Solve {l}{(1+1/2+1/2)x_{1}+y-10=0}{(1/6+1/6+frac{1}{3})x_{2}-frac{2}{3}+y=0}{frac{x_{1}-x_{2}}{2}=y}

$$\left. \begin{array} { l } { ( 1 + \frac { 1 } { 2 } + \frac { 1 } { 2 } ) x _ { 1 } + y - 10 = 0 } \\ { ( \frac { 1 } { 6 } + \frac { 1 } { 6 } + \frac { 1 } { 3 } ) x _ { 2 } - \frac { 2 } { 3 } + y = 0 } \\ { \frac { x _ { 1 } - x _ { 2 } } { 2 } = y } \end{array} \right.$$

Solve for x_1, y, x_2

$$y=4$$

$$x_{1}=3$$

$$x_{2}=-5$$

Short Steps Using Substitution

Multiply each equation by the least common multiple of denominators in it. Simplify.

$$4x_{1}+2y-20=0 4x_{2}-4+6y=0 x_{1}-x_{2}=2y$$

Reorder the equations.

$$x_{1}-x_{2}=2y 4x_{2}-4+6y=0 4x_{1}+2y-20=0$$

Solve x_{1}-x_{2}=2y for x_{1}.

$$x_{1}=x_{2}+2y$$

Substitute x_{2}+2y for x_{1} in the equation 4x_{1}+2y-20=0.

$$4\left(x_{2}+2y\right)+2y-20=0$$

Solve the second equation for y and the third equation for x_{2}.

$$y=\frac{2}{3}-\frac{2}{3}x_{2} x_{2}=5-\frac{5}{2}y$$

Substitute \frac{2}{3}-\frac{2}{3}x_{2} for y in the equation x_{2}=5-\frac{5}{2}y.

$$x_{2}=5-\frac{5}{2}\left(\frac{2}{3}-\frac{2}{3}x_{2}\right)$$

Solve x_{2}=5-\frac{5}{2}\left(\frac{2}{3}-\frac{2}{3}x_{2}\right) for x_{2}.

$$x_{2}=-5$$

Substitute -5 for x_{2} in the equation y=\frac{2}{3}-\frac{2}{3}x_{2}.

$$y=\frac{2}{3}-\frac{2}{3}\left(-5\right)$$

Calculate y from y=\frac{2}{3}-\frac{2}{3}\left(-5\right).

$$y=4$$

Substitute 4 for y and -5 for x_{2} in the equation x_{1}=x_{2}+2y.

$$x_{1}=-5+2\times 4$$

Calculate x_{1} from x_{1}=-5+2\times 4.

$$x_{1}=3$$

The system is now solved.

$$x_{1}=3 y=4 x_{2}=-5$$