Solve 6x+8/5x=x

$$\frac { 6 x + 8 } { 5 x } = x$$

Solve for x

$$x=-\frac{4}{5}=-0.8$$

$$x=2$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 5x.

$$6x+8=x\times 5x$$

Multiply x and x to get x^{2}.

$$6x+8=x^{2}\times 5$$

Subtract x^{2}\times 5 from both sides.

$$6x+8-x^{2}\times 5=0$$

Multiply -1 and 5 to get -5.

$$6x+8-5x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-5x^{2}+6x+8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -5x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=6 ab=-5\times 8=-40$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -40.

$$-1,40 -2,20 -4,10 -5,8$$

Calculate the sum for each pair.

$$-1+40=39 -2+20=18 -4+10=6 -5+8=3$$

The solution is the pair that gives sum 6.

$$a=10 b=-4$$

Rewrite -5x^{2}+6x+8 as \left(-5x^{2}+10x\right)+\left(-4x+8\right).

$$\left(-5x^{2}+10x\right)+\left(-4x+8\right)$$

Factor out 5x in the first and 4 in the second group.

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

Factor out common term -x+2 by using distributive property.

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

To find equation solutions, solve -x+2=0 and 5x+4=0.

$$x=2 x=-\frac{4}{5}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 5x.

$$6x+8=x\times 5x$$

Multiply x and x to get x^{2}.

$$6x+8=x^{2}\times 5$$

Subtract x^{2}\times 5 from both sides.

$$6x+8-x^{2}\times 5=0$$

Multiply -1 and 5 to get -5.

$$6x+8-5x^{2}=0$$

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.

$$-5x^{2}+6x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 6 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-5\right)\times 8}}{2\left(-5\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-5\right)\times 8}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-6±\sqrt{36+20\times 8}}{2\left(-5\right)}$$

Multiply 20 times 8.

$$x=\frac{-6±\sqrt{36+160}}{2\left(-5\right)}$$

Add 36 to 160.

$$x=\frac{-6±\sqrt{196}}{2\left(-5\right)}$$

Take the square root of 196.

$$x=\frac{-6±14}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-6±14}{-10}$$

Now solve the equation x=\frac{-6±14}{-10} when ± is plus. Add -6 to 14.

$$x=\frac{8}{-10}$$

Reduce the fraction \frac{8}{-10} to lowest terms by extracting and canceling out 2.

$$x=-\frac{4}{5}$$

Now solve the equation x=\frac{-6±14}{-10} when ± is minus. Subtract 14 from -6.

$$x=-\frac{20}{-10}$$

Divide -20 by -10.

$$x=2$$

The equation is now solved.

$$x=-\frac{4}{5} x=2$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 5x.

$$6x+8=x\times 5x$$

Multiply x and x to get x^{2}.

$$6x+8=x^{2}\times 5$$

Subtract x^{2}\times 5 from both sides.

$$6x+8-x^{2}\times 5=0$$

Multiply -1 and 5 to get -5.

$$6x+8-5x^{2}=0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$6x-5x^{2}=-8$$

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.

$$-5x^{2}+6x=-8$$

Divide both sides by -5.

$$\frac{-5x^{2}+6x}{-5}=-\frac{8}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{6}{-5}x=-\frac{8}{-5}$$

Divide 6 by -5.

$$x^{2}-\frac{6}{5}x=-\frac{8}{-5}$$

Divide -8 by -5.

$$x^{2}-\frac{6}{5}x=\frac{8}{5}$$

Divide -\frac{6}{5}, the coefficient of the x term, by 2 to get -\frac{3}{5}. Then add the square of -\frac{3}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{6}{5}x+\left(-\frac{3}{5}\right)^{2}=\frac{8}{5}+\left(-\frac{3}{5}\right)^{2}$$

Square -\frac{3}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{6}{5}x+\frac{9}{25}=\frac{8}{5}+\frac{9}{25}$$

Add \frac{8}{5} to \frac{9}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{6}{5}x+\frac{9}{25}=\frac{49}{25}$$

Factor x^{2}-\frac{6}{5}x+\frac{9}{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}.

$$\left(x-\frac{3}{5}\right)^{2}=\frac{49}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{5}\right)^{2}}=\sqrt{\frac{49}{25}}$$

Simplify.

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

Add \frac{3}{5} to both sides of the equation.

$$x=2 x=-\frac{4}{5}$$