Solve 12[1+frac{v_{1}}{c}]=11[1+frac{v_{2}}{c}]

$$12 [ 1 + \frac { v _ { 1 } } { c } ] = 11 [ 1 + \frac { v _ { 2 } } { c } ]$$

Solve for v_1

$$v_{1}=\frac{11v_{2}-c}{12}$$

$$c\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by c.

$$12\left(1+\frac{v_{1}}{c}\right)c=11\left(1+\frac{v_{2}}{c}\right)c$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{c}{c}.

$$12\left(\frac{c}{c}+\frac{v_{1}}{c}\right)c=11\left(1+\frac{v_{2}}{c}\right)c$$

Since \frac{c}{c} and \frac{v_{1}}{c} have the same denominator, add them by adding their numerators.

$$12\times \frac{c+v_{1}}{c}c=11\left(1+\frac{v_{2}}{c}\right)c$$

Express 12\times \frac{c+v_{1}}{c} as a single fraction.

$$\frac{12\left(c+v_{1}\right)}{c}c=11\left(1+\frac{v_{2}}{c}\right)c$$

Express \frac{12\left(c+v_{1}\right)}{c}c as a single fraction.

$$\frac{12\left(c+v_{1}\right)c}{c}=11\left(1+\frac{v_{2}}{c}\right)c$$

Cancel out c in both numerator and denominator.

$$12\left(v_{1}+c\right)=11\left(1+\frac{v_{2}}{c}\right)c$$

Use the distributive property to multiply 12 by v_{1}+c.

$$12v_{1}+12c=11\left(1+\frac{v_{2}}{c}\right)c$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{c}{c}.

$$12v_{1}+12c=11\left(\frac{c}{c}+\frac{v_{2}}{c}\right)c$$

Since \frac{c}{c} and \frac{v_{2}}{c} have the same denominator, add them by adding their numerators.

$$12v_{1}+12c=11\times \frac{c+v_{2}}{c}c$$

Express 11\times \frac{c+v_{2}}{c} as a single fraction.

$$12v_{1}+12c=\frac{11\left(c+v_{2}\right)}{c}c$$

Express \frac{11\left(c+v_{2}\right)}{c}c as a single fraction.

$$12v_{1}+12c=\frac{11\left(c+v_{2}\right)c}{c}$$

Cancel out c in both numerator and denominator.

$$12v_{1}+12c=11\left(v_{2}+c\right)$$

Use the distributive property to multiply 11 by v_{2}+c.

$$12v_{1}+12c=11v_{2}+11c$$

Subtract 12c from both sides.

$$12v_{1}=11v_{2}+11c-12c$$

Combine 11c and -12c to get -c.

$$12v_{1}=11v_{2}-c$$

Divide both sides by 12.

$$\frac{12v_{1}}{12}=\frac{11v_{2}-c}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$v_{1}=\frac{11v_{2}-c}{12}$$

Solve for c

$$c=11v_{2}-12v_{1}$$

$$v_{1}\neq \frac{11v_{2}}{12}$$