-10x+25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 25}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 25}}{2}$$

Multiply -4 times 25.

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

Add 100 to -100.

$$x=\frac{-\left(-10\right)±\sqrt{0}}{2}$$

Take the square root of 0.

$$x=-\frac{-10}{2}$$

The opposite of -10 is 10.

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

Steps for Completing the Square

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

$$\left(x-4\right)\left(x-7\right)=\left(x-3\right)\left(x-5\right)+\left(x-4\right)\left(x-3\right)\left(-1\right)$$

Use the distributive property to multiply x-4 by x-7 and combine like terms.

$$x^{2}-11x+28=\left(x-3\right)\left(x-5\right)+\left(x-4\right)\left(x-3\right)\left(-1\right)$$

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

$$x^{2}-11x+28=x^{2}-8x+15+\left(x-4\right)\left(x-3\right)\left(-1\right)$$

Use the distributive property to multiply x-4 by x-3 and combine like terms.

$$x^{2}-11x+28=x^{2}-8x+15+\left(x^{2}-7x+12\right)\left(-1\right)$$

Use the distributive property to multiply x^{2}-7x+12 by -1.

$$x^{2}-11x+28=x^{2}-8x+15-x^{2}+7x-12$$

Combine x^{2} and -x^{2} to get 0.

$$x^{2}-11x+28=-8x+15+7x-12$$

Combine -8x and 7x to get -x.

$$x^{2}-11x+28=-x+15-12$$

Subtract 12 from 15 to get 3.

$$x^{2}-11x+28=-x+3$$

Add x to both sides.

$$x^{2}-11x+28+x=3$$

Combine -11x and x to get -10x.

$$x^{2}-10x+28=3$$

Subtract 28 from both sides.

$$x^{2}-10x=3-28$$

Subtract 28 from 3 to get -25.

$$x^{2}-10x=-25$$

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

$$x^{2}-10x+\left(-5\right)^{2}=-25+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-25+25$$

Add -25 to 25.

$$x^{2}-10x+25=0$$

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

$$\left(x-5\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-5=0 x-5=0$$

Add 5 to both sides of the equation.

$$x=5 x=5$$

The equation is now solved. Solutions are the same.

$$x=5$$