Solve {l}{a+b+c=5}{4a-b+c=-4}{32a+8b+c=9}

$$\left. \begin{array} { l } { a + b + c = 5 } \\ { 4 a - b + c = - 4 } \\ { 32 a + 8 b + c = 9 } \end{array} \right.$$

Solve for a, b, c

$$a=-\frac{55}{83}\approx -0.662650602$$

$$b = \frac{291}{83} = 3\frac{42}{83} \approx 3.506024096$$

$$c = \frac{179}{83} = 2\frac{13}{83} \approx 2.156626506$$

Short Steps Using Substitution

Solve a+b+c=5 for a.

$$a=-b-c+5$$

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

$$4\left(-b-c+5\right)-b+c=-4 32\left(-b-c+5\right)+8b+c=9$$

Solve these equations for b and c respectively.

$$b=\frac{24}{5}-\frac{3}{5}c c=-\frac{24}{31}b+\frac{151}{31}$$

Substitute \frac{24}{5}-\frac{3}{5}c for b in the equation c=-\frac{24}{31}b+\frac{151}{31}.

$$c=-\frac{24}{31}\left(\frac{24}{5}-\frac{3}{5}c\right)+\frac{151}{31}$$

Solve c=-\frac{24}{31}\left(\frac{24}{5}-\frac{3}{5}c\right)+\frac{151}{31} for c.

$$c=\frac{179}{83}$$

Substitute \frac{179}{83} for c in the equation b=\frac{24}{5}-\frac{3}{5}c.

$$b=\frac{24}{5}-\frac{3}{5}\times \frac{179}{83}$$

Calculate b from b=\frac{24}{5}-\frac{3}{5}\times \frac{179}{83}.

$$b=\frac{291}{83}$$

Substitute \frac{291}{83} for b and \frac{179}{83} for c in the equation a=-b-c+5.

$$a=-\frac{291}{83}-\frac{179}{83}+5$$

Calculate a from a=-\frac{291}{83}-\frac{179}{83}+5.

$$a=-\frac{55}{83}$$

The system is now solved.

$$a=-\frac{55}{83} b=\frac{291}{83} c=\frac{179}{83}$$