Solve 6x-4/x-1+5x-2=3x+1

$$\frac { 6 x - 4 } { x - 1 } + 5 x - 2 = 3 x + 1$$

Solve for x

$$x=-1$$

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

Steps Using the Quadratic Formula

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by x-1.

$$6x-4+5x\left(x-1\right)+\left(x-1\right)\left(-2\right)=3x\left(x-1\right)+x-1$$

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

$$6x-4+5x^{2}-5x+\left(x-1\right)\left(-2\right)=3x\left(x-1\right)+x-1$$

Combine 6x and -5x to get x.

$$x-4+5x^{2}+\left(x-1\right)\left(-2\right)=3x\left(x-1\right)+x-1$$

Use the distributive property to multiply x-1 by -2.

$$x-4+5x^{2}-2x+2=3x\left(x-1\right)+x-1$$

Combine x and -2x to get -x.

$$-x-4+5x^{2}+2=3x\left(x-1\right)+x-1$$

Add -4 and 2 to get -2.

$$-x-2+5x^{2}=3x\left(x-1\right)+x-1$$

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

$$-x-2+5x^{2}=3x^{2}-3x+x-1$$

Combine -3x and x to get -2x.

$$-x-2+5x^{2}=3x^{2}-2x-1$$

Subtract 3x^{2} from both sides.

$$-x-2+5x^{2}-3x^{2}=-2x-1$$

Combine 5x^{2} and -3x^{2} to get 2x^{2}.

$$-x-2+2x^{2}=-2x-1$$

Add 2x to both sides.

$$-x-2+2x^{2}+2x=-1$$

Combine -x and 2x to get x.

$$x-2+2x^{2}=-1$$

Add 1 to both sides.

$$x-2+2x^{2}+1=0$$

Add -2 and 1 to get -1.

$$x-1+2x^{2}=0$$

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.

$$2x^{2}+x-1=0$$

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

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

Square 1.

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

Multiply -4 times 2.

$$x=\frac{-1±\sqrt{1-8\left(-1\right)}}{2\times 2}$$

Multiply -8 times -1.

$$x=\frac{-1±\sqrt{1+8}}{2\times 2}$$

Add 1 to 8.

$$x=\frac{-1±\sqrt{9}}{2\times 2}$$

Take the square root of 9.

$$x=\frac{-1±3}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-1±3}{4}$$

Now solve the equation x=\frac{-1±3}{4} when ± is plus. Add -1 to 3.

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

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-1±3}{4} when ± is minus. Subtract 3 from -1.

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

Divide -4 by 4.

$$x=-1$$

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by x-1.

$$6x-4+5x\left(x-1\right)+\left(x-1\right)\left(-2\right)=3x\left(x-1\right)+x-1$$

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

$$6x-4+5x^{2}-5x+\left(x-1\right)\left(-2\right)=3x\left(x-1\right)+x-1$$

Combine 6x and -5x to get x.

$$x-4+5x^{2}+\left(x-1\right)\left(-2\right)=3x\left(x-1\right)+x-1$$

Use the distributive property to multiply x-1 by -2.

$$x-4+5x^{2}-2x+2=3x\left(x-1\right)+x-1$$

Combine x and -2x to get -x.

$$-x-4+5x^{2}+2=3x\left(x-1\right)+x-1$$

Add -4 and 2 to get -2.

$$-x-2+5x^{2}=3x\left(x-1\right)+x-1$$

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

$$-x-2+5x^{2}=3x^{2}-3x+x-1$$

Combine -3x and x to get -2x.

$$-x-2+5x^{2}=3x^{2}-2x-1$$

Subtract 3x^{2} from both sides.

$$-x-2+5x^{2}-3x^{2}=-2x-1$$

Combine 5x^{2} and -3x^{2} to get 2x^{2}.

$$-x-2+2x^{2}=-2x-1$$

Add 2x to both sides.

$$-x-2+2x^{2}+2x=-1$$

Combine -x and 2x to get x.

$$x-2+2x^{2}=-1$$

Add 2 to both sides.

$$x+2x^{2}=-1+2$$

Add -1 and 2 to get 1.

$$x+2x^{2}=1$$

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.

$$2x^{2}+x=1$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

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

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

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

Factor x^{2}+\frac{1}{2}x+\frac{1}{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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

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

Subtract \frac{1}{4} from both sides of the equation.

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