Solve -3/4-(-1)/1+(-frac{3{4})x(-1)}=x-(-3/4)/1+(-frac{3{4})x}

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

Solve for x

$$x=-\frac{1}{3}\approx -0.333333333$$

$$x=-2$$

Steps Using Factoring By Grouping

The opposite of -1 is 1.

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

Add -\frac{3}{4} and 1 to get \frac{1}{4}.

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

Multiply -\frac{3}{4} and -1 to get \frac{3}{4}.

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

Express \frac{\frac{1}{4}}{1+\frac{3}{4}x} as a single fraction.

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

The opposite of -\frac{3}{4} is \frac{3}{4}.

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

Factor the expressions that are not already factored in \frac{x+\frac{3}{4}}{1-\frac{3}{4}x}.

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

To divide powers of the same base, subtract the numerator's exponent from the denominator's exponent.

$$\frac{1}{4\left(1+\frac{3}{4}x\right)}=\frac{4x+3}{\left(\frac{1}{4}\right)^{0}\left(-3x+4\right)}$$

Calculate \frac{1}{4} to the power of 0 and get 1.

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

Use the distributive property to multiply 4 by 1+\frac{3}{4}x.

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

Use the distributive property to multiply 1 by -3x+4.

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

Subtract \frac{4x+3}{-3x+4} from both sides.

$$\frac{1}{4+3x}-\frac{4x+3}{-3x+4}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4+3x and -3x+4 is \left(-3x+4\right)\left(3x+4\right). Multiply \frac{1}{4+3x} times \frac{-3x+4}{-3x+4}. Multiply \frac{4x+3}{-3x+4} times \frac{3x+4}{3x+4}.

$$\frac{-3x+4}{\left(-3x+4\right)\left(3x+4\right)}-\frac{\left(4x+3\right)\left(3x+4\right)}{\left(-3x+4\right)\left(3x+4\right)}=0$$

Since \frac{-3x+4}{\left(-3x+4\right)\left(3x+4\right)} and \frac{\left(4x+3\right)\left(3x+4\right)}{\left(-3x+4\right)\left(3x+4\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-3x+4-\left(4x+3\right)\left(3x+4\right)}{\left(-3x+4\right)\left(3x+4\right)}=0$$

Do the multiplications in -3x+4-\left(4x+3\right)\left(3x+4\right).

$$\frac{-3x+4-12x^{2}-16x-9x-12}{\left(-3x+4\right)\left(3x+4\right)}=0$$

Combine like terms in -3x+4-12x^{2}-16x-9x-12.

$$\frac{-28x-8-12x^{2}}{\left(-3x+4\right)\left(3x+4\right)}=0$$

Variable x cannot be equal to any of the values -\frac{4}{3},\frac{4}{3} since division by zero is not defined. Multiply both sides of the equation by \left(-3x-4\right)\left(3x-4\right).

$$-28x-8-12x^{2}=0$$

Divide both sides by 4.

$$-7x-2-3x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-3x^{2}-7x-2=0$$

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

$$a+b=-7 ab=-3\left(-2\right)=6$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -7.

$$a=-1 b=-6$$

Rewrite -3x^{2}-7x-2 as \left(-3x^{2}-x\right)+\left(-6x-2\right).

$$\left(-3x^{2}-x\right)+\left(-6x-2\right)$$

Factor out -x in the first and -2 in the second group.

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

Factor out common term 3x+1 by using distributive property.

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

To find equation solutions, solve 3x+1=0 and -x-2=0.

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

Steps Using the Quadratic Formula

The opposite of -1 is 1.

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

Add -\frac{3}{4} and 1 to get \frac{1}{4}.

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

Multiply -\frac{3}{4} and -1 to get \frac{3}{4}.

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

Express \frac{\frac{1}{4}}{1+\frac{3}{4}x} as a single fraction.

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

The opposite of -\frac{3}{4} is \frac{3}{4}.

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

Factor the expressions that are not already factored in \frac{x+\frac{3}{4}}{1-\frac{3}{4}x}.

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

To divide powers of the same base, subtract the numerator's exponent from the denominator's exponent.

$$\frac{1}{4\left(1+\frac{3}{4}x\right)}=\frac{4x+3}{\left(\frac{1}{4}\right)^{0}\left(-3x+4\right)}$$

Calculate \frac{1}{4} to the power of 0 and get 1.

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

Use the distributive property to multiply 4 by 1+\frac{3}{4}x.

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

Use the distributive property to multiply 1 by -3x+4.

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

Subtract \frac{4x+3}{-3x+4} from both sides.

$$\frac{1}{4+3x}-\frac{4x+3}{-3x+4}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4+3x and -3x+4 is \left(-3x+4\right)\left(3x+4\right). Multiply \frac{1}{4+3x} times \frac{-3x+4}{-3x+4}. Multiply \frac{4x+3}{-3x+4} times \frac{3x+4}{3x+4}.

$$\frac{-3x+4}{\left(-3x+4\right)\left(3x+4\right)}-\frac{\left(4x+3\right)\left(3x+4\right)}{\left(-3x+4\right)\left(3x+4\right)}=0$$

Since \frac{-3x+4}{\left(-3x+4\right)\left(3x+4\right)} and \frac{\left(4x+3\right)\left(3x+4\right)}{\left(-3x+4\right)\left(3x+4\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-3x+4-\left(4x+3\right)\left(3x+4\right)}{\left(-3x+4\right)\left(3x+4\right)}=0$$

Do the multiplications in -3x+4-\left(4x+3\right)\left(3x+4\right).

$$\frac{-3x+4-12x^{2}-16x-9x-12}{\left(-3x+4\right)\left(3x+4\right)}=0$$

Combine like terms in -3x+4-12x^{2}-16x-9x-12.

$$\frac{-28x-8-12x^{2}}{\left(-3x+4\right)\left(3x+4\right)}=0$$

Variable x cannot be equal to any of the values -\frac{4}{3},\frac{4}{3} since division by zero is not defined. Multiply both sides of the equation by \left(-3x-4\right)\left(3x-4\right).

$$-28x-8-12x^{2}=0$$

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.

$$-12x^{2}-28x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -12 for a, -28 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\left(-12\right)\left(-8\right)}}{2\left(-12\right)}$$

Square -28.

$$x=\frac{-\left(-28\right)±\sqrt{784-4\left(-12\right)\left(-8\right)}}{2\left(-12\right)}$$

Multiply -4 times -12.

$$x=\frac{-\left(-28\right)±\sqrt{784+48\left(-8\right)}}{2\left(-12\right)}$$

Multiply 48 times -8.

$$x=\frac{-\left(-28\right)±\sqrt{784-384}}{2\left(-12\right)}$$

Add 784 to -384.

$$x=\frac{-\left(-28\right)±\sqrt{400}}{2\left(-12\right)}$$

Take the square root of 400.

$$x=\frac{-\left(-28\right)±20}{2\left(-12\right)}$$

The opposite of -28 is 28.

$$x=\frac{28±20}{2\left(-12\right)}$$

Multiply 2 times -12.

$$x=\frac{28±20}{-24}$$

Now solve the equation x=\frac{28±20}{-24} when ± is plus. Add 28 to 20.

$$x=\frac{48}{-24}$$

Divide 48 by -24.

$$x=-2$$

Now solve the equation x=\frac{28±20}{-24} when ± is minus. Subtract 20 from 28.

$$x=\frac{8}{-24}$$

Reduce the fraction \frac{8}{-24} to lowest terms by extracting and canceling out 8.

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

The equation is now solved.

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

Steps for Completing the Square

The opposite of -1 is 1.

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

Add -\frac{3}{4} and 1 to get \frac{1}{4}.

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

Multiply -\frac{3}{4} and -1 to get \frac{3}{4}.

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

Express \frac{\frac{1}{4}}{1+\frac{3}{4}x} as a single fraction.

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

The opposite of -\frac{3}{4} is \frac{3}{4}.

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

Factor the expressions that are not already factored in \frac{x+\frac{3}{4}}{1-\frac{3}{4}x}.

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

To divide powers of the same base, subtract the numerator's exponent from the denominator's exponent.

$$\frac{1}{4\left(1+\frac{3}{4}x\right)}=\frac{4x+3}{\left(\frac{1}{4}\right)^{0}\left(-3x+4\right)}$$

Calculate \frac{1}{4} to the power of 0 and get 1.

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

Use the distributive property to multiply 4 by 1+\frac{3}{4}x.

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

Use the distributive property to multiply 1 by -3x+4.

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

Subtract \frac{4x+3}{-3x+4} from both sides.

$$\frac{1}{4+3x}-\frac{4x+3}{-3x+4}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4+3x and -3x+4 is \left(-3x+4\right)\left(3x+4\right). Multiply \frac{1}{4+3x} times \frac{-3x+4}{-3x+4}. Multiply \frac{4x+3}{-3x+4} times \frac{3x+4}{3x+4}.

$$\frac{-3x+4}{\left(-3x+4\right)\left(3x+4\right)}-\frac{\left(4x+3\right)\left(3x+4\right)}{\left(-3x+4\right)\left(3x+4\right)}=0$$

Since \frac{-3x+4}{\left(-3x+4\right)\left(3x+4\right)} and \frac{\left(4x+3\right)\left(3x+4\right)}{\left(-3x+4\right)\left(3x+4\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-3x+4-\left(4x+3\right)\left(3x+4\right)}{\left(-3x+4\right)\left(3x+4\right)}=0$$

Do the multiplications in -3x+4-\left(4x+3\right)\left(3x+4\right).

$$\frac{-3x+4-12x^{2}-16x-9x-12}{\left(-3x+4\right)\left(3x+4\right)}=0$$

Combine like terms in -3x+4-12x^{2}-16x-9x-12.

$$\frac{-28x-8-12x^{2}}{\left(-3x+4\right)\left(3x+4\right)}=0$$

Variable x cannot be equal to any of the values -\frac{4}{3},\frac{4}{3} since division by zero is not defined. Multiply both sides of the equation by \left(-3x-4\right)\left(3x-4\right).

$$-28x-8-12x^{2}=0$$

Add 8 to both sides. Anything plus zero gives itself.

$$-28x-12x^{2}=8$$

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.

$$-12x^{2}-28x=8$$

Divide both sides by -12.

$$\frac{-12x^{2}-28x}{-12}=\frac{8}{-12}$$

Dividing by -12 undoes the multiplication by -12.

$$x^{2}+\left(-\frac{28}{-12}\right)x=\frac{8}{-12}$$

Reduce the fraction \frac{-28}{-12} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{7}{3}x=\frac{8}{-12}$$

Reduce the fraction \frac{8}{-12} to lowest terms by extracting and canceling out 4.

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

Divide \frac{7}{3}, the coefficient of the x term, by 2 to get \frac{7}{6}. Then add the square of \frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{3}x+\left(\frac{7}{6}\right)^{2}=-\frac{2}{3}+\left(\frac{7}{6}\right)^{2}$$

Square \frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=-\frac{2}{3}+\frac{49}{36}$$

Add -\frac{2}{3} to \frac{49}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=\frac{25}{36}$$

Factor x^{2}+\frac{7}{3}x+\frac{49}{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}.

$$\left(x+\frac{7}{6}\right)^{2}=\frac{25}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{6}\right)^{2}}=\sqrt{\frac{25}{36}}$$

Simplify.

$$x+\frac{7}{6}=\frac{5}{6} x+\frac{7}{6}=-\frac{5}{6}$$

Subtract \frac{7}{6} from both sides of the equation.

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