Solve {l}{3x+4y-8=0}{4y+12=0}

$$\left\{ \begin{array} { l } { 3 x + 4 y - 8 = 0 } \\ { 4 y + 12 = 0 } \end{array} \right.$$

Solve for x, y

$$x = \frac{20}{3} = 6\frac{2}{3} \approx 6.666666667$$

$$y=-3$$

Solution Steps

Consider the second equation. Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$4y=-12$$

Divide both sides by 4.

$$y=\frac{-12}{4}$$

Divide -12 by 4 to get -3.

$$y=-3$$

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

$$3x+4\left(-3\right)-8=0$$

Multiply 4 and -3 to get -12.

$$3x-12-8=0$$

Subtract 8 from -12 to get -20.

$$3x-20=0$$

Add 20 to both sides. Anything plus zero gives itself.

$$3x=20$$

Divide both sides by 3.

$$x=\frac{20}{3}$$

The system is now solved.

$$x=\frac{20}{3} y=-3$$