Solve {l}{0=a-b+c}{0=9a+3b+c}{c=3/2}

$$\left\{ \begin{array} { l } { 0 = a - b + c } \\ { 0 = 9 a + 3 b + c } \\ { c = \frac { 3 } { 2 } } \end{array} \right.$$

Solve for a, b, c

$$a=-\frac{1}{2}=-0.5$$

$$b=1$$

$$c = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Short Steps Using Substitution

Reorder the equations.

$$c=\frac{3}{2} 0=9a+3b+c 0=a-b+c$$

Substitute \frac{3}{2} for c in the second and third equation.

$$0=9a+3b+\frac{3}{2} 0=a-b+\frac{3}{2}$$

Solve these equations for b and a respectively.

$$b=-\frac{1}{2}-3a a=b-\frac{3}{2}$$

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

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

Solve a=-\frac{1}{2}-3a-\frac{3}{2} for a.

$$a=-\frac{1}{2}$$

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

$$b=-\frac{1}{2}-3\left(-\frac{1}{2}\right)$$

Calculate b from b=-\frac{1}{2}-3\left(-\frac{1}{2}\right).

$$b=1$$

The system is now solved.

$$a=-\frac{1}{2} b=1 c=\frac{3}{2}$$