Solve {l}{2a+b=1}{1+3b=0}

$$\left\{ \begin{array} { l } { 2 a + b = 1 } \\ { 1 + 3 b = 0 } \end{array} \right.$$

Solve for a, b

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

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

Solution Steps

Consider the second equation. Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$3b=-1$$

Divide both sides by 3.

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

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

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

Add \frac{1}{3} to both sides.

$$2a=1+\frac{1}{3}$$

Add 1 and \frac{1}{3} to get \frac{4}{3}.

$$2a=\frac{4}{3}$$

Divide both sides by 2.

$$a=\frac{\frac{4}{3}}{2}$$

Express \frac{\frac{4}{3}}{2} as a single fraction.

$$a=\frac{4}{3\times 2}$$

Multiply 3 and 2 to get 6.

$$a=\frac{4}{6}$$

Reduce the fraction \frac{4}{6} to lowest terms by extracting and canceling out 2.

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

The system is now solved.

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