Solve {l}{4x+3=0}{2x+3y=12}

$$\left. \begin{array} { l } { 4 x + 3 = 0 } \\ { 2 x + 3 y = 12 } \end{array} \right.$$

Solve for x, y

$$x=-\frac{3}{4}=-0.75$$

$$y = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Solution Steps

Consider the first equation. Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$4x=-3$$

Divide both sides by 4.

$$x=-\frac{3}{4}$$

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

$$2\left(-\frac{3}{4}\right)+3y=12$$

Multiply 2 and -\frac{3}{4} to get -\frac{3}{2}.

$$-\frac{3}{2}+3y=12$$

Add \frac{3}{2} to both sides.

$$3y=12+\frac{3}{2}$$

Add 12 and \frac{3}{2} to get \frac{27}{2}.

$$3y=\frac{27}{2}$$

Divide both sides by 3.

$$y=\frac{\frac{27}{2}}{3}$$

Express \frac{\frac{27}{2}}{3} as a single fraction.

$$y=\frac{27}{2\times 3}$$

Multiply 2 and 3 to get 6.

$$y=\frac{27}{6}$$

Reduce the fraction \frac{27}{6} to lowest terms by extracting and canceling out 3.

$$y=\frac{9}{2}$$

The system is now solved.

$$x=-\frac{3}{4} y=\frac{9}{2}$$