Solve {l}{25a+5b+c=60}{a-b+c=0}{4a+2b+c=3}

$$\left\{ \begin{array} { l } { 25 a + 5 b + c = 60 } \\ { a - b + c = 0 } \\ { 4 a + 2 b + c = 3 } \end{array} \right.$$

Solve for a, b, c

$$a=3$$

$$b=-2$$

$$c=-5$$

Short Steps Using Substitution

Solve 25a+5b+c=60 for c.

$$c=60-25a-5b$$

Substitute 60-25a-5b for c in the second and third equation.

$$a-b+60-25a-5b=0 4a+2b+60-25a-5b=3$$

Solve these equations for b and a respectively.

$$b=10-4a a=\frac{19}{7}-\frac{1}{7}b$$

Substitute 10-4a for b in the equation a=\frac{19}{7}-\frac{1}{7}b.

$$a=\frac{19}{7}-\frac{1}{7}\left(10-4a\right)$$

Solve a=\frac{19}{7}-\frac{1}{7}\left(10-4a\right) for a.

$$a=3$$

Substitute 3 for a in the equation b=10-4a.

$$b=10-4\times 3$$

Calculate b from b=10-4\times 3.

$$b=-2$$

Substitute -2 for b and 3 for a in the equation c=60-25a-5b.

$$c=60-25\times 3-5\left(-2\right)$$

Calculate c from c=60-25\times 3-5\left(-2\right).

$$c=-5$$

The system is now solved.

$$a=3 b=-2 c=-5$$