Solve {l}{R_{2}+R_{1}=18}{R_{2}+R_{3}=10}{R_{3}+R_{1}=7}

$$\left. \begin{array} { l } { R _ { 2 } + R _ { 1 } = 18 } \\ { R _ { 2 } + R _ { 3 } = 10 } \\ { R _ { 3 } + R _ { 1 } = 7 } \end{array} \right.$$

Solve for R_2, R_1, R_3

$$R_{2} = \frac{21}{2} = 10\frac{1}{2} = 10.5$$

$$R_{1} = \frac{15}{2} = 7\frac{1}{2} = 7.5$$

$$R_{3}=-\frac{1}{2}=-0.5$$

Short Steps Using Substitution

Solve R_{2}+R_{1}=18 for R_{2}.

$$R_{2}=-R_{1}+18$$

Substitute -R_{1}+18 for R_{2} in the equation R_{2}+R_{3}=10.

$$-R_{1}+18+R_{3}=10$$

Solve the second equation for R_{1} and the third equation for R_{3}.

$$R_{1}=8+R_{3} R_{3}=-R_{1}+7$$

Substitute 8+R_{3} for R_{1} in the equation R_{3}=-R_{1}+7.

$$R_{3}=-\left(8+R_{3}\right)+7$$

Solve R_{3}=-\left(8+R_{3}\right)+7 for R_{3}.

$$R_{3}=-\frac{1}{2}$$

Substitute -\frac{1}{2} for R_{3} in the equation R_{1}=8+R_{3}.

$$R_{1}=8-\frac{1}{2}$$

Calculate R_{1} from R_{1}=8-\frac{1}{2}.

$$R_{1}=\frac{15}{2}$$

Substitute \frac{15}{2} for R_{1} in the equation R_{2}=-R_{1}+18.

$$R_{2}=-\frac{15}{2}+18$$

Calculate R_{2} from R_{2}=-\frac{15}{2}+18.

$$R_{2}=\frac{21}{2}$$

The system is now solved.

$$R_{2}=\frac{21}{2} R_{1}=\frac{15}{2} R_{3}=-\frac{1}{2}$$