Solve 40/a-30-80/a=2

$$\frac { 40 } { a - 30 } - \frac { 80 } { a } = 2$$

Solve for a

$$a=-30$$

$$a=40$$

Steps Using Factoring By Grouping

Variable a cannot be equal to any of the values 0,30 since division by zero is not defined. Multiply both sides of the equation by a\left(a-30\right), the least common multiple of a-30,a.

$$a\times 40-\left(a-30\right)\times 80=2a\left(a-30\right)$$

Use the distributive property to multiply a-30 by 80.

$$a\times 40-\left(80a-2400\right)=2a\left(a-30\right)$$

To find the opposite of 80a-2400, find the opposite of each term.

$$a\times 40-80a+2400=2a\left(a-30\right)$$

Combine a\times 40 and -80a to get -40a.

$$-40a+2400=2a\left(a-30\right)$$

Use the distributive property to multiply 2a by a-30.

$$-40a+2400=2a^{2}-60a$$

Subtract 2a^{2} from both sides.

$$-40a+2400-2a^{2}=-60a$$

Add 60a to both sides.

$$-40a+2400-2a^{2}+60a=0$$

Combine -40a and 60a to get 20a.

$$20a+2400-2a^{2}=0$$

Divide both sides by 2.

$$10a+1200-a^{2}=0$$

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

$$-a^{2}+10a+1200=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -a^{2}+aa+ba+1200. To find a and b, set up a system to be solved.

$$a+b=10 ab=-1200=-1200$$

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 -1200.

$$-1,1200 -2,600 -3,400 -4,300 -5,240 -6,200 -8,150 -10,120 -12,100 -15,80 -16,75 -20,60 -24,50 -25,48 -30,40$$

Calculate the sum for each pair.

$$-1+1200=1199 -2+600=598 -3+400=397 -4+300=296 -5+240=235 -6+200=194 -8+150=142 -10+120=110 -12+100=88 -15+80=65 -16+75=59 -20+60=40 -24+50=26 -25+48=23 -30+40=10$$

The solution is the pair that gives sum 10.

$$a=40 b=-30$$

Rewrite -a^{2}+10a+1200 as \left(-a^{2}+40a\right)+\left(-30a+1200\right).

$$\left(-a^{2}+40a\right)+\left(-30a+1200\right)$$

Factor out -a in the first and -30 in the second group.

$$-a\left(a-40\right)-30\left(a-40\right)$$

Factor out common term a-40 by using distributive property.

$$\left(a-40\right)\left(-a-30\right)$$

To find equation solutions, solve a-40=0 and -a-30=0.

$$a=40 a=-30$$

Steps Using the Quadratic Formula

Variable a cannot be equal to any of the values 0,30 since division by zero is not defined. Multiply both sides of the equation by a\left(a-30\right), the least common multiple of a-30,a.

$$a\times 40-\left(a-30\right)\times 80=2a\left(a-30\right)$$

Use the distributive property to multiply a-30 by 80.

$$a\times 40-\left(80a-2400\right)=2a\left(a-30\right)$$

To find the opposite of 80a-2400, find the opposite of each term.

$$a\times 40-80a+2400=2a\left(a-30\right)$$

Combine a\times 40 and -80a to get -40a.

$$-40a+2400=2a\left(a-30\right)$$

Use the distributive property to multiply 2a by a-30.

$$-40a+2400=2a^{2}-60a$$

Subtract 2a^{2} from both sides.

$$-40a+2400-2a^{2}=-60a$$

Add 60a to both sides.

$$-40a+2400-2a^{2}+60a=0$$

Combine -40a and 60a to get 20a.

$$20a+2400-2a^{2}=0$$

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.

$$-2a^{2}+20a+2400=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 20 for b, and 2400 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-20±\sqrt{20^{2}-4\left(-2\right)\times 2400}}{2\left(-2\right)}$$

Square 20.

$$a=\frac{-20±\sqrt{400-4\left(-2\right)\times 2400}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$a=\frac{-20±\sqrt{400+8\times 2400}}{2\left(-2\right)}$$

Multiply 8 times 2400.

$$a=\frac{-20±\sqrt{400+19200}}{2\left(-2\right)}$$

Add 400 to 19200.

$$a=\frac{-20±\sqrt{19600}}{2\left(-2\right)}$$

Take the square root of 19600.

$$a=\frac{-20±140}{2\left(-2\right)}$$

Multiply 2 times -2.

$$a=\frac{-20±140}{-4}$$

Now solve the equation a=\frac{-20±140}{-4} when ± is plus. Add -20 to 140.

$$a=\frac{120}{-4}$$

Divide 120 by -4.

$$a=-30$$

Now solve the equation a=\frac{-20±140}{-4} when ± is minus. Subtract 140 from -20.

$$a=-\frac{160}{-4}$$

Divide -160 by -4.

$$a=40$$

The equation is now solved.

$$a=-30 a=40$$

Steps for Completing the Square

Variable a cannot be equal to any of the values 0,30 since division by zero is not defined. Multiply both sides of the equation by a\left(a-30\right), the least common multiple of a-30,a.

$$a\times 40-\left(a-30\right)\times 80=2a\left(a-30\right)$$

Use the distributive property to multiply a-30 by 80.

$$a\times 40-\left(80a-2400\right)=2a\left(a-30\right)$$

To find the opposite of 80a-2400, find the opposite of each term.

$$a\times 40-80a+2400=2a\left(a-30\right)$$

Combine a\times 40 and -80a to get -40a.

$$-40a+2400=2a\left(a-30\right)$$

Use the distributive property to multiply 2a by a-30.

$$-40a+2400=2a^{2}-60a$$

Subtract 2a^{2} from both sides.

$$-40a+2400-2a^{2}=-60a$$

Add 60a to both sides.

$$-40a+2400-2a^{2}+60a=0$$

Combine -40a and 60a to get 20a.

$$20a+2400-2a^{2}=0$$

Subtract 2400 from both sides. Anything subtracted from zero gives its negation.

$$20a-2a^{2}=-2400$$

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.

$$-2a^{2}+20a=-2400$$

Divide both sides by -2.

$$\frac{-2a^{2}+20a}{-2}=-\frac{2400}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$a^{2}+\frac{20}{-2}a=-\frac{2400}{-2}$$

Divide 20 by -2.

$$a^{2}-10a=-\frac{2400}{-2}$$

Divide -2400 by -2.

$$a^{2}-10a=1200$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-10a+\left(-5\right)^{2}=1200+\left(-5\right)^{2}$$

Square -5.

$$a^{2}-10a+25=1200+25$$

Add 1200 to 25.

$$a^{2}-10a+25=1225$$

Factor a^{2}-10a+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-5\right)^{2}=1225$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-5\right)^{2}}=\sqrt{1225}$$

Simplify.

$$a-5=35 a-5=-35$$

Add 5 to both sides of the equation.

$$a=40 a=-30$$