Solve {l}{34+4y=8}{5u+2y=8}

$$\left. \begin{array} { l } { 34 + 4 y = 8 } \\ { 5 u + 2 y = 8 } \end{array} \right.$$

Solve for y, u

$$y = -\frac{13}{2} = -6\frac{1}{2} = -6.5$$

$$u = \frac{21}{5} = 4\frac{1}{5} = 4.2$$

Solution Steps

Consider the first equation. Subtract 34 from both sides.

$$4y=8-34$$

Subtract 34 from 8 to get -26.

$$4y=-26$$

Divide both sides by 4.

$$y=\frac{-26}{4}$$

Reduce the fraction \frac{-26}{4} to lowest terms by extracting and canceling out 2.

$$y=-\frac{13}{2}$$

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

$$5u+2\left(-\frac{13}{2}\right)=8$$

Multiply 2 and -\frac{13}{2} to get -13.

$$5u-13=8$$

Add 13 to both sides.

$$5u=8+13$$

Add 8 and 13 to get 21.

$$5u=21$$

Divide both sides by 5.

$$u=\frac{21}{5}$$

The system is now solved.

$$y=-\frac{13}{2} u=\frac{21}{5}$$