Solve 8x-1/10x^2-19x+6+x-1/10x-4=2x+1/4x-6

$$\frac { 8 x - 1 } { 10 x ^ { 2 } - 19 x + 6 } + \frac { x - 1 } { 10 x - 4 } = \frac { 2 x + 1 } { 4 x - 6 }$$

Solve for x

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

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values \frac{2}{5},\frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by 2\left(2x-3\right)\left(5x-2\right), the least common multiple of 10x^{2}-19x+6,10x-4,4x-6.

$$2\left(8x-1\right)+\left(2x-3\right)\left(x-1\right)=\left(5x-2\right)\left(2x+1\right)$$

Use the distributive property to multiply 2 by 8x-1.

$$16x-2+\left(2x-3\right)\left(x-1\right)=\left(5x-2\right)\left(2x+1\right)$$

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

$$16x-2+2x^{2}-5x+3=\left(5x-2\right)\left(2x+1\right)$$

Combine 16x and -5x to get 11x.

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

Add -2 and 3 to get 1.

$$11x+1+2x^{2}=\left(5x-2\right)\left(2x+1\right)$$

Use the distributive property to multiply 5x-2 by 2x+1 and combine like terms.

$$11x+1+2x^{2}=10x^{2}+x-2$$

Subtract 10x^{2} from both sides.

$$11x+1+2x^{2}-10x^{2}=x-2$$

Combine 2x^{2} and -10x^{2} to get -8x^{2}.

$$11x+1-8x^{2}=x-2$$

Subtract x from both sides.

$$11x+1-8x^{2}-x=-2$$

Combine 11x and -x to get 10x.

$$10x+1-8x^{2}=-2$$

Add 2 to both sides.

$$10x+1-8x^{2}+2=0$$

Add 1 and 2 to get 3.

$$10x+3-8x^{2}=0$$

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

$$-8x^{2}+10x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -8x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=10 ab=-8\times 3=-24$$

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 -24.

$$-1,24 -2,12 -3,8 -4,6$$

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 10.

$$a=12 b=-2$$

Rewrite -8x^{2}+10x+3 as \left(-8x^{2}+12x\right)+\left(-2x+3\right).

$$\left(-8x^{2}+12x\right)+\left(-2x+3\right)$$

Factor out -4x in the first and -1 in the second group.

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

Factor out common term 2x-3 by using distributive property.

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

To find equation solutions, solve 2x-3=0 and -4x-1=0.

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

Variable x cannot be equal to \frac{3}{2}.

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

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values \frac{2}{5},\frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by 2\left(2x-3\right)\left(5x-2\right), the least common multiple of 10x^{2}-19x+6,10x-4,4x-6.

$$2\left(8x-1\right)+\left(2x-3\right)\left(x-1\right)=\left(5x-2\right)\left(2x+1\right)$$

Use the distributive property to multiply 2 by 8x-1.

$$16x-2+\left(2x-3\right)\left(x-1\right)=\left(5x-2\right)\left(2x+1\right)$$

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

$$16x-2+2x^{2}-5x+3=\left(5x-2\right)\left(2x+1\right)$$

Combine 16x and -5x to get 11x.

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

Add -2 and 3 to get 1.

$$11x+1+2x^{2}=\left(5x-2\right)\left(2x+1\right)$$

Use the distributive property to multiply 5x-2 by 2x+1 and combine like terms.

$$11x+1+2x^{2}=10x^{2}+x-2$$

Subtract 10x^{2} from both sides.

$$11x+1+2x^{2}-10x^{2}=x-2$$

Combine 2x^{2} and -10x^{2} to get -8x^{2}.

$$11x+1-8x^{2}=x-2$$

Subtract x from both sides.

$$11x+1-8x^{2}-x=-2$$

Combine 11x and -x to get 10x.

$$10x+1-8x^{2}=-2$$

Add 2 to both sides.

$$10x+1-8x^{2}+2=0$$

Add 1 and 2 to get 3.

$$10x+3-8x^{2}=0$$

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.

$$-8x^{2}+10x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, 10 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-8\right)\times 3}}{2\left(-8\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-8\right)\times 3}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$x=\frac{-10±\sqrt{100+32\times 3}}{2\left(-8\right)}$$

Multiply 32 times 3.

$$x=\frac{-10±\sqrt{100+96}}{2\left(-8\right)}$$

Add 100 to 96.

$$x=\frac{-10±\sqrt{196}}{2\left(-8\right)}$$

Take the square root of 196.

$$x=\frac{-10±14}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{-10±14}{-16}$$

Now solve the equation x=\frac{-10±14}{-16} when ± is plus. Add -10 to 14.

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

Reduce the fraction \frac{4}{-16} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-10±14}{-16} when ± is minus. Subtract 14 from -10.

$$x=-\frac{24}{-16}$$

Reduce the fraction \frac{-24}{-16} to lowest terms by extracting and canceling out 8.

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

The equation is now solved.

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

Variable x cannot be equal to \frac{3}{2}.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values \frac{2}{5},\frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by 2\left(2x-3\right)\left(5x-2\right), the least common multiple of 10x^{2}-19x+6,10x-4,4x-6.

$$2\left(8x-1\right)+\left(2x-3\right)\left(x-1\right)=\left(5x-2\right)\left(2x+1\right)$$

Use the distributive property to multiply 2 by 8x-1.

$$16x-2+\left(2x-3\right)\left(x-1\right)=\left(5x-2\right)\left(2x+1\right)$$

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

$$16x-2+2x^{2}-5x+3=\left(5x-2\right)\left(2x+1\right)$$

Combine 16x and -5x to get 11x.

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

Add -2 and 3 to get 1.

$$11x+1+2x^{2}=\left(5x-2\right)\left(2x+1\right)$$

Use the distributive property to multiply 5x-2 by 2x+1 and combine like terms.

$$11x+1+2x^{2}=10x^{2}+x-2$$

Subtract 10x^{2} from both sides.

$$11x+1+2x^{2}-10x^{2}=x-2$$

Combine 2x^{2} and -10x^{2} to get -8x^{2}.

$$11x+1-8x^{2}=x-2$$

Subtract x from both sides.

$$11x+1-8x^{2}-x=-2$$

Combine 11x and -x to get 10x.

$$10x+1-8x^{2}=-2$$

Subtract 1 from both sides.

$$10x-8x^{2}=-2-1$$

Subtract 1 from -2 to get -3.

$$10x-8x^{2}=-3$$

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.

$$-8x^{2}+10x=-3$$

Divide both sides by -8.

$$\frac{-8x^{2}+10x}{-8}=-\frac{3}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$x^{2}+\frac{10}{-8}x=-\frac{3}{-8}$$

Reduce the fraction \frac{10}{-8} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{5}{4}x=-\frac{3}{-8}$$

Divide -3 by -8.

$$x^{2}-\frac{5}{4}x=\frac{3}{8}$$

Divide -\frac{5}{4}, the coefficient of the x term, by 2 to get -\frac{5}{8}. Then add the square of -\frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{4}x+\left(-\frac{5}{8}\right)^{2}=\frac{3}{8}+\left(-\frac{5}{8}\right)^{2}$$

Square -\frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{4}x+\frac{25}{64}=\frac{3}{8}+\frac{25}{64}$$

Add \frac{3}{8} to \frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{4}x+\frac{25}{64}=\frac{49}{64}$$

Factor x^{2}-\frac{5}{4}x+\frac{25}{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}.

$$\left(x-\frac{5}{8}\right)^{2}=\frac{49}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{8}\right)^{2}}=\sqrt{\frac{49}{64}}$$

Simplify.

$$x-\frac{5}{8}=\frac{7}{8} x-\frac{5}{8}=-\frac{7}{8}$$

Add \frac{5}{8} to both sides of the equation.

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

Variable x cannot be equal to \frac{3}{2}.

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