Solve {r}{4y-14=14}{7m+8=78}{adiv3+5=14}{tdiv3-2=12}{7h-11=45}

$$\left. \begin{array} { r } { 4 y - 14 = 14 } \\ { 7 m + 8 = 78 } \\ { a \div 3 + 5 = 14 } \\ { t \div 3 - 2 = 12 } \\ { 7 h - 11 = 45 } \end{array} \right.$$

Solve for y, m, a, t, h

$$y=7$$

$$t=42$$

$$m=10$$

$$a=27$$

$$h=8$$

Solution Steps

Consider the first equation. Add 14 to both sides.

$$4y=14+14$$

Add 14 and 14 to get 28.

$$4y=28$$

Divide both sides by 4.

$$y=\frac{28}{4}$$

Divide 28 by 4 to get 7.

$$y=7$$

Consider the second equation. Subtract 8 from both sides.

$$7m=78-8$$

Subtract 8 from 78 to get 70.

$$7m=70$$

Divide both sides by 7.

$$m=\frac{70}{7}$$

Divide 70 by 7 to get 10.

$$m=10$$

Consider the third equation. Multiply both sides of the equation by 3.

$$a+15=42$$

Subtract 15 from both sides.

$$a=42-15$$

Subtract 15 from 42 to get 27.

$$a=27$$

Consider the fourth equation. Multiply both sides of the equation by 3.

$$t-6=36$$

Add 6 to both sides.

$$t=36+6$$

Add 36 and 6 to get 42.

$$t=42$$

Consider the fifth equation. Add 11 to both sides.

$$7h=45+11$$

Add 45 and 11 to get 56.

$$7h=56$$

Divide both sides by 7.

$$h=\frac{56}{7}$$

Divide 56 by 7 to get 8.

$$h=8$$

The system is now solved.

$$y=7 m=10 a=27 t=42 h=8$$