Solve 2x-2x+4/5=2x/5quad(9)x/3-x+1/5

$$2 x - \frac { 2 x + 4 } { 5 } = \frac { 2 x } { 5 } \quad ( 9 ) \frac { x } { 3 } - \frac { x + 1 } { 5 }$$

Solve for x

$$x=1$$

$$x=\frac{1}{2}=0.5$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 15, the least common multiple of 5,3.

$$30x-3\left(2x+4\right)=2x\times 9x-3\left(x+1\right)$$

Use the distributive property to multiply -3 by 2x+4.

$$30x-6x-12=2x\times 9x-3\left(x+1\right)$$

Combine 30x and -6x to get 24x.

$$24x-12=2x\times 9x-3\left(x+1\right)$$

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

$$24x-12=2x^{2}\times 9-3\left(x+1\right)$$

Multiply 2 and 9 to get 18.

$$24x-12=18x^{2}-3\left(x+1\right)$$

Use the distributive property to multiply -3 by x+1.

$$24x-12=18x^{2}-3x-3$$

Subtract 18x^{2} from both sides.

$$24x-12-18x^{2}=-3x-3$$

Add 3x to both sides.

$$24x-12-18x^{2}+3x=-3$$

Combine 24x and 3x to get 27x.

$$27x-12-18x^{2}=-3$$

Add 3 to both sides.

$$27x-12-18x^{2}+3=0$$

Add -12 and 3 to get -9.

$$27x-9-18x^{2}=0$$

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.

$$-18x^{2}+27x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -18 for a, 27 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-27±\sqrt{27^{2}-4\left(-18\right)\left(-9\right)}}{2\left(-18\right)}$$

Square 27.

$$x=\frac{-27±\sqrt{729-4\left(-18\right)\left(-9\right)}}{2\left(-18\right)}$$

Multiply -4 times -18.

$$x=\frac{-27±\sqrt{729+72\left(-9\right)}}{2\left(-18\right)}$$

Multiply 72 times -9.

$$x=\frac{-27±\sqrt{729-648}}{2\left(-18\right)}$$

Add 729 to -648.

$$x=\frac{-27±\sqrt{81}}{2\left(-18\right)}$$

Take the square root of 81.

$$x=\frac{-27±9}{2\left(-18\right)}$$

Multiply 2 times -18.

$$x=\frac{-27±9}{-36}$$

Now solve the equation x=\frac{-27±9}{-36} when ± is plus. Add -27 to 9.

$$x=-\frac{18}{-36}$$

Reduce the fraction \frac{-18}{-36} to lowest terms by extracting and canceling out 18.

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{-27±9}{-36} when ± is minus. Subtract 9 from -27.

$$x=-\frac{36}{-36}$$

Divide -36 by -36.

$$x=1$$

The equation is now solved.

$$x=\frac{1}{2} x=1$$

Steps for Completing the Square

Multiply both sides of the equation by 15, the least common multiple of 5,3.

$$30x-3\left(2x+4\right)=2x\times 9x-3\left(x+1\right)$$

Use the distributive property to multiply -3 by 2x+4.

$$30x-6x-12=2x\times 9x-3\left(x+1\right)$$

Combine 30x and -6x to get 24x.

$$24x-12=2x\times 9x-3\left(x+1\right)$$

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

$$24x-12=2x^{2}\times 9-3\left(x+1\right)$$

Multiply 2 and 9 to get 18.

$$24x-12=18x^{2}-3\left(x+1\right)$$

Use the distributive property to multiply -3 by x+1.

$$24x-12=18x^{2}-3x-3$$

Subtract 18x^{2} from both sides.

$$24x-12-18x^{2}=-3x-3$$

Add 3x to both sides.

$$24x-12-18x^{2}+3x=-3$$

Combine 24x and 3x to get 27x.

$$27x-12-18x^{2}=-3$$

Add 12 to both sides.

$$27x-18x^{2}=-3+12$$

Add -3 and 12 to get 9.

$$27x-18x^{2}=9$$

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.

$$-18x^{2}+27x=9$$

Divide both sides by -18.

$$\frac{-18x^{2}+27x}{-18}=\frac{9}{-18}$$

Dividing by -18 undoes the multiplication by -18.

$$x^{2}+\frac{27}{-18}x=\frac{9}{-18}$$

Reduce the fraction \frac{27}{-18} to lowest terms by extracting and canceling out 9.

$$x^{2}-\frac{3}{2}x=\frac{9}{-18}$$

Reduce the fraction \frac{9}{-18} to lowest terms by extracting and canceling out 9.

$$x^{2}-\frac{3}{2}x=-\frac{1}{2}$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=-\frac{1}{2}+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{1}{2}+\frac{9}{16}$$

Add -\frac{1}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{1}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{1}{4} x-\frac{3}{4}=-\frac{1}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=1 x=\frac{1}{2}$$