Solve {l}{+74+5*a+10b=0}{-30+5c+10b=0}{b=a+c}

$$\left. \begin{array}{l}{ + 74 + 5 \times a + 10 b = 0 }\\{ - 30 + 5 c + 10 b = 0 }\\{ b = a + c }\end{array} \right.$$

Solve for a, b, c

$$a = -\frac{282}{25} = -11\frac{7}{25} = -11.28$$

$$b = -\frac{44}{25} = -1\frac{19}{25} = -1.76$$

$$c = \frac{238}{25} = 9\frac{13}{25} = 9.52$$

Short Steps Using Substitution

Reorder the equations.

$$b=a+c -30+5c+10b=0 74+5a+10b=0$$

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

$$-30+5c+10\left(a+c\right)=0 74+5a+10\left(a+c\right)=0$$

Solve these equations for a and c respectively.

$$a=3-\frac{3}{2}c c=-\frac{3}{2}a-\frac{37}{5}$$

Substitute 3-\frac{3}{2}c for a in the equation c=-\frac{3}{2}a-\frac{37}{5}.

$$c=-\frac{3}{2}\left(3-\frac{3}{2}c\right)-\frac{37}{5}$$

Solve c=-\frac{3}{2}\left(3-\frac{3}{2}c\right)-\frac{37}{5} for c.

$$c=\frac{238}{25}$$

Substitute \frac{238}{25} for c in the equation a=3-\frac{3}{2}c.

$$a=3-\frac{3}{2}\times \frac{238}{25}$$

Calculate a from a=3-\frac{3}{2}\times \frac{238}{25}.

$$a=-\frac{282}{25}$$

Substitute -\frac{282}{25} for a and \frac{238}{25} for c in the equation b=a+c.

$$b=-\frac{282}{25}+\frac{238}{25}$$

Calculate b from b=-\frac{282}{25}+\frac{238}{25}.

$$b=-\frac{44}{25}$$

The system is now solved.

$$a=-\frac{282}{25} b=-\frac{44}{25} c=\frac{238}{25}$$