Solve {l}{4a-b+5c=-6}{3a+3b-4c=30}{6a+2b-3c=33}

$$\left\{ \begin{array} { l } { 4 a - b + 5 c = - 6 } \\ { 3 a + 3 b - 4 c = 30 } \\ { 6 a + 2 b - 3 c = 33 } \end{array} \right.$$

Solve for a, b, c

$$a=3$$

$$b=3$$

$$c=-3$$

Short Steps Using Substitution

Solve 4a-b+5c=-6 for b.

$$b=4a+5c+6$$

Substitute 4a+5c+6 for b in the second and third equation.

$$3a+3\left(4a+5c+6\right)-4c=30 6a+2\left(4a+5c+6\right)-3c=33$$

Solve these equations for a and c respectively.

$$a=-\frac{11}{15}c+\frac{4}{5} c=3-2a$$

Substitute -\frac{11}{15}c+\frac{4}{5} for a in the equation c=3-2a.

$$c=3-2\left(-\frac{11}{15}c+\frac{4}{5}\right)$$

Solve c=3-2\left(-\frac{11}{15}c+\frac{4}{5}\right) for c.

$$c=-3$$

Substitute -3 for c in the equation a=-\frac{11}{15}c+\frac{4}{5}.

$$a=-\frac{11}{15}\left(-3\right)+\frac{4}{5}$$

Calculate a from a=-\frac{11}{15}\left(-3\right)+\frac{4}{5}.

$$a=3$$

Substitute 3 for a and -3 for c in the equation b=4a+5c+6.

$$b=4\times 3+5\left(-3\right)+6$$

Calculate b from b=4\times 3+5\left(-3\right)+6.

$$b=3$$

The system is now solved.

$$a=3 b=3 c=-3$$