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

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

Solve for x

$$x=-9$$

Steps Using the Quadratic Formula

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

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

Use the distributive property to multiply x-4 by 5.

$$5x-20-\left(-\left(4+x\right)\left(x+3\right)\right)=\left(x+4\right)\times 7$$

Use the distributive property to multiply -1 by 4+x.

$$5x-20-\left(-4-x\right)\left(x+3\right)=\left(x+4\right)\times 7$$

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

$$5x-20-\left(-7x-12-x^{2}\right)=\left(x+4\right)\times 7$$

To find the opposite of -7x-12-x^{2}, find the opposite of each term.

$$5x-20+7x+12+x^{2}=\left(x+4\right)\times 7$$

Combine 5x and 7x to get 12x.

$$12x-20+12+x^{2}=\left(x+4\right)\times 7$$

Add -20 and 12 to get -8.

$$12x-8+x^{2}=\left(x+4\right)\times 7$$

Use the distributive property to multiply x+4 by 7.

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

Subtract 7x from both sides.

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

Combine 12x and -7x to get 5x.

$$5x-8+x^{2}=28$$

Subtract 28 from both sides.

$$5x-8+x^{2}-28=0$$

Subtract 28 from -8 to get -36.

$$5x-36+x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x^{2}+5x-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-36\right)}}{2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-36\right)}}{2}$$

Multiply -4 times -36.

$$x=\frac{-5±\sqrt{25+144}}{2}$$

Add 25 to 144.

$$x=\frac{-5±\sqrt{169}}{2}$$

Take the square root of 169.

$$x=\frac{-5±13}{2}$$

Now solve the equation x=\frac{-5±13}{2} when ± is plus. Add -5 to 13.

$$x=\frac{8}{2}$$

Divide 8 by 2.

$$x=4$$

Now solve the equation x=\frac{-5±13}{2} when ± is minus. Subtract 13 from -5.

$$x=-\frac{18}{2}$$

Divide -18 by 2.

$$x=-9$$

The equation is now solved.

$$x=4 x=-9$$

Variable x cannot be equal to 4.

$$x=-9$$

Steps for Completing the Square

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

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

Use the distributive property to multiply x-4 by 5.

$$5x-20-\left(-\left(4+x\right)\left(x+3\right)\right)=\left(x+4\right)\times 7$$

Use the distributive property to multiply -1 by 4+x.

$$5x-20-\left(-4-x\right)\left(x+3\right)=\left(x+4\right)\times 7$$

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

$$5x-20-\left(-7x-12-x^{2}\right)=\left(x+4\right)\times 7$$

To find the opposite of -7x-12-x^{2}, find the opposite of each term.

$$5x-20+7x+12+x^{2}=\left(x+4\right)\times 7$$

Combine 5x and 7x to get 12x.

$$12x-20+12+x^{2}=\left(x+4\right)\times 7$$

Add -20 and 12 to get -8.

$$12x-8+x^{2}=\left(x+4\right)\times 7$$

Use the distributive property to multiply x+4 by 7.

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

Subtract 7x from both sides.

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

Combine 12x and -7x to get 5x.

$$5x-8+x^{2}=28$$

Add 8 to both sides.

$$5x+x^{2}=28+8$$

Add 28 and 8 to get 36.

$$5x+x^{2}=36$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$x^{2}+5x=36$$

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

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=36+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=36+\frac{25}{4}$$

Add 36 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{169}{4}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{13}{2} x+\frac{5}{2}=-\frac{13}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=4 x=-9$$

Variable x cannot be equal to 4.

$$x=-9$$