Solve 4x=4(9-2x/4-x)

$$4x = 4( \frac{ 9-2x }{ 4-x } )$$

Solve for x

$$x=3$$

Steps Using Factoring By Grouping

Cancel out 4 on both sides.

$$x=\frac{9-2x}{4-x}$$

Subtract \frac{9-2x}{4-x} from both sides.

$$x-\frac{9-2x}{4-x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{4-x}{4-x}.

$$\frac{x\left(4-x\right)}{4-x}-\frac{9-2x}{4-x}=0$$

Since \frac{x\left(4-x\right)}{4-x} and \frac{9-2x}{4-x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(4-x\right)-\left(9-2x\right)}{4-x}=0$$

Do the multiplications in x\left(4-x\right)-\left(9-2x\right).

$$\frac{4x-x^{2}-9+2x}{4-x}=0$$

Combine like terms in 4x-x^{2}-9+2x.

$$\frac{6x-x^{2}-9}{4-x}=0$$

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by -x+4.

$$6x-x^{2}-9=0$$

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

$$-x^{2}+6x-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=6 ab=-\left(-9\right)=9$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 6.

$$a=3 b=3$$

Rewrite -x^{2}+6x-9 as \left(-x^{2}+3x\right)+\left(3x-9\right).

$$\left(-x^{2}+3x\right)+\left(3x-9\right)$$

Factor out -x in the first and 3 in the second group.

$$-x\left(x-3\right)+3\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(-x+3\right)$$

To find equation solutions, solve x-3=0 and -x+3=0.

$$x=3 x=3$$

Steps Using the Quadratic Formula

Cancel out 4 on both sides.

$$x=\frac{9-2x}{4-x}$$

Subtract \frac{9-2x}{4-x} from both sides.

$$x-\frac{9-2x}{4-x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{4-x}{4-x}.

$$\frac{x\left(4-x\right)}{4-x}-\frac{9-2x}{4-x}=0$$

Since \frac{x\left(4-x\right)}{4-x} and \frac{9-2x}{4-x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(4-x\right)-\left(9-2x\right)}{4-x}=0$$

Do the multiplications in x\left(4-x\right)-\left(9-2x\right).

$$\frac{4x-x^{2}-9+2x}{4-x}=0$$

Combine like terms in 4x-x^{2}-9+2x.

$$\frac{6x-x^{2}-9}{4-x}=0$$

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by -x+4.

$$6x-x^{2}-9=0$$

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.

$$-x^{2}+6x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 6 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-1\right)\left(-9\right)}}{2\left(-1\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-1\right)\left(-9\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-6±\sqrt{36+4\left(-9\right)}}{2\left(-1\right)}$$

Multiply 4 times -9.

$$x=\frac{-6±\sqrt{36-36}}{2\left(-1\right)}$$

Add 36 to -36.

$$x=\frac{-6±\sqrt{0}}{2\left(-1\right)}$$

Take the square root of 0.

$$x=-\frac{6}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=-\frac{6}{-2}$$

Divide -6 by -2.

$$x=3$$

Steps for Completing the Square

Cancel out 4 on both sides.

$$x=\frac{9-2x}{4-x}$$

Subtract \frac{9-2x}{4-x} from both sides.

$$x-\frac{9-2x}{4-x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{4-x}{4-x}.

$$\frac{x\left(4-x\right)}{4-x}-\frac{9-2x}{4-x}=0$$

Since \frac{x\left(4-x\right)}{4-x} and \frac{9-2x}{4-x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(4-x\right)-\left(9-2x\right)}{4-x}=0$$

Do the multiplications in x\left(4-x\right)-\left(9-2x\right).

$$\frac{4x-x^{2}-9+2x}{4-x}=0$$

Combine like terms in 4x-x^{2}-9+2x.

$$\frac{6x-x^{2}-9}{4-x}=0$$

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by -x+4.

$$6x-x^{2}-9=0$$

Add 9 to both sides. Anything plus zero gives itself.

$$6x-x^{2}=9$$

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.

$$-x^{2}+6x=9$$

Divide both sides by -1.

$$\frac{-x^{2}+6x}{-1}=\frac{9}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{6}{-1}x=\frac{9}{-1}$$

Divide 6 by -1.

$$x^{2}-6x=\frac{9}{-1}$$

Divide 9 by -1.

$$x^{2}-6x=-9$$

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

$$x^{2}-6x+\left(-3\right)^{2}=-9+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-9+9$$

Add -9 to 9.

$$x^{2}-6x+9=0$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-3=0 x-3=0$$

Add 3 to both sides of the equation.

$$x=3 x=3$$

The equation is now solved. Solutions are the same.

$$x=3$$