Solve {l}{a+2b+c=1}{2a-b=3}{a+2c=+5}

$$\left\{ \begin{array} { l } { a + 2 b + c = 1 } \\ { 2 a - b = 3 } \\ { a + 2 c = + 5 } \end{array} \right.$$

Solve for a, b, c

$$a=1$$

$$b=-1$$

$$c=2$$

Short Steps Using Substitution

Solve a+2b+c=1 for a.

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

Substitute -2b-c+1 for a in the second and third equation.

$$2\left(-2b-c+1\right)-b=3 -2b-c+1+2c=5$$

Solve these equations for b and c respectively.

$$b=-\frac{2}{5}c-\frac{1}{5} c=4+2b$$

Substitute -\frac{2}{5}c-\frac{1}{5} for b in the equation c=4+2b.

$$c=4+2\left(-\frac{2}{5}c-\frac{1}{5}\right)$$

Solve c=4+2\left(-\frac{2}{5}c-\frac{1}{5}\right) for c.

$$c=2$$

Substitute 2 for c in the equation b=-\frac{2}{5}c-\frac{1}{5}.

$$b=-\frac{2}{5}\times 2-\frac{1}{5}$$

Calculate b from b=-\frac{2}{5}\times 2-\frac{1}{5}.

$$b=-1$$

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

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

Calculate a from a=-2\left(-1\right)-2+1.

$$a=1$$

The system is now solved.

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