Solve 5-4/v-2=-4/(v-1)(v-2)

$$5 - \frac { 4 } { v - 2 } = - \frac { 4 } { ( v - 1 ) ( v - 2 ) }$$

Solve for v

$$v = \frac{9}{5} = 1\frac{4}{5} = 1.8$$

Steps Using the Quadratic Formula

Variable v cannot be equal to any of the values 1,2 since division by zero is not defined. Multiply both sides of the equation by \left(v-2\right)\left(v-1\right), the least common multiple of v-2,\left(v-1\right)\left(v-2\right).

$$\left(v-2\right)\left(v-1\right)\times 5-\left(v-1\right)\times 4=-4$$

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

$$\left(v^{2}-3v+2\right)\times 5-\left(v-1\right)\times 4=-4$$

Use the distributive property to multiply v^{2}-3v+2 by 5.

$$5v^{2}-15v+10-\left(v-1\right)\times 4=-4$$

Use the distributive property to multiply v-1 by 4.

$$5v^{2}-15v+10-\left(4v-4\right)=-4$$

To find the opposite of 4v-4, find the opposite of each term.

$$5v^{2}-15v+10-4v+4=-4$$

Combine -15v and -4v to get -19v.

$$5v^{2}-19v+10+4=-4$$

Add 10 and 4 to get 14.

$$5v^{2}-19v+14=-4$$

Add 4 to both sides.

$$5v^{2}-19v+14+4=0$$

Add 14 and 4 to get 18.

$$5v^{2}-19v+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -19 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$v=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 5\times 18}}{2\times 5}$$

Square -19.

$$v=\frac{-\left(-19\right)±\sqrt{361-4\times 5\times 18}}{2\times 5}$$

Multiply -4 times 5.

$$v=\frac{-\left(-19\right)±\sqrt{361-20\times 18}}{2\times 5}$$

Multiply -20 times 18.

$$v=\frac{-\left(-19\right)±\sqrt{361-360}}{2\times 5}$$

Add 361 to -360.

$$v=\frac{-\left(-19\right)±\sqrt{1}}{2\times 5}$$

Take the square root of 1.

$$v=\frac{-\left(-19\right)±1}{2\times 5}$$

The opposite of -19 is 19.

$$v=\frac{19±1}{2\times 5}$$

Multiply 2 times 5.

$$v=\frac{19±1}{10}$$

Now solve the equation v=\frac{19±1}{10} when ± is plus. Add 19 to 1.

$$v=\frac{20}{10}$$

Divide 20 by 10.

$$v=2$$

Now solve the equation v=\frac{19±1}{10} when ± is minus. Subtract 1 from 19.

$$v=\frac{18}{10}$$

Reduce the fraction \frac{18}{10} to lowest terms by extracting and canceling out 2.

$$v=\frac{9}{5}$$

The equation is now solved.

$$v=2 v=\frac{9}{5}$$

Variable v cannot be equal to 2.

$$v=\frac{9}{5}$$

Steps for Completing the Square

Variable v cannot be equal to any of the values 1,2 since division by zero is not defined. Multiply both sides of the equation by \left(v-2\right)\left(v-1\right), the least common multiple of v-2,\left(v-1\right)\left(v-2\right).

$$\left(v-2\right)\left(v-1\right)\times 5-\left(v-1\right)\times 4=-4$$

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

$$\left(v^{2}-3v+2\right)\times 5-\left(v-1\right)\times 4=-4$$

Use the distributive property to multiply v^{2}-3v+2 by 5.

$$5v^{2}-15v+10-\left(v-1\right)\times 4=-4$$

Use the distributive property to multiply v-1 by 4.

$$5v^{2}-15v+10-\left(4v-4\right)=-4$$

To find the opposite of 4v-4, find the opposite of each term.

$$5v^{2}-15v+10-4v+4=-4$$

Combine -15v and -4v to get -19v.

$$5v^{2}-19v+10+4=-4$$

Add 10 and 4 to get 14.

$$5v^{2}-19v+14=-4$$

Subtract 14 from both sides.

$$5v^{2}-19v=-4-14$$

Subtract 14 from -4 to get -18.

$$5v^{2}-19v=-18$$

Divide both sides by 5.

$$\frac{5v^{2}-19v}{5}=-\frac{18}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$v^{2}-\frac{19}{5}v=-\frac{18}{5}$$

Divide -\frac{19}{5}, the coefficient of the x term, by 2 to get -\frac{19}{10}. Then add the square of -\frac{19}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$v^{2}-\frac{19}{5}v+\left(-\frac{19}{10}\right)^{2}=-\frac{18}{5}+\left(-\frac{19}{10}\right)^{2}$$

Square -\frac{19}{10} by squaring both the numerator and the denominator of the fraction.

$$v^{2}-\frac{19}{5}v+\frac{361}{100}=-\frac{18}{5}+\frac{361}{100}$$

Add -\frac{18}{5} to \frac{361}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$v^{2}-\frac{19}{5}v+\frac{361}{100}=\frac{1}{100}$$

Factor v^{2}-\frac{19}{5}v+\frac{361}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(v-\frac{19}{10}\right)^{2}=\frac{1}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(v-\frac{19}{10}\right)^{2}}=\sqrt{\frac{1}{100}}$$

Simplify.

$$v-\frac{19}{10}=\frac{1}{10} v-\frac{19}{10}=-\frac{1}{10}$$

Add \frac{19}{10} to both sides of the equation.

$$v=2 v=\frac{9}{5}$$

Variable v cannot be equal to 2.

$$v=\frac{9}{5}$$