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

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

Solve for x

$$x=\frac{1}{4}=0.25$$

$$x=-2$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values \frac{1}{5},\frac{3}{4} since division by zero is not defined. Multiply both sides of the equation by \left(4x-3\right)\left(5x-1\right), the least common multiple of 4x-3,5x-1.

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

Use the distributive property to multiply 5x-1 by 2.

$$-\left(10x-2\right)=\left(4x-3\right)x$$

To find the opposite of 10x-2, find the opposite of each term.

$$-10x+2=\left(4x-3\right)x$$

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

$$-10x+2=4x^{2}-3x$$

Subtract 4x^{2} from both sides.

$$-10x+2-4x^{2}=-3x$$

Add 3x to both sides.

$$-10x+2-4x^{2}+3x=0$$

Combine -10x and 3x to get -7x.

$$-7x+2-4x^{2}=0$$

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

$$-4x^{2}-7x+2=0$$

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

$$a+b=-7 ab=-4\times 2=-8$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -8.

$$1,-8 2,-4$$

Calculate the sum for each pair.

$$1-8=-7 2-4=-2$$

The solution is the pair that gives sum -7.

$$a=1 b=-8$$

Rewrite -4x^{2}-7x+2 as \left(-4x^{2}+x\right)+\left(-8x+2\right).

$$\left(-4x^{2}+x\right)+\left(-8x+2\right)$$

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

$$-x\left(4x-1\right)-2\left(4x-1\right)$$

Factor out common term 4x-1 by using distributive property.

$$\left(4x-1\right)\left(-x-2\right)$$

To find equation solutions, solve 4x-1=0 and -x-2=0.

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

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values \frac{1}{5},\frac{3}{4} since division by zero is not defined. Multiply both sides of the equation by \left(4x-3\right)\left(5x-1\right), the least common multiple of 4x-3,5x-1.

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

Use the distributive property to multiply 5x-1 by 2.

$$-\left(10x-2\right)=\left(4x-3\right)x$$

To find the opposite of 10x-2, find the opposite of each term.

$$-10x+2=\left(4x-3\right)x$$

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

$$-10x+2=4x^{2}-3x$$

Subtract 4x^{2} from both sides.

$$-10x+2-4x^{2}=-3x$$

Add 3x to both sides.

$$-10x+2-4x^{2}+3x=0$$

Combine -10x and 3x to get -7x.

$$-7x+2-4x^{2}=0$$

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.

$$-4x^{2}-7x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -7 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-4\right)\times 2}}{2\left(-4\right)}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-4\right)\times 2}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-7\right)±\sqrt{49+16\times 2}}{2\left(-4\right)}$$

Multiply 16 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49+32}}{2\left(-4\right)}$$

Add 49 to 32.

$$x=\frac{-\left(-7\right)±\sqrt{81}}{2\left(-4\right)}$$

Take the square root of 81.

$$x=\frac{-\left(-7\right)±9}{2\left(-4\right)}$$

The opposite of -7 is 7.

$$x=\frac{7±9}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{7±9}{-8}$$

Now solve the equation x=\frac{7±9}{-8} when ± is plus. Add 7 to 9.

$$x=\frac{16}{-8}$$

Divide 16 by -8.

$$x=-2$$

Now solve the equation x=\frac{7±9}{-8} when ± is minus. Subtract 9 from 7.

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

Reduce the fraction \frac{-2}{-8} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values \frac{1}{5},\frac{3}{4} since division by zero is not defined. Multiply both sides of the equation by \left(4x-3\right)\left(5x-1\right), the least common multiple of 4x-3,5x-1.

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

Use the distributive property to multiply 5x-1 by 2.

$$-\left(10x-2\right)=\left(4x-3\right)x$$

To find the opposite of 10x-2, find the opposite of each term.

$$-10x+2=\left(4x-3\right)x$$

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

$$-10x+2=4x^{2}-3x$$

Subtract 4x^{2} from both sides.

$$-10x+2-4x^{2}=-3x$$

Add 3x to both sides.

$$-10x+2-4x^{2}+3x=0$$

Combine -10x and 3x to get -7x.

$$-7x+2-4x^{2}=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$-7x-4x^{2}=-2$$

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.

$$-4x^{2}-7x=-2$$

Divide both sides by -4.

$$\frac{-4x^{2}-7x}{-4}=-\frac{2}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\left(-\frac{7}{-4}\right)x=-\frac{2}{-4}$$

Divide -7 by -4.

$$x^{2}+\frac{7}{4}x=-\frac{2}{-4}$$

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

$$x^{2}+\frac{7}{4}x=\frac{1}{2}$$

Divide \frac{7}{4}, the coefficient of the x term, by 2 to get \frac{7}{8}. Then add the square of \frac{7}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{4}x+\left(\frac{7}{8}\right)^{2}=\frac{1}{2}+\left(\frac{7}{8}\right)^{2}$$

Square \frac{7}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{4}x+\frac{49}{64}=\frac{1}{2}+\frac{49}{64}$$

Add \frac{1}{2} to \frac{49}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{4}x+\frac{49}{64}=\frac{81}{64}$$

Factor x^{2}+\frac{7}{4}x+\frac{49}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{8}\right)^{2}=\frac{81}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{8}\right)^{2}}=\sqrt{\frac{81}{64}}$$

Simplify.

$$x+\frac{7}{8}=\frac{9}{8} x+\frac{7}{8}=-\frac{9}{8}$$

Subtract \frac{7}{8} from both sides of the equation.

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