Solve 1/v+1/v-10=1/12

$$\frac { 1 } { v } + \frac { 1 } { v - 10 } = \frac { 1 } { 12 }$$

Solve for v

$$v=4$$

$$v=30$$

Steps Using Factoring By Grouping

Variable v cannot be equal to any of the values 0,10 since division by zero is not defined. Multiply both sides of the equation by 12v\left(v-10\right), the least common multiple of v,v-10,12.

$$12v-120+12v=v\left(v-10\right)$$

Combine 12v and 12v to get 24v.

$$24v-120=v\left(v-10\right)$$

Use the distributive property to multiply v by v-10.

$$24v-120=v^{2}-10v$$

Subtract v^{2} from both sides.

$$24v-120-v^{2}=-10v$$

Add 10v to both sides.

$$24v-120-v^{2}+10v=0$$

Combine 24v and 10v to get 34v.

$$34v-120-v^{2}=0$$

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

$$-v^{2}+34v-120=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -v^{2}+av+bv-120. To find a and b, set up a system to be solved.

$$a+b=34 ab=-\left(-120\right)=120$$

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 120.

$$1,120 2,60 3,40 4,30 5,24 6,20 8,15 10,12$$

Calculate the sum for each pair.

$$1+120=121 2+60=62 3+40=43 4+30=34 5+24=29 6+20=26 8+15=23 10+12=22$$

The solution is the pair that gives sum 34.

$$a=30 b=4$$

Rewrite -v^{2}+34v-120 as \left(-v^{2}+30v\right)+\left(4v-120\right).

$$\left(-v^{2}+30v\right)+\left(4v-120\right)$$

Factor out -v in the first and 4 in the second group.

$$-v\left(v-30\right)+4\left(v-30\right)$$

Factor out common term v-30 by using distributive property.

$$\left(v-30\right)\left(-v+4\right)$$

To find equation solutions, solve v-30=0 and -v+4=0.

$$v=30 v=4$$

Steps Using the Quadratic Formula

Variable v cannot be equal to any of the values 0,10 since division by zero is not defined. Multiply both sides of the equation by 12v\left(v-10\right), the least common multiple of v,v-10,12.

$$12v-120+12v=v\left(v-10\right)$$

Combine 12v and 12v to get 24v.

$$24v-120=v\left(v-10\right)$$

Use the distributive property to multiply v by v-10.

$$24v-120=v^{2}-10v$$

Subtract v^{2} from both sides.

$$24v-120-v^{2}=-10v$$

Add 10v to both sides.

$$24v-120-v^{2}+10v=0$$

Combine 24v and 10v to get 34v.

$$34v-120-v^{2}=0$$

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.

$$-v^{2}+34v-120=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 34 for b, and -120 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$v=\frac{-34±\sqrt{34^{2}-4\left(-1\right)\left(-120\right)}}{2\left(-1\right)}$$

Square 34.

$$v=\frac{-34±\sqrt{1156-4\left(-1\right)\left(-120\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$v=\frac{-34±\sqrt{1156+4\left(-120\right)}}{2\left(-1\right)}$$

Multiply 4 times -120.

$$v=\frac{-34±\sqrt{1156-480}}{2\left(-1\right)}$$

Add 1156 to -480.

$$v=\frac{-34±\sqrt{676}}{2\left(-1\right)}$$

Take the square root of 676.

$$v=\frac{-34±26}{2\left(-1\right)}$$

Multiply 2 times -1.

$$v=\frac{-34±26}{-2}$$

Now solve the equation v=\frac{-34±26}{-2} when ± is plus. Add -34 to 26.

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

Divide -8 by -2.

$$v=4$$

Now solve the equation v=\frac{-34±26}{-2} when ± is minus. Subtract 26 from -34.

$$v=-\frac{60}{-2}$$

Divide -60 by -2.

$$v=30$$

The equation is now solved.

$$v=4 v=30$$

Steps for Completing the Square

Variable v cannot be equal to any of the values 0,10 since division by zero is not defined. Multiply both sides of the equation by 12v\left(v-10\right), the least common multiple of v,v-10,12.

$$12v-120+12v=v\left(v-10\right)$$

Combine 12v and 12v to get 24v.

$$24v-120=v\left(v-10\right)$$

Use the distributive property to multiply v by v-10.

$$24v-120=v^{2}-10v$$

Subtract v^{2} from both sides.

$$24v-120-v^{2}=-10v$$

Add 10v to both sides.

$$24v-120-v^{2}+10v=0$$

Combine 24v and 10v to get 34v.

$$34v-120-v^{2}=0$$

Add 120 to both sides. Anything plus zero gives itself.

$$34v-v^{2}=120$$

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.

$$-v^{2}+34v=120$$

Divide both sides by -1.

$$\frac{-v^{2}+34v}{-1}=\frac{120}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$v^{2}+\frac{34}{-1}v=\frac{120}{-1}$$

Divide 34 by -1.

$$v^{2}-34v=\frac{120}{-1}$$

Divide 120 by -1.

$$v^{2}-34v=-120$$

Divide -34, the coefficient of the x term, by 2 to get -17. Then add the square of -17 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$v^{2}-34v+\left(-17\right)^{2}=-120+\left(-17\right)^{2}$$

Square -17.

$$v^{2}-34v+289=-120+289$$

Add -120 to 289.

$$v^{2}-34v+289=169$$

Factor v^{2}-34v+289. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(v-17\right)^{2}=169$$

Take the square root of both sides of the equation.

$$\sqrt{\left(v-17\right)^{2}}=\sqrt{169}$$

Simplify.

$$v-17=13 v-17=-13$$

Add 17 to both sides of the equation.

$$v=30 v=4$$