Solve {l}{text{StationI:Solvetheratio}}{2x}{2x/5-x^2-x-10/x^2+8x+15=3/x+3}

$$\left. \begin{array} { l } { \text { Station I: Solve the ratio } } \\ { 2 x } \\ { \frac { 2 x } { 5 } - \frac { x ^ { 2 } - x - 10 } { x ^ { 2 } + 8 x + 15 } = \frac { 3 } { x + 3 } } \end{array} \right.$$

Solve for x

$$x=-7$$

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -5,-3 since division by zero is not defined. Multiply both sides of the equation by \left(x+3\right)\left(x+5\right), the least common multiple of x^{2}+8x+15,x+3.

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

Use the distributive property to multiply x+3 by x+5 and combine like terms.

$$\left(x^{2}+8x+15\right)\times 5-\left(x^{2}-x-10\right)=\left(x+5\right)\times 3$$

Use the distributive property to multiply x^{2}+8x+15 by 5.

$$5x^{2}+40x+75-\left(x^{2}-x-10\right)=\left(x+5\right)\times 3$$

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

$$5x^{2}+40x+75-x^{2}+x+10=\left(x+5\right)\times 3$$

Combine 5x^{2} and -x^{2} to get 4x^{2}.

$$4x^{2}+40x+75+x+10=\left(x+5\right)\times 3$$

Combine 40x and x to get 41x.

$$4x^{2}+41x+75+10=\left(x+5\right)\times 3$$

Add 75 and 10 to get 85.

$$4x^{2}+41x+85=\left(x+5\right)\times 3$$

Use the distributive property to multiply x+5 by 3.

$$4x^{2}+41x+85=3x+15$$

Subtract 3x from both sides.

$$4x^{2}+41x+85-3x=15$$

Combine 41x and -3x to get 38x.

$$4x^{2}+38x+85=15$$

Subtract 15 from both sides.

$$4x^{2}+38x+85-15=0$$

Subtract 15 from 85 to get 70.

$$4x^{2}+38x+70=0$$

Divide both sides by 2.

$$2x^{2}+19x+35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+35. To find a and b, set up a system to be solved.

$$a+b=19 ab=2\times 35=70$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 70.

$$1,70 2,35 5,14 7,10$$

Calculate the sum for each pair.

$$1+70=71 2+35=37 5+14=19 7+10=17$$

The solution is the pair that gives sum 19.

$$a=5 b=14$$

Rewrite 2x^{2}+19x+35 as \left(2x^{2}+5x\right)+\left(14x+35\right).

$$\left(2x^{2}+5x\right)+\left(14x+35\right)$$

Factor out x in the first and 7 in the second group.

$$x\left(2x+5\right)+7\left(2x+5\right)$$

Factor out common term 2x+5 by using distributive property.

$$\left(2x+5\right)\left(x+7\right)$$

To find equation solutions, solve 2x+5=0 and x+7=0.

$$x=-\frac{5}{2} x=-7$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -5,-3 since division by zero is not defined. Multiply both sides of the equation by \left(x+3\right)\left(x+5\right), the least common multiple of x^{2}+8x+15,x+3.

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

Use the distributive property to multiply x+3 by x+5 and combine like terms.

$$\left(x^{2}+8x+15\right)\times 5-\left(x^{2}-x-10\right)=\left(x+5\right)\times 3$$

Use the distributive property to multiply x^{2}+8x+15 by 5.

$$5x^{2}+40x+75-\left(x^{2}-x-10\right)=\left(x+5\right)\times 3$$

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

$$5x^{2}+40x+75-x^{2}+x+10=\left(x+5\right)\times 3$$

Combine 5x^{2} and -x^{2} to get 4x^{2}.

$$4x^{2}+40x+75+x+10=\left(x+5\right)\times 3$$

Combine 40x and x to get 41x.

$$4x^{2}+41x+75+10=\left(x+5\right)\times 3$$

Add 75 and 10 to get 85.

$$4x^{2}+41x+85=\left(x+5\right)\times 3$$

Use the distributive property to multiply x+5 by 3.

$$4x^{2}+41x+85=3x+15$$

Subtract 3x from both sides.

$$4x^{2}+41x+85-3x=15$$

Combine 41x and -3x to get 38x.

$$4x^{2}+38x+85=15$$

Subtract 15 from both sides.

$$4x^{2}+38x+85-15=0$$

Subtract 15 from 85 to get 70.

$$4x^{2}+38x+70=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 38 for b, and 70 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-38±\sqrt{38^{2}-4\times 4\times 70}}{2\times 4}$$

Square 38.

$$x=\frac{-38±\sqrt{1444-4\times 4\times 70}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-38±\sqrt{1444-16\times 70}}{2\times 4}$$

Multiply -16 times 70.

$$x=\frac{-38±\sqrt{1444-1120}}{2\times 4}$$

Add 1444 to -1120.

$$x=\frac{-38±\sqrt{324}}{2\times 4}$$

Take the square root of 324.

$$x=\frac{-38±18}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-38±18}{8}$$

Now solve the equation x=\frac{-38±18}{8} when ± is plus. Add -38 to 18.

$$x=-\frac{20}{8}$$

Reduce the fraction \frac{-20}{8} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-38±18}{8} when ± is minus. Subtract 18 from -38.

$$x=-\frac{56}{8}$$

Divide -56 by 8.

$$x=-7$$

The equation is now solved.

$$x=-\frac{5}{2} x=-7$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -5,-3 since division by zero is not defined. Multiply both sides of the equation by \left(x+3\right)\left(x+5\right), the least common multiple of x^{2}+8x+15,x+3.

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

Use the distributive property to multiply x+3 by x+5 and combine like terms.

$$\left(x^{2}+8x+15\right)\times 5-\left(x^{2}-x-10\right)=\left(x+5\right)\times 3$$

Use the distributive property to multiply x^{2}+8x+15 by 5.

$$5x^{2}+40x+75-\left(x^{2}-x-10\right)=\left(x+5\right)\times 3$$

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

$$5x^{2}+40x+75-x^{2}+x+10=\left(x+5\right)\times 3$$

Combine 5x^{2} and -x^{2} to get 4x^{2}.

$$4x^{2}+40x+75+x+10=\left(x+5\right)\times 3$$

Combine 40x and x to get 41x.

$$4x^{2}+41x+75+10=\left(x+5\right)\times 3$$

Add 75 and 10 to get 85.

$$4x^{2}+41x+85=\left(x+5\right)\times 3$$

Use the distributive property to multiply x+5 by 3.

$$4x^{2}+41x+85=3x+15$$

Subtract 3x from both sides.

$$4x^{2}+41x+85-3x=15$$

Combine 41x and -3x to get 38x.

$$4x^{2}+38x+85=15$$

Subtract 85 from both sides.

$$4x^{2}+38x=15-85$$

Subtract 85 from 15 to get -70.

$$4x^{2}+38x=-70$$

Divide both sides by 4.

$$\frac{4x^{2}+38x}{4}=-\frac{70}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{38}{4}x=-\frac{70}{4}$$

Reduce the fraction \frac{38}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{19}{2}x=-\frac{70}{4}$$

Reduce the fraction \frac{-70}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{19}{2}x=-\frac{35}{2}$$

Divide \frac{19}{2}, the coefficient of the x term, by 2 to get \frac{19}{4}. Then add the square of \frac{19}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{19}{2}x+\left(\frac{19}{4}\right)^{2}=-\frac{35}{2}+\left(\frac{19}{4}\right)^{2}$$

Square \frac{19}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{19}{2}x+\frac{361}{16}=-\frac{35}{2}+\frac{361}{16}$$

Add -\frac{35}{2} to \frac{361}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{19}{2}x+\frac{361}{16}=\frac{81}{16}$$

Factor x^{2}+\frac{19}{2}x+\frac{361}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{19}{4}\right)^{2}=\frac{81}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{19}{4}\right)^{2}}=\sqrt{\frac{81}{16}}$$

Simplify.

$$x+\frac{19}{4}=\frac{9}{4} x+\frac{19}{4}=-\frac{9}{4}$$

Subtract \frac{19}{4} from both sides of the equation.

$$x=-\frac{5}{2} x=-7$$