Solve 5/x+2=4-3x/(x-1)

$$\frac { 5 } { x + 2 } = \frac { 4 - 3 x } { ( x - 1 ) }$$

Solve for x

$$x = \frac{\sqrt{205} - 7}{6} \approx 1.219636844$$

$$x=\frac{-\sqrt{205}-7}{6}\approx -3.552970177$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(x+2\right), the least common multiple of x+2,x-1.

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

Use the distributive property to multiply x-1 by 5.

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

Use the distributive property to multiply x+2 by 4-3x and combine like terms.

$$5x-5=-2x-3x^{2}+8$$

Add 2x to both sides.

$$5x-5+2x=-3x^{2}+8$$

Combine 5x and 2x to get 7x.

$$7x-5=-3x^{2}+8$$

Add 3x^{2} to both sides.

$$7x-5+3x^{2}=8$$

Subtract 8 from both sides.

$$7x-5+3x^{2}-8=0$$

Subtract 8 from -5 to get -13.

$$7x-13+3x^{2}=0$$

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.

$$3x^{2}+7x-13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 7 for b, and -13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 3\left(-13\right)}}{2\times 3}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 3\left(-13\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-7±\sqrt{49-12\left(-13\right)}}{2\times 3}$$

Multiply -12 times -13.

$$x=\frac{-7±\sqrt{49+156}}{2\times 3}$$

Add 49 to 156.

$$x=\frac{-7±\sqrt{205}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-7±\sqrt{205}}{6}$$

Now solve the equation x=\frac{-7±\sqrt{205}}{6} when ± is plus. Add -7 to \sqrt{205}.

$$x=\frac{\sqrt{205}-7}{6}$$

Now solve the equation x=\frac{-7±\sqrt{205}}{6} when ± is minus. Subtract \sqrt{205} from -7.

$$x=\frac{-\sqrt{205}-7}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{205}-7}{6} x=\frac{-\sqrt{205}-7}{6}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(x+2\right), the least common multiple of x+2,x-1.

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

Use the distributive property to multiply x-1 by 5.

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

Use the distributive property to multiply x+2 by 4-3x and combine like terms.

$$5x-5=-2x-3x^{2}+8$$

Add 2x to both sides.

$$5x-5+2x=-3x^{2}+8$$

Combine 5x and 2x to get 7x.

$$7x-5=-3x^{2}+8$$

Add 3x^{2} to both sides.

$$7x-5+3x^{2}=8$$

Add 5 to both sides.

$$7x+3x^{2}=8+5$$

Add 8 and 5 to get 13.

$$7x+3x^{2}=13$$

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.

$$3x^{2}+7x=13$$

Divide both sides by 3.

$$\frac{3x^{2}+7x}{3}=\frac{13}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{7}{3}x=\frac{13}{3}$$

Divide \frac{7}{3}, the coefficient of the x term, by 2 to get \frac{7}{6}. Then add the square of \frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{3}x+\left(\frac{7}{6}\right)^{2}=\frac{13}{3}+\left(\frac{7}{6}\right)^{2}$$

Square \frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=\frac{13}{3}+\frac{49}{36}$$

Add \frac{13}{3} to \frac{49}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=\frac{205}{36}$$

Factor x^{2}+\frac{7}{3}x+\frac{49}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{6}\right)^{2}=\frac{205}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{6}\right)^{2}}=\sqrt{\frac{205}{36}}$$

Simplify.

$$x+\frac{7}{6}=\frac{\sqrt{205}}{6} x+\frac{7}{6}=-\frac{\sqrt{205}}{6}$$

Subtract \frac{7}{6} from both sides of the equation.

$$x=\frac{\sqrt{205}-7}{6} x=\frac{-\sqrt{205}-7}{6}$$