Solve {l}{4a+b=1}{4a+2b+c=-1}{a+b+c=1}

$$\left. \begin{array} { l } { 4 a + b = 1 } \\ { 4 a + 2 b + c = - 1 } \\ { a + b + c = 1 } \end{array} \right.$$

Solve for a, b, c

$$a=3$$

$$b=-11$$

$$c=9$$

Short Steps Using Substitution

Solve 4a+b=1 for b.

$$b=-4a+1$$

Substitute -4a+1 for b in the second and third equation.

$$4a+2\left(-4a+1\right)+c=-1 a-4a+1+c=1$$

Solve these equations for a and c respectively.

$$a=\frac{3}{4}+\frac{1}{4}c c=3a$$

Substitute \frac{3}{4}+\frac{1}{4}c for a in the equation c=3a.

$$c=3\left(\frac{3}{4}+\frac{1}{4}c\right)$$

Solve c=3\left(\frac{3}{4}+\frac{1}{4}c\right) for c.

$$c=9$$

Substitute 9 for c in the equation a=\frac{3}{4}+\frac{1}{4}c.

$$a=\frac{3}{4}+\frac{1}{4}\times 9$$

Calculate a from a=\frac{3}{4}+\frac{1}{4}\times 9.

$$a=3$$

Substitute 3 for a in the equation b=-4a+1.

$$b=-4\times 3+1$$

Calculate b from b=-4\times 3+1.

$$b=-11$$

The system is now solved.

$$a=3 b=-11 c=9$$