Solve (3x+1)(x+3)=1/3(1-x)(7x+9)

$$( 3 x + 1 ) ( x + 3 ) = \frac { 1 } { 3 } ( 1 - x ) ( 7 x + 9 )$$

Solve for x

$$x=-2$$

$$x=0$$

Steps Using the Quadratic Formula

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

$$3x^{2}+10x+3=\frac{1}{3}\left(1-x\right)\left(7x+9\right)$$

Use the distributive property to multiply \frac{1}{3} by 1-x.

$$3x^{2}+10x+3=\left(\frac{1}{3}-\frac{1}{3}x\right)\left(7x+9\right)$$

Use the distributive property to multiply \frac{1}{3}-\frac{1}{3}x by 7x+9 and combine like terms.

$$3x^{2}+10x+3=-\frac{2}{3}x+3-\frac{7}{3}x^{2}$$

Add \frac{2}{3}x to both sides.

$$3x^{2}+10x+3+\frac{2}{3}x=3-\frac{7}{3}x^{2}$$

Combine 10x and \frac{2}{3}x to get \frac{32}{3}x.

$$3x^{2}+\frac{32}{3}x+3=3-\frac{7}{3}x^{2}$$

Subtract 3 from both sides.

$$3x^{2}+\frac{32}{3}x+3-3=-\frac{7}{3}x^{2}$$

Subtract 3 from 3 to get 0.

$$3x^{2}+\frac{32}{3}x=-\frac{7}{3}x^{2}$$

Add \frac{7}{3}x^{2} to both sides.

$$3x^{2}+\frac{32}{3}x+\frac{7}{3}x^{2}=0$$

Combine 3x^{2} and \frac{7}{3}x^{2} to get \frac{16}{3}x^{2}.

$$\frac{16}{3}x^{2}+\frac{32}{3}x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{16}{3} for a, \frac{32}{3} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{32}{3}±\sqrt{\left(\frac{32}{3}\right)^{2}}}{2\times \frac{16}{3}}$$

Take the square root of \left(\frac{32}{3}\right)^{2}.

$$x=\frac{-\frac{32}{3}±\frac{32}{3}}{2\times \frac{16}{3}}$$

Multiply 2 times \frac{16}{3}.

$$x=\frac{-\frac{32}{3}±\frac{32}{3}}{\frac{32}{3}}$$

Now solve the equation x=\frac{-\frac{32}{3}±\frac{32}{3}}{\frac{32}{3}} when ± is plus. Add -\frac{32}{3} to \frac{32}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{0}{\frac{32}{3}}$$

Divide 0 by \frac{32}{3} by multiplying 0 by the reciprocal of \frac{32}{3}.

$$x=0$$

Now solve the equation x=\frac{-\frac{32}{3}±\frac{32}{3}}{\frac{32}{3}} when ± is minus. Subtract \frac{32}{3} from -\frac{32}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{64}{3}}{\frac{32}{3}}$$

Divide -\frac{64}{3} by \frac{32}{3} by multiplying -\frac{64}{3} by the reciprocal of \frac{32}{3}.

$$x=-2$$

The equation is now solved.

$$x=0 x=-2$$

Steps for Completing the Square

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

$$3x^{2}+10x+3=\frac{1}{3}\left(1-x\right)\left(7x+9\right)$$

Use the distributive property to multiply \frac{1}{3} by 1-x.

$$3x^{2}+10x+3=\left(\frac{1}{3}-\frac{1}{3}x\right)\left(7x+9\right)$$

Use the distributive property to multiply \frac{1}{3}-\frac{1}{3}x by 7x+9 and combine like terms.

$$3x^{2}+10x+3=-\frac{2}{3}x+3-\frac{7}{3}x^{2}$$

Add \frac{2}{3}x to both sides.

$$3x^{2}+10x+3+\frac{2}{3}x=3-\frac{7}{3}x^{2}$$

Combine 10x and \frac{2}{3}x to get \frac{32}{3}x.

$$3x^{2}+\frac{32}{3}x+3=3-\frac{7}{3}x^{2}$$

Add \frac{7}{3}x^{2} to both sides.

$$3x^{2}+\frac{32}{3}x+3+\frac{7}{3}x^{2}=3$$

Combine 3x^{2} and \frac{7}{3}x^{2} to get \frac{16}{3}x^{2}.

$$\frac{16}{3}x^{2}+\frac{32}{3}x+3=3$$

Subtract 3 from both sides.

$$\frac{16}{3}x^{2}+\frac{32}{3}x=3-3$$

Subtract 3 from 3 to get 0.

$$\frac{16}{3}x^{2}+\frac{32}{3}x=0$$

Divide both sides of the equation by \frac{16}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{16}{3}x^{2}+\frac{32}{3}x}{\frac{16}{3}}=\frac{0}{\frac{16}{3}}$$

Dividing by \frac{16}{3} undoes the multiplication by \frac{16}{3}.

$$x^{2}+\frac{\frac{32}{3}}{\frac{16}{3}}x=\frac{0}{\frac{16}{3}}$$

Divide \frac{32}{3} by \frac{16}{3} by multiplying \frac{32}{3} by the reciprocal of \frac{16}{3}.

$$x^{2}+2x=\frac{0}{\frac{16}{3}}$$

Divide 0 by \frac{16}{3} by multiplying 0 by the reciprocal of \frac{16}{3}.

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

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

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

Square 1.

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

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+1=1 x+1=-1$$

Subtract 1 from both sides of the equation.

$$x=0 x=-2$$