Solve 4x+3y=15quadtext{II}3-4y=-8

$$4 x + 3 y = 15 \quad \text { II } 3 - 4 y = - 8$$

Solve for x, y

$$x = \frac{27}{16} = 1\frac{11}{16} = 1.6875$$

$$y = \frac{11}{4} = 2\frac{3}{4} = 2.75$$

Solution Steps

Consider the second equation. Subtract 3 from both sides.

$$-4y=-8-3$$

Subtract 3 from -8 to get -11.

$$-4y=-11$$

Divide both sides by -4.

$$y=\frac{-11}{-4}$$

Fraction \frac{-11}{-4} can be simplified to \frac{11}{4} by removing the negative sign from both the numerator and the denominator.

$$y=\frac{11}{4}$$

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

$$4x+3\times \frac{11}{4}=15$$

Multiply 3 and \frac{11}{4} to get \frac{33}{4}.

$$4x+\frac{33}{4}=15$$

Subtract \frac{33}{4} from both sides.

$$4x=15-\frac{33}{4}$$

Subtract \frac{33}{4} from 15 to get \frac{27}{4}.

$$4x=\frac{27}{4}$$

Divide both sides by 4.

$$x=\frac{\frac{27}{4}}{4}$$

Express \frac{\frac{27}{4}}{4} as a single fraction.

$$x=\frac{27}{4\times 4}$$

Multiply 4 and 4 to get 16.

$$x=\frac{27}{16}$$

The system is now solved.

$$x=\frac{27}{16} y=\frac{11}{4}$$