Solve {l}{5a-b-2c=7}{4b-4-2c=9}{7c-2a-2b=0}

$$\left. \begin{array} { l } { 5 a - b - 2 c = 7 } \\ { 4 b - 4 - 2 c = 9 } \\ { 7 c - 2 a - 2 b = 0 } \end{array} \right.$$

Solve for a, b, c

$$a = \frac{311}{100} = 3\frac{11}{100} = 3.11$$

$$b = \frac{431}{100} = 4\frac{31}{100} = 4.31$$

$$c = \frac{53}{25} = 2\frac{3}{25} = 2.12$$

Short Steps Using Substitution

Solve 5a-b-2c=7 for b.

$$b=5a-2c-7$$

Substitute 5a-2c-7 for b in the second and third equation.

$$4\left(5a-2c-7\right)-4-2c=9 7c-2a-2\left(5a-2c-7\right)=0$$

Solve these equations for a and c respectively.

$$a=\frac{41}{20}+\frac{1}{2}c c=-\frac{14}{11}+\frac{12}{11}a$$

Substitute \frac{41}{20}+\frac{1}{2}c for a in the equation c=-\frac{14}{11}+\frac{12}{11}a.

$$c=-\frac{14}{11}+\frac{12}{11}\left(\frac{41}{20}+\frac{1}{2}c\right)$$

Solve c=-\frac{14}{11}+\frac{12}{11}\left(\frac{41}{20}+\frac{1}{2}c\right) for c.

$$c=\frac{53}{25}$$

Substitute \frac{53}{25} for c in the equation a=\frac{41}{20}+\frac{1}{2}c.

$$a=\frac{41}{20}+\frac{1}{2}\times \frac{53}{25}$$

Calculate a from a=\frac{41}{20}+\frac{1}{2}\times \frac{53}{25}.

$$a=\frac{311}{100}$$

Substitute \frac{311}{100} for a and \frac{53}{25} for c in the equation b=5a-2c-7.

$$b=5\times \frac{311}{100}-2\times \frac{53}{25}-7$$

Calculate b from b=5\times \frac{311}{100}-2\times \frac{53}{25}-7.

$$b=\frac{431}{100}$$

The system is now solved.

$$a=\frac{311}{100} b=\frac{431}{100} c=\frac{53}{25}$$