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

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

Solve for a, b, c

$$a=3$$

$$b=-2$$

$$c=-5$$

Short Steps Using Substitution

Solve a-b+c=0 for a.

$$a=b-c$$

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

$$4\left(b-c\right)+2b+c=3 25\left(b-c\right)+5b+c=60$$

Solve these equations for b and c respectively.

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

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

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

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

$$c=-5$$

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

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

Calculate b from b=\frac{1}{2}+\frac{1}{2}\left(-5\right).

$$b=-2$$

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

$$a=-2-\left(-5\right)$$

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

$$a=3$$

The system is now solved.

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