Solve {c}{-08x+.13y=.09(310)}{x+y=310}

$$\left. \begin{array} { c } { - 08 x + .13 y = .09 ( 310 ) } \\ { x + y = 310 } \end{array} \right.$$

Solve for x, y

$$x = \frac{1240}{13} = 95\frac{5}{13} \approx 95.384615385$$

$$y = \frac{2790}{13} = 214\frac{8}{13} \approx 214.615384615$$

Solution Steps

Consider the first equation. Multiply 0 and 8 to get 0.

$$0x+0.13y=0.09\times 310$$

Anything times zero gives zero.

$$0+0.13y=0.09\times 310$$

Anything plus zero gives itself.

$$0.13y=0.09\times 310$$

Multiply 0.09 and 310 to get 27.9.

$$0.13y=27.9$$

Divide both sides by 0.13.

$$y=\frac{27.9}{0.13}$$

Expand \frac{27.9}{0.13} by multiplying both numerator and the denominator by 100.

$$y=\frac{2790}{13}$$

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

$$x+\frac{2790}{13}=310$$

Subtract \frac{2790}{13} from both sides.

$$x=310-\frac{2790}{13}$$

Subtract \frac{2790}{13} from 310 to get \frac{1240}{13}.

$$x=\frac{1240}{13}$$

The system is now solved.

$$x=\frac{1240}{13} y=\frac{2790}{13}$$