Solve {l}{3x+4y+z=0}{2x+2y+1=0}{4x+1=3}

$$\left\{ \begin{array} { l } { 3 x + 4 y + z = 0 } \\ { 2 x + 2 y + 1 = 0 } \\ { 4 x + 1 = 3 } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{1}{2}=0.5$$

$$y=-1$$

$$z = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Solution Steps

Consider the third equation. Subtract 1 from both sides.

$$4x=3-1$$

Subtract 1 from 3 to get 2.

$$4x=2$$

Divide both sides by 4.

$$x=\frac{2}{4}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{2}$$

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

$$2\times \frac{1}{2}+2y+1=0$$

Multiply 2 and \frac{1}{2} to get 1.

$$1+2y+1=0$$

Add 1 and 1 to get 2.

$$2+2y=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$2y=-2$$

Divide both sides by 2.

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

Divide -2 by 2 to get -1.

$$y=-1$$

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

$$3\times \frac{1}{2}+4\left(-1\right)+z=0$$

Multiply 3 and \frac{1}{2} to get \frac{3}{2}.

$$\frac{3}{2}+4\left(-1\right)+z=0$$

Multiply 4 and -1 to get -4.

$$\frac{3}{2}-4+z=0$$

Subtract 4 from \frac{3}{2} to get -\frac{5}{2}.

$$-\frac{5}{2}+z=0$$

Add \frac{5}{2} to both sides. Anything plus zero gives itself.

$$z=\frac{5}{2}$$

The system is now solved.

$$x=\frac{1}{2} y=-1 z=\frac{5}{2}$$