Solve {l}{7=R-r_{1}}{5=R-r_{2}}{6=r_{1}+r_{2}}

$$\left\{ \begin{array} { l } { 7 = R - r _ { 1 } } \\ { 5 = R - r _ { 2 } } \\ { 6 = r _ { 1 } + r _ { 2 } } \end{array} \right.$$

Solve for R, r_1, r_2

$$R=9$$

$$r_{1}=2$$

$$r_{2}=4$$

Short Steps Using Substitution

Solve 7=R-r_{1} for R.

$$R=7+r_{1}$$

Substitute 7+r_{1} for R in the equation 5=R-r_{2}.

$$5=7+r_{1}-r_{2}$$

Solve the second equation for r_{1} and the third equation for r_{2}.

$$r_{1}=-2+r_{2} r_{2}=6-r_{1}$$

Substitute -2+r_{2} for r_{1} in the equation r_{2}=6-r_{1}.

$$r_{2}=6-\left(-2+r_{2}\right)$$

Solve r_{2}=6-\left(-2+r_{2}\right) for r_{2}.

$$r_{2}=4$$

Substitute 4 for r_{2} in the equation r_{1}=-2+r_{2}.

$$r_{1}=-2+4$$

Calculate r_{1} from r_{1}=-2+4.

$$r_{1}=2$$

Substitute 2 for r_{1} in the equation R=7+r_{1}.

$$R=7+2$$

Calculate R from R=7+2.

$$R=9$$

The system is now solved.

$$R=9 r_{1}=2 r_{2}=4$$