Solve {l}{a+2b+c=10}{2a+3b+5c=22}{3a+4b+7c=36}

$$\left. \begin{array} { l } { a + 2 b + c = 10 } \\ { 2 a + 3 b + 5 c = 22 } \\ { 3 a + 4 b + 7 c = 36 } \end{array} \right.$$

Solve for a, b, c

$$a=21$$

$$b=-5$$

$$c=-1$$

Short Steps Using Substitution

Solve a+2b+c=10 for a.

$$a=-2b-c+10$$

Substitute -2b-c+10 for a in the second and third equation.

$$2\left(-2b-c+10\right)+3b+5c=22 3\left(-2b-c+10\right)+4b+7c=36$$

Solve these equations for b and c respectively.

$$b=3c-2 c=\frac{3}{2}+\frac{1}{2}b$$

Substitute 3c-2 for b in the equation c=\frac{3}{2}+\frac{1}{2}b.

$$c=\frac{3}{2}+\frac{1}{2}\left(3c-2\right)$$

Solve c=\frac{3}{2}+\frac{1}{2}\left(3c-2\right) for c.

$$c=-1$$

Substitute -1 for c in the equation b=3c-2.

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

Calculate b from b=3\left(-1\right)-2.

$$b=-5$$

Substitute -5 for b and -1 for c in the equation a=-2b-c+10.

$$a=-2\left(-5\right)-\left(-1\right)+10$$

Calculate a from a=-2\left(-5\right)-\left(-1\right)+10.

$$a=21$$

The system is now solved.

$$a=21 b=-5 c=-1$$