Solve {c}{2a+3b+c=22}{a+b+c=15}{3a+2b+2c=33}

$$\left. \begin{array} { c } { 2 a + 3 b + c = 22 } \\ { a + b + c = 15 } \\ { 3 a + 2 b + 2 c = 33 } \end{array} \right.$$

Solve for a, b, c

$$a=3$$

$$b=2$$

$$c=10$$

Short Steps Using Substitution

Solve 2a+3b+c=22 for c.

$$c=-2a-3b+22$$

Substitute -2a-3b+22 for c in the second and third equation.

$$a+b-2a-3b+22=15 3a+2b+2\left(-2a-3b+22\right)=33$$

Solve these equations for b and a respectively.

$$b=-\frac{1}{2}a+\frac{7}{2} a=-4b+11$$

Substitute -\frac{1}{2}a+\frac{7}{2} for b in the equation a=-4b+11.

$$a=-4\left(-\frac{1}{2}a+\frac{7}{2}\right)+11$$

Solve a=-4\left(-\frac{1}{2}a+\frac{7}{2}\right)+11 for a.

$$a=3$$

Substitute 3 for a in the equation b=-\frac{1}{2}a+\frac{7}{2}.

$$b=-\frac{1}{2}\times 3+\frac{7}{2}$$

Calculate b from b=-\frac{1}{2}\times 3+\frac{7}{2}.

$$b=2$$

Substitute 2 for b and 3 for a in the equation c=-2a-3b+22.

$$c=-2\times 3-3\times 2+22$$

Calculate c from c=-2\times 3-3\times 2+22.

$$c=10$$

The system is now solved.

$$a=3 b=2 c=10$$