Solve {l}{2=16a-8b+4c}{0=-32a+12b-45}{2=-80a+16b}

$$\left\{ \begin{array} { l } { 2 = 16 a - 8 b + 4 c } \\ { 0 = - 32 a + 12 b - 45 } \\ { 2 = - 80 a + 16 b } \end{array} \right.$$

Solve for a, b, c

$$a = \frac{87}{56} = 1\frac{31}{56} \approx 1.553571429$$

$$b = \frac{221}{28} = 7\frac{25}{28} \approx 7.892857143$$

$$c = \frac{141}{14} = 10\frac{1}{14} \approx 10.071428571$$

Short Steps Using Substitution

Solve 2=16a-8b+4c for a.

$$a=\frac{1}{8}+\frac{1}{2}b-\frac{1}{4}c$$

Substitute \frac{1}{8}+\frac{1}{2}b-\frac{1}{4}c for a in the second and third equation.

$$0=-32\left(\frac{1}{8}+\frac{1}{2}b-\frac{1}{4}c\right)+12b-45 2=-80\left(\frac{1}{8}+\frac{1}{2}b-\frac{1}{4}c\right)+16b$$

Solve these equations for b and c respectively.

$$b=-\frac{49}{4}+2c c=\frac{3}{5}+\frac{6}{5}b$$

Substitute -\frac{49}{4}+2c for b in the equation c=\frac{3}{5}+\frac{6}{5}b.

$$c=\frac{3}{5}+\frac{6}{5}\left(-\frac{49}{4}+2c\right)$$

Solve c=\frac{3}{5}+\frac{6}{5}\left(-\frac{49}{4}+2c\right) for c.

$$c=\frac{141}{14}$$

Substitute \frac{141}{14} for c in the equation b=-\frac{49}{4}+2c.

$$b=-\frac{49}{4}+2\times \frac{141}{14}$$

Calculate b from b=-\frac{49}{4}+2\times \frac{141}{14}.

$$b=\frac{221}{28}$$

Substitute \frac{221}{28} for b and \frac{141}{14} for c in the equation a=\frac{1}{8}+\frac{1}{2}b-\frac{1}{4}c.

$$a=\frac{1}{8}+\frac{1}{2}\times \frac{221}{28}-\frac{1}{4}\times \frac{141}{14}$$

Calculate a from a=\frac{1}{8}+\frac{1}{2}\times \frac{221}{28}-\frac{1}{4}\times \frac{141}{14}.

$$a=\frac{87}{56}$$

The system is now solved.

$$a=\frac{87}{56} b=\frac{221}{28} c=\frac{141}{14}$$