Solve {l}{3x+2y=5}{4y-3c=7}{9x+7c=21}

$$\left. \begin{array} { l } { 3 x + 2 y = 5 } \\ { 4 y - 3 c = 7 } \\ { 9 x + 7 c = 21 } \end{array} \right.$$

Solve for x, y, c

$$x = -\frac{14}{5} = -2\frac{4}{5} = -2.8$$

$$y = \frac{67}{10} = 6\frac{7}{10} = 6.7$$

$$c = \frac{33}{5} = 6\frac{3}{5} = 6.6$$

Short Steps Using Substitution

Solve 3x+2y=5 for x.

$$x=-\frac{2}{3}y+\frac{5}{3}$$

Substitute -\frac{2}{3}y+\frac{5}{3} for x in the equation 9x+7c=21.

$$9\left(-\frac{2}{3}y+\frac{5}{3}\right)+7c=21$$

Solve the second equation for y and the third equation for c.

$$y=\frac{3}{4}c+\frac{7}{4} c=\frac{6}{7}+\frac{6}{7}y$$

Substitute \frac{3}{4}c+\frac{7}{4} for y in the equation c=\frac{6}{7}+\frac{6}{7}y.

$$c=\frac{6}{7}+\frac{6}{7}\left(\frac{3}{4}c+\frac{7}{4}\right)$$

Solve c=\frac{6}{7}+\frac{6}{7}\left(\frac{3}{4}c+\frac{7}{4}\right) for c.

$$c=\frac{33}{5}$$

Substitute \frac{33}{5} for c in the equation y=\frac{3}{4}c+\frac{7}{4}.

$$y=\frac{3}{4}\times \frac{33}{5}+\frac{7}{4}$$

Calculate y from y=\frac{3}{4}\times \frac{33}{5}+\frac{7}{4}.

$$y=\frac{67}{10}$$

Substitute \frac{67}{10} for y in the equation x=-\frac{2}{3}y+\frac{5}{3}.

$$x=-\frac{2}{3}\times \frac{67}{10}+\frac{5}{3}$$

Calculate x from x=-\frac{2}{3}\times \frac{67}{10}+\frac{5}{3}.

$$x=-\frac{14}{5}$$

The system is now solved.

$$x=-\frac{14}{5} y=\frac{67}{10} c=\frac{33}{5}$$