Solve {l}{5=4a-2b+c}{0=a-b+c}{0=9a+3b+c}

$$\left\{ \begin{array} { l } { 5 = 4 a - 2 b + c } \\ { 0 = a - b + c } \\ { 0 = 9 a + 3 b + c } \end{array} \right.$$

Solve for a, b, c

$$a=1$$

$$b=-2$$

$$c=-3$$

Short Steps Using Substitution

Solve 5=4a-2b+c for c.

$$c=-4a+5+2b$$

Substitute -4a+5+2b for c in the second and third equation.

$$0=a-b-4a+5+2b 0=9a+3b-4a+5+2b$$

Solve these equations for b and a respectively.

$$b=3a-5 a=-1-b$$

Substitute 3a-5 for b in the equation a=-1-b.

$$a=-1-\left(3a-5\right)$$

Solve a=-1-\left(3a-5\right) for a.

$$a=1$$

Substitute 1 for a in the equation b=3a-5.

$$b=3\times 1-5$$

Calculate b from b=3\times 1-5.

$$b=-2$$

Substitute -2 for b and 1 for a in the equation c=-4a+5+2b.

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

Calculate c from c=-4+5+2\left(-2\right).

$$c=-3$$

The system is now solved.

$$a=1 b=-2 c=-3$$