Solve {l}{2a-b=1}{2b-c=1}{2a+c=1}

$$\left\{ \begin{array} { l } { 2 a - b = 1 } \\ { 2 b - c = 1 } \\ { 2 a + c = 1 } \end{array} \right.$$

Solve for a, b, c

$$a=\frac{2}{3}\approx 0.666666667$$

$$b=\frac{1}{3}\approx 0.333333333$$

$$c=-\frac{1}{3}\approx -0.333333333$$

Short Steps Using Substitution

Solve 2a-b=1 for b.

$$b=2a-1$$

Substitute 2a-1 for b in the equation 2b-c=1.

$$2\left(2a-1\right)-c=1$$

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

$$a=\frac{3}{4}+\frac{1}{4}c c=-2a+1$$

Substitute \frac{3}{4}+\frac{1}{4}c for a in the equation c=-2a+1.

$$c=-2\left(\frac{3}{4}+\frac{1}{4}c\right)+1$$

Solve c=-2\left(\frac{3}{4}+\frac{1}{4}c\right)+1 for c.

$$c=-\frac{1}{3}$$

Substitute -\frac{1}{3} for c in the equation a=\frac{3}{4}+\frac{1}{4}c.

$$a=\frac{3}{4}+\frac{1}{4}\left(-\frac{1}{3}\right)$$

Calculate a from a=\frac{3}{4}+\frac{1}{4}\left(-\frac{1}{3}\right).

$$a=\frac{2}{3}$$

Substitute \frac{2}{3} for a in the equation b=2a-1.

$$b=2\times \frac{2}{3}-1$$

Calculate b from b=2\times \frac{2}{3}-1.

$$b=\frac{1}{3}$$

The system is now solved.

$$a=\frac{2}{3} b=\frac{1}{3} c=-\frac{1}{3}$$