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

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

Solve for x

$$x=-1$$

$$x=4$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -3,\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by \left(2x-1\right)\left(x+3\right), the least common multiple of x+3,2x-1.

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

Use the distributive property to multiply 2x-1 by 4.

$$8x-4+\left(x+3\right)\times 3=\left(2x-1\right)\left(x+3\right)$$

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

$$8x-4+3x+9=\left(2x-1\right)\left(x+3\right)$$

Combine 8x and 3x to get 11x.

$$11x-4+9=\left(2x-1\right)\left(x+3\right)$$

Add -4 and 9 to get 5.

$$11x+5=\left(2x-1\right)\left(x+3\right)$$

Use the distributive property to multiply 2x-1 by x+3 and combine like terms.

$$11x+5=2x^{2}+5x-3$$

Subtract 2x^{2} from both sides.

$$11x+5-2x^{2}=5x-3$$

Subtract 5x from both sides.

$$11x+5-2x^{2}-5x=-3$$

Combine 11x and -5x to get 6x.

$$6x+5-2x^{2}=-3$$

Add 3 to both sides.

$$6x+5-2x^{2}+3=0$$

Add 5 and 3 to get 8.

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

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.

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

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

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

Square 6.

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

Multiply -4 times -2.

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

Multiply 8 times 8.

$$x=\frac{-6±\sqrt{36+64}}{2\left(-2\right)}$$

Add 36 to 64.

$$x=\frac{-6±\sqrt{100}}{2\left(-2\right)}$$

Take the square root of 100.

$$x=\frac{-6±10}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-6±10}{-4}$$

Now solve the equation x=\frac{-6±10}{-4} when ± is plus. Add -6 to 10.

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

Divide 4 by -4.

$$x=-1$$

Now solve the equation x=\frac{-6±10}{-4} when ± is minus. Subtract 10 from -6.

$$x=-\frac{16}{-4}$$

Divide -16 by -4.

$$x=4$$

The equation is now solved.

$$x=-1 x=4$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -3,\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by \left(2x-1\right)\left(x+3\right), the least common multiple of x+3,2x-1.

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

Use the distributive property to multiply 2x-1 by 4.

$$8x-4+\left(x+3\right)\times 3=\left(2x-1\right)\left(x+3\right)$$

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

$$8x-4+3x+9=\left(2x-1\right)\left(x+3\right)$$

Combine 8x and 3x to get 11x.

$$11x-4+9=\left(2x-1\right)\left(x+3\right)$$

Add -4 and 9 to get 5.

$$11x+5=\left(2x-1\right)\left(x+3\right)$$

Use the distributive property to multiply 2x-1 by x+3 and combine like terms.

$$11x+5=2x^{2}+5x-3$$

Subtract 2x^{2} from both sides.

$$11x+5-2x^{2}=5x-3$$

Subtract 5x from both sides.

$$11x+5-2x^{2}-5x=-3$$

Combine 11x and -5x to get 6x.

$$6x+5-2x^{2}=-3$$

Subtract 5 from both sides.

$$6x-2x^{2}=-3-5$$

Subtract 5 from -3 to get -8.

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

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.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 6 by -2.

$$x^{2}-3x=-\frac{8}{-2}$$

Divide -8 by -2.

$$x^{2}-3x=4$$

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=4+\frac{9}{4}$$

Add 4 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{25}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{5}{2} x-\frac{3}{2}=-\frac{5}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=4 x=-1$$