Solve 10x^2+59x-43/15x^2-19x+6=20/5x-3

$$\frac { 10 x ^ { 2 } + 59 x - 43 } { 15 x ^ { 2 } - 19 x + 6 } = \frac { 20 } { 5 x - 3 }$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values \frac{3}{5},\frac{2}{3} since division by zero is not defined. Multiply both sides of the equation by \left(3x-2\right)\left(5x-3\right), the least common multiple of 15x^{2}-19x+6,5x-3.

$$10x^{2}+59x-43=\left(3x-2\right)\times 20$$

Use the distributive property to multiply 3x-2 by 20.

$$10x^{2}+59x-43=60x-40$$

Subtract 60x from both sides.

$$10x^{2}+59x-43-60x=-40$$

Combine 59x and -60x to get -x.

$$10x^{2}-x-43=-40$$

Add 40 to both sides.

$$10x^{2}-x-43+40=0$$

Add -43 and 40 to get -3.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 10x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-1 ab=10\left(-3\right)=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -1.

$$a=-6 b=5$$

Rewrite 10x^{2}-x-3 as \left(10x^{2}-6x\right)+\left(5x-3\right).

$$\left(10x^{2}-6x\right)+\left(5x-3\right)$$

Factor out 2x in 10x^{2}-6x.

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

Factor out common term 5x-3 by using distributive property.

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

To find equation solutions, solve 5x-3=0 and 2x+1=0.

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

Variable x cannot be equal to \frac{3}{5}.

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

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values \frac{3}{5},\frac{2}{3} since division by zero is not defined. Multiply both sides of the equation by \left(3x-2\right)\left(5x-3\right), the least common multiple of 15x^{2}-19x+6,5x-3.

$$10x^{2}+59x-43=\left(3x-2\right)\times 20$$

Use the distributive property to multiply 3x-2 by 20.

$$10x^{2}+59x-43=60x-40$$

Subtract 60x from both sides.

$$10x^{2}+59x-43-60x=-40$$

Combine 59x and -60x to get -x.

$$10x^{2}-x-43=-40$$

Add 40 to both sides.

$$10x^{2}-x-43+40=0$$

Add -43 and 40 to get -3.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -1 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 10\left(-3\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-1\right)±\sqrt{1-40\left(-3\right)}}{2\times 10}$$

Multiply -40 times -3.

$$x=\frac{-\left(-1\right)±\sqrt{1+120}}{2\times 10}$$

Add 1 to 120.

$$x=\frac{-\left(-1\right)±\sqrt{121}}{2\times 10}$$

Take the square root of 121.

$$x=\frac{-\left(-1\right)±11}{2\times 10}$$

The opposite of -1 is 1.

$$x=\frac{1±11}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{1±11}{20}$$

Now solve the equation x=\frac{1±11}{20} when ± is plus. Add 1 to 11.

$$x=\frac{12}{20}$$

Reduce the fraction \frac{12}{20} to lowest terms by extracting and canceling out 4.

$$x=\frac{3}{5}$$

Now solve the equation x=\frac{1±11}{20} when ± is minus. Subtract 11 from 1.

$$x=-\frac{10}{20}$$

Reduce the fraction \frac{-10}{20} to lowest terms by extracting and canceling out 10.

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

The equation is now solved.

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

Variable x cannot be equal to \frac{3}{5}.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values \frac{3}{5},\frac{2}{3} since division by zero is not defined. Multiply both sides of the equation by \left(3x-2\right)\left(5x-3\right), the least common multiple of 15x^{2}-19x+6,5x-3.

$$10x^{2}+59x-43=\left(3x-2\right)\times 20$$

Use the distributive property to multiply 3x-2 by 20.

$$10x^{2}+59x-43=60x-40$$

Subtract 60x from both sides.

$$10x^{2}+59x-43-60x=-40$$

Combine 59x and -60x to get -x.

$$10x^{2}-x-43=-40$$

Add 43 to both sides.

$$10x^{2}-x=-40+43$$

Add -40 and 43 to get 3.

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

Divide both sides by 10.

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

Dividing by 10 undoes the multiplication by 10.

$$x^{2}-\frac{1}{10}x=\frac{3}{10}$$

Divide -\frac{1}{10}, the coefficient of the x term, by 2 to get -\frac{1}{20}. Then add the square of -\frac{1}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{10}x+\left(-\frac{1}{20}\right)^{2}=\frac{3}{10}+\left(-\frac{1}{20}\right)^{2}$$

Square -\frac{1}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{10}x+\frac{1}{400}=\frac{3}{10}+\frac{1}{400}$$

Add \frac{3}{10} to \frac{1}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{10}x+\frac{1}{400}=\frac{121}{400}$$

Factor x^{2}-\frac{1}{10}x+\frac{1}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{20}\right)^{2}=\frac{121}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{20}\right)^{2}}=\sqrt{\frac{121}{400}}$$

Simplify.

$$x-\frac{1}{20}=\frac{11}{20} x-\frac{1}{20}=-\frac{11}{20}$$

Add \frac{1}{20} to both sides of the equation.

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

Variable x cannot be equal to \frac{3}{5}.

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