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

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

Solve for x

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

$$x=1$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 4x, the least common multiple of x,4.

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

Multiply 3 and 4 to get 12.

$$12xx+4x-4\times 3=x\left(1+3x\right)$$

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

$$12x^{2}+4x-4\times 3=x\left(1+3x\right)$$

Multiply -4 and 3 to get -12.

$$12x^{2}+4x-12=x\left(1+3x\right)$$

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

$$12x^{2}+4x-12=x+3x^{2}$$

Subtract x from both sides.

$$12x^{2}+4x-12-x=3x^{2}$$

Combine 4x and -x to get 3x.

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

Subtract 3x^{2} from both sides.

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

Combine 12x^{2} and -3x^{2} to get 9x^{2}.

$$9x^{2}+3x-12=0$$

Divide both sides by 3.

$$3x^{2}+x-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=1 ab=3\left(-4\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=-3 b=4$$

Rewrite 3x^{2}+x-4 as \left(3x^{2}-3x\right)+\left(4x-4\right).

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

Factor out 3x in the first and 4 in the second group.

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

Factor out common term x-1 by using distributive property.

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

To find equation solutions, solve x-1=0 and 3x+4=0.

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

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 4x, the least common multiple of x,4.

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

Multiply 3 and 4 to get 12.

$$12xx+4x-4\times 3=x\left(1+3x\right)$$

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

$$12x^{2}+4x-4\times 3=x\left(1+3x\right)$$

Multiply -4 and 3 to get -12.

$$12x^{2}+4x-12=x\left(1+3x\right)$$

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

$$12x^{2}+4x-12=x+3x^{2}$$

Subtract x from both sides.

$$12x^{2}+4x-12-x=3x^{2}$$

Combine 4x and -x to get 3x.

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

Subtract 3x^{2} from both sides.

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

Combine 12x^{2} and -3x^{2} to get 9x^{2}.

$$9x^{2}+3x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 3 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times 9\left(-12\right)}}{2\times 9}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 9\left(-12\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-3±\sqrt{9-36\left(-12\right)}}{2\times 9}$$

Multiply -36 times -12.

$$x=\frac{-3±\sqrt{9+432}}{2\times 9}$$

Add 9 to 432.

$$x=\frac{-3±\sqrt{441}}{2\times 9}$$

Take the square root of 441.

$$x=\frac{-3±21}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-3±21}{18}$$

Now solve the equation x=\frac{-3±21}{18} when ± is plus. Add -3 to 21.

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

Divide 18 by 18.

$$x=1$$

Now solve the equation x=\frac{-3±21}{18} when ± is minus. Subtract 21 from -3.

$$x=-\frac{24}{18}$$

Reduce the fraction \frac{-24}{18} to lowest terms by extracting and canceling out 6.

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

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 4x, the least common multiple of x,4.

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

Multiply 3 and 4 to get 12.

$$12xx+4x-4\times 3=x\left(1+3x\right)$$

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

$$12x^{2}+4x-4\times 3=x\left(1+3x\right)$$

Multiply -4 and 3 to get -12.

$$12x^{2}+4x-12=x\left(1+3x\right)$$

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

$$12x^{2}+4x-12=x+3x^{2}$$

Subtract x from both sides.

$$12x^{2}+4x-12-x=3x^{2}$$

Combine 4x and -x to get 3x.

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

Subtract 3x^{2} from both sides.

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

Combine 12x^{2} and -3x^{2} to get 9x^{2}.

$$9x^{2}+3x-12=0$$

Add 12 to both sides. Anything plus zero gives itself.

$$9x^{2}+3x=12$$

Divide both sides by 9.

$$\frac{9x^{2}+3x}{9}=\frac{12}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{3}{9}x=\frac{12}{9}$$

Reduce the fraction \frac{3}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{1}{3}x=\frac{12}{9}$$

Reduce the fraction \frac{12}{9} to lowest terms by extracting and canceling out 3.

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

Divide \frac{1}{3}, the coefficient of the x term, by 2 to get \frac{1}{6}. Then add the square of \frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{4}{3}+\frac{1}{36}$$

Add \frac{4}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{49}{36}$$

Factor x^{2}+\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{6}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$x+\frac{1}{6}=\frac{7}{6} x+\frac{1}{6}=-\frac{7}{6}$$

Subtract \frac{1}{6} from both sides of the equation.

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