Solve {l}{3(y+8)=10(y-4)}{b+(b+1)+(b+2)=2}

$$\left. \begin{array} { l } { 3 ( y + 8 ) = 10 ( y - 4 ) } \\ { b + ( b + 1 ) + ( b + 2 ) = 2 } \end{array} \right.$$

Solve for y, b

$$y = \frac{64}{7} = 9\frac{1}{7} \approx 9.142857143$$

$$b=-\frac{1}{3}\approx -0.333333333$$

Solution Steps

Consider the first equation. Use the distributive property to multiply 3 by y+8.

$$3y+24=10\left(y-4\right)$$

Use the distributive property to multiply 10 by y-4.

$$3y+24=10y-40$$

Subtract 10y from both sides.

$$3y+24-10y=-40$$

Combine 3y and -10y to get -7y.

$$-7y+24=-40$$

Subtract 24 from both sides.

$$-7y=-40-24$$

Subtract 24 from -40 to get -64.

$$-7y=-64$$

Divide both sides by -7.

$$y=\frac{-64}{-7}$$

Fraction \frac{-64}{-7} can be simplified to \frac{64}{7} by removing the negative sign from both the numerator and the denominator.

$$y=\frac{64}{7}$$

Consider the second equation. Combine b and b to get 2b.

$$2b+1+b+2=2$$

Combine 2b and b to get 3b.

$$3b+1+2=2$$

Add 1 and 2 to get 3.

$$3b+3=2$$

Subtract 3 from both sides.

$$3b=2-3$$

Subtract 3 from 2 to get -1.

$$3b=-1$$

Divide both sides by 3.

$$b=-\frac{1}{3}$$

The system is now solved.

$$y=\frac{64}{7} b=-\frac{1}{3}$$