Solve {l}{4/9A+B=14/3-0}{9A-21=B-B}

$$\left. \begin{array} { l } { \frac { 4 } { 9 } A + B = \frac { 14 } { 3 } - 0 } \\ { 9 A - 21 = B - B } \end{array} \right.$$

Solve for A, B

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

$$B = \frac{98}{27} = 3\frac{17}{27} \approx 3.62962963$$

Solution Steps

Consider the first equation. Subtract 0 from \frac{14}{3} to get \frac{14}{3}.

$$\frac{4}{9}A+B=\frac{14}{3}$$

Consider the second equation. Combine B and -B to get 0.

$$9A-21=0$$

Add 21 to both sides. Anything plus zero gives itself.

$$9A=21$$

Divide both sides by 9.

$$A=\frac{21}{9}$$

Reduce the fraction \frac{21}{9} to lowest terms by extracting and canceling out 3.

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

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

$$\frac{4}{9}\times \frac{7}{3}+B=\frac{14}{3}$$

Multiply \frac{4}{9} and \frac{7}{3} to get \frac{28}{27}.

$$\frac{28}{27}+B=\frac{14}{3}$$

Subtract \frac{28}{27} from both sides.

$$B=\frac{14}{3}-\frac{28}{27}$$

Subtract \frac{28}{27} from \frac{14}{3} to get \frac{98}{27}.

$$B=\frac{98}{27}$$

The system is now solved.

$$A=\frac{7}{3} B=\frac{98}{27}$$