Solve {l}{a+b+c=-2}{a-b+c=20}{9/4a+3/2b+c=a/9+b/3+c}

$$\left\{ \begin{array} { l } { a + b + c = - 2 } \\ { a - b + c = 20 } \\ { \frac { 9 } { 4 } a + \frac { 3 } { 2 } b + c = \frac { a } { 9 } + \frac { b } { 3 } + c } \end{array} \right.$$

Solve for a, b, c

$$a=6$$

$$b=-11$$

$$c=3$$

Short Steps Using Substitution

Multiply each equation by the least common multiple of denominators in it. Simplify.

$$a+b+c=-2 a-b+c=20 81a+54b+36c=4a+12b+36c$$

Solve a+b+c=-2 for a.

$$a=-b-c-2$$

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

$$-b-c-2-b+c=20 81\left(-b-c-2\right)+54b+36c=4\left(-b-c-2\right)+12b+36c$$

Solve these equations for b and c respectively.

$$b=-11 c=-2-\frac{5}{11}b$$

Substitute -11 for b in the equation c=-2-\frac{5}{11}b.

$$c=-2-\frac{5}{11}\left(-11\right)$$

Calculate c from c=-2-\frac{5}{11}\left(-11\right).

$$c=3$$

Substitute -11 for b and 3 for c in the equation a=-b-c-2.

$$a=-\left(-11\right)-3-2$$

Calculate a from a=-\left(-11\right)-3-2.

$$a=6$$

The system is now solved.

$$a=6 b=-11 c=3$$