Solve {c}{-12+3/5x+(7/15+R)y=0}{-12+3/5x+14/15z+6=0}{x=y+z}{z=3}

$$\left. \begin{array} { c } { - 12 + \frac { 3 } { 5 } x + ( \frac { 7 } { 15 } + R ) y = 0 } \\ { - 12 + \frac { 3 } { 5 } x + \frac { 14 } { 15 } z + 6 = 0 } \\ { x = y + z } \\ { z = 3 } \end{array} \right.$$

Solve for x, R, y, z

$$x = \frac{16}{3} = 5\frac{1}{3} \approx 5.333333333$$

$$y = \frac{7}{3} = 2\frac{1}{3} \approx 2.333333333$$

$$z=3$$

$$R = \frac{347}{105} = 3\frac{32}{105} \approx 3.304761905$$

Solution Steps

Consider the second equation. Insert the known values of variables into the equation.

$$-12+\frac{3}{5}x+\frac{14}{15}\times 3+6=0$$

Multiply \frac{14}{15} and 3 to get \frac{14}{5}.

$$-12+\frac{3}{5}x+\frac{14}{5}+6=0$$

Add -12 and \frac{14}{5} to get -\frac{46}{5}.

$$-\frac{46}{5}+\frac{3}{5}x+6=0$$

Add -\frac{46}{5} and 6 to get -\frac{16}{5}.

$$-\frac{16}{5}+\frac{3}{5}x=0$$

Add \frac{16}{5} to both sides. Anything plus zero gives itself.

$$\frac{3}{5}x=\frac{16}{5}$$

Multiply both sides by \frac{5}{3}, the reciprocal of \frac{3}{5}.

$$x=\frac{16}{5}\times \frac{5}{3}$$

Multiply \frac{16}{5} and \frac{5}{3} to get \frac{16}{3}.

$$x=\frac{16}{3}$$

Consider the third equation. Insert the known values of variables into the equation.

$$\frac{16}{3}=y+3$$

Swap sides so that all variable terms are on the left hand side.

$$y+3=\frac{16}{3}$$

Subtract 3 from both sides.

$$y=\frac{16}{3}-3$$

Subtract 3 from \frac{16}{3} to get \frac{7}{3}.

$$y=\frac{7}{3}$$

Consider the first equation. Insert the known values of variables into the equation.

$$-12+\frac{3}{5}\times \frac{16}{3}+\left(\frac{7}{15}+R\right)\times \frac{7}{3}=0$$

Multiply \frac{3}{5} and \frac{16}{3} to get \frac{16}{5}.

$$-12+\frac{16}{5}+\left(\frac{7}{15}+R\right)\times \frac{7}{3}=0$$

Add -12 and \frac{16}{5} to get -\frac{44}{5}.

$$-\frac{44}{5}+\left(\frac{7}{15}+R\right)\times \frac{7}{3}=0$$

Use the distributive property to multiply \frac{7}{15}+R by \frac{7}{3}.

$$-\frac{44}{5}+\frac{49}{45}+\frac{7}{3}R=0$$

Add -\frac{44}{5} and \frac{49}{45} to get -\frac{347}{45}.

$$-\frac{347}{45}+\frac{7}{3}R=0$$

Add \frac{347}{45} to both sides. Anything plus zero gives itself.

$$\frac{7}{3}R=\frac{347}{45}$$

Multiply both sides by \frac{3}{7}, the reciprocal of \frac{7}{3}.

$$R=\frac{347}{45}\times \frac{3}{7}$$

Multiply \frac{347}{45} and \frac{3}{7} to get \frac{347}{105}.

$$R=\frac{347}{105}$$

The system is now solved.

$$x=\frac{16}{3} R=\frac{347}{105} y=\frac{7}{3} z=3$$