Solve {l}{5=a-b+c}{0=0+0+c}{-4=4a+2b+c}

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

Solve for a, b, c

$$a=1$$

$$b=-4$$

$$c=0$$

Short Steps Using Substitution

Solve 5=a-b+c for a.

$$a=b+5-c$$

Substitute b+5-c for a in the equation -4=4a+2b+c.

$$-4=4\left(b+5-c\right)+2b+c$$

Solve the second equation for c and the third equation for b.

$$c=0 b=-4+\frac{1}{2}c$$

Substitute 0 for c in the equation b=-4+\frac{1}{2}c.

$$b=-4+\frac{1}{2}\times 0$$

Calculate b from b=-4+\frac{1}{2}\times 0.

$$b=-4$$

Substitute 0 for c and -4 for b in the equation a=b+5-c.

$$a=-4+5-0$$

Calculate a from a=-4+5-0.

$$a=1$$

The system is now solved.

$$a=1 b=-4 c=0$$