Solve {l}{2a-3b+2c=20}{-4a+2b+4c=-8}{3a-b-3c=5}

$$\left. \begin{array} { l } { 2 a - 3 b + 2 c = 20 } \\ { - 4 a + 2 b + 4 c = - 8 } \\ { 3 a - b - 3 c = 5 } \end{array} \right.$$

Solve for a, b, c

$$a=4$$

$$b=-2$$

$$c=3$$

Short Steps Using Substitution

Reorder the equations.

$$3a-b-3c=5 -4a+2b+4c=-8 2a-3b+2c=20$$

Solve 3a-b-3c=5 for b.

$$b=3a-3c-5$$

Substitute 3a-3c-5 for b in the second and third equation.

$$-4a+2\left(3a-3c-5\right)+4c=-8 2a-3\left(3a-3c-5\right)+2c=20$$

Solve these equations for a and c respectively.

$$a=1+c c=\frac{7}{11}a+\frac{5}{11}$$

Substitute 1+c for a in the equation c=\frac{7}{11}a+\frac{5}{11}.

$$c=\frac{7}{11}\left(1+c\right)+\frac{5}{11}$$

Solve c=\frac{7}{11}\left(1+c\right)+\frac{5}{11} for c.

$$c=3$$

Substitute 3 for c in the equation a=1+c.

$$a=1+3$$

Calculate a from a=1+3.

$$a=4$$

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

$$b=3\times 4-3\times 3-5$$

Calculate b from b=3\times 4-3\times 3-5.

$$b=-2$$

The system is now solved.

$$a=4 b=-2 c=3$$