Solve 35/x-3=x-7/4

$$\frac{ 35 }{ x-3 } = \frac{ x-7 }{ 4 }$$

Solve for x

$$x=-7$$

$$x=17$$

Steps Using the Quadratic Formula

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

$$4\times 35=\left(x-3\right)\left(x-7\right)$$

Multiply 4 and 35 to get 140.

$$140=\left(x-3\right)\left(x-7\right)$$

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

$$140=x^{2}-10x+21$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-10x+21=140$$

Subtract 140 from both sides.

$$x^{2}-10x+21-140=0$$

Subtract 140 from 21 to get -119.

$$x^{2}-10x-119=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and -119 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-119\right)}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-119\right)}}{2}$$

Multiply -4 times -119.

$$x=\frac{-\left(-10\right)±\sqrt{100+476}}{2}$$

Add 100 to 476.

$$x=\frac{-\left(-10\right)±\sqrt{576}}{2}$$

Take the square root of 576.

$$x=\frac{-\left(-10\right)±24}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±24}{2}$$

Now solve the equation x=\frac{10±24}{2} when ± is plus. Add 10 to 24.

$$x=\frac{34}{2}$$

Divide 34 by 2.

$$x=17$$

Now solve the equation x=\frac{10±24}{2} when ± is minus. Subtract 24 from 10.

$$x=-\frac{14}{2}$$

Divide -14 by 2.

$$x=-7$$

The equation is now solved.

$$x=17 x=-7$$

Steps for Completing the Square

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

$$4\times 35=\left(x-3\right)\left(x-7\right)$$

Multiply 4 and 35 to get 140.

$$140=\left(x-3\right)\left(x-7\right)$$

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

$$140=x^{2}-10x+21$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-10x+21=140$$

Subtract 21 from both sides.

$$x^{2}-10x=140-21$$

Subtract 21 from 140 to get 119.

$$x^{2}-10x=119$$

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.

$$x^{2}-10x+\left(-5\right)^{2}=119+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=119+25$$

Add 119 to 25.

$$x^{2}-10x+25=144$$

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}.

$$\left(x-5\right)^{2}=144$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{144}$$

Simplify.

$$x-5=12 x-5=-12$$

Add 5 to both sides of the equation.

$$x=17 x=-7$$