Solve {l}{2a+3b-2c=18}{5a-6b+c=21}{4b-2c-6=0}

$$\left\{ \begin{array} { l } { 2 a + 3 b - 2 c = 18 } \\ { 5 a - 6 b + c = 21 } \\ { 4 b - 2 c - 6 = 0 } \end{array} \right.$$

Solve for a, b, c

$$a=8$$

$$b=4$$

$$c=5$$

Short Steps Using Substitution

Reorder the equations.

$$5a-6b+c=21 2a+3b-2c=18 4b-2c-6=0$$

Solve 5a-6b+c=21 for c.

$$c=-5a+6b+21$$

Substitute -5a+6b+21 for c in the second and third equation.

$$2a+3b-2\left(-5a+6b+21\right)=18 4b-2\left(-5a+6b+21\right)-6=0$$

Solve these equations for b and a respectively.

$$b=-\frac{20}{3}+\frac{4}{3}a a=\frac{24}{5}+\frac{4}{5}b$$

Substitute -\frac{20}{3}+\frac{4}{3}a for b in the equation a=\frac{24}{5}+\frac{4}{5}b.

$$a=\frac{24}{5}+\frac{4}{5}\left(-\frac{20}{3}+\frac{4}{3}a\right)$$

Solve a=\frac{24}{5}+\frac{4}{5}\left(-\frac{20}{3}+\frac{4}{3}a\right) for a.

$$a=8$$

Substitute 8 for a in the equation b=-\frac{20}{3}+\frac{4}{3}a.

$$b=-\frac{20}{3}+\frac{4}{3}\times 8$$

Calculate b from b=-\frac{20}{3}+\frac{4}{3}\times 8.

$$b=4$$

Substitute 4 for b and 8 for a in the equation c=-5a+6b+21.

$$c=-5\times 8+6\times 4+21$$

Calculate c from c=-5\times 8+6\times 4+21.

$$c=5$$

The system is now solved.

$$a=8 b=4 c=5$$