Question

variable in each equation.
2 $\frac{4}{q}(q - 10)=8$
4 $3b + 4 = -31$
$\frac{-4quad -4}{3b = -35}$
6 $\frac{1}{6}(b + 12)= -4$
8 $-9 = -9(t + 1)$
$-9=-9t-9$
$\frac{+9quad +9}{0=-9t}$
10 $\frac{c + 14}{3}= -12$

Explanation:

Step1: Solve for q (Eq2)

Multiply both sides by $q$:
$$4(q-10)=8q$$
Expand left side:
$$4q-40=8q$$
Subtract $4q$:
$$-40=4q$$
Divide by 4:
$$q=-10$$

Step2: Solve for b (Eq4)

Subtract 4 from both sides:
$$3b=-31-4$$
Simplify right side:
$$3b=-35$$
Divide by 3:
$$b=-\frac{35}{3}$$

Step3: Solve for b (Eq6)

Multiply both sides by 6:
$$b+12=-24$$
Subtract 12:
$$b=-24-12$$
$$b=-36$$

Step4: Solve for t (Eq8)

Divide both sides by -9:
$$1=t+1$$
Subtract 1:
$$t=1-1$$
$$t=0$$

Step5: Solve for c (Eq10)

Multiply both sides by 3:
$$c+14=-36$$
Subtract 14:
$$c=-36-14$$
$$c=-50$$

Answer:

1. $q=-10$
2. $b=-\frac{35}{3}$
3. $b=-36$
4. $t=0$
5. $c=-50$