In ten years, Mark will be $\displaystyle 43$ years old, so Mark is $\displaystyle 43-10 = 33$ years old now, and Brian is one-third of this, or $\displaystyle 33 \div 3 = 11$ years old. 

Let $\displaystyle N$ be the number of years in which Mark will be twice Brian's age. Then Brian will be $\displaystyle N + 11$, and Mark will be $\displaystyle N + 33$. Since Mark will be twice Brian's age, we can set up and solve the equation:

$\displaystyle 2 (N + 11) = N + 33$

$\displaystyle 2N + 22 = N + 33$

$\displaystyle 2N + 22-N - 22 = N + 33 -N - 22$

$\displaystyle N = 11$

Mark will be twice Brian's age in $\displaystyle 11$ years.

Since Gary will be 37 in five years, he is $\displaystyle 37 - 5 = 32$ years old now. He is twice as old as Cathy, so she is $\displaystyle 32 \div 2 =16$ years old, and in five years, she will be $\displaystyle 16 + 5 = 21$ years old.

Solve for b when f is equal to 0

$\displaystyle b^3=\frac{4}{f}+2$

Let's start by plugging in 0 for f

$\displaystyle b^3=\frac{4}{(0)}+2$

Now, we cannot go any further, because we cannot divide by zero. 

This means that there is no value of b for which f will equal zero.

This means we have no real solutions.

Solve the following equation for u when y is equal to 6.

$\displaystyle \frac{4y}{u}=12$

To begin, let's plug in 6 for y.

$\displaystyle \frac{4*6}{u}=12$

Next, bring the u over to the other side by multiplying both sides by u

$\displaystyle 24=12u$

Finally, divide both sides by 12 to see what u is

$\displaystyle \frac{24}{12}=u=2$

So we get u=2 as our answer.

To solve for x, we want x to stand alone or to be by itself.  So in the equation

$\displaystyle 2x = 10$

we want x to stand alone.  To do that, we need to cancel the $\displaystyle 2$ that is next to it.  The $\displaystyle 2$ is being multiplied to x, so to cancel it out, we will need to divide by $\displaystyle 2$.  If we divide by $\displaystyle 2$ on the left side of the equal sign, we have to divide by $\displaystyle 2$ on the right side of the equal sign.  So,

$\displaystyle \frac{2x}{2} = \frac{10}{2}$

Now, we simplify.

$\displaystyle x = 5$

$\displaystyle A + B = 100$

To evaluate, substitute -40 in for A.

$\displaystyle -40 + B = 100$

Add 40 to each side to isolate and solve for B.

$\displaystyle -40 + B + 40 = 100 + 40$

$\displaystyle B = 140$

To solve the equation, isolate the variable on one side of the equation and all other constants on the other side. To accomplish this, perform the opposite operation to manipulate the equation.

$\displaystyle 5x+ 7 = 89$

First add seven to both sides.

$\displaystyle 5x+ 7 - 7 = 89 - 7$

$\displaystyle 5x = 82$

Now divide by five on both sides.

$\displaystyle 5x \div 5 = 82 \div 5$

$\displaystyle x = 16.4$

Solve the following equation for y when $\displaystyle z=3$

$\displaystyle 12y+6=2z^3-26$

Let's begin by rearranging the equation to get y by itself.

$\displaystyle 12y+6(-6)=2z^3-26(-6)$

$\displaystyle 12y(\div12)=(2z^3-32)(\div12)$

$\displaystyle y=\frac{2z^3-32}{12}$

Next, just plug in 3 for z and solve.

$\displaystyle y=\frac{2(3)^3-32}{12}=\frac{54-32}{12}=\frac{22}{12}=1.8\bar{3}$

So our answer is 

$\displaystyle 1.8\bar{3}$

$\displaystyle -0.04 z + (-7) = - 121$

$\displaystyle -0.04 z + (-7) - (-7) = - 121 - (-7)$

$\displaystyle -0.04 z = - 121+ 7$

$\displaystyle -0.04 z = - (121- 7)$

$\displaystyle -0.04 z = -114$

$\displaystyle -0.04 z \div (-0.04 ) = -114 \div (-0.04 )$

$\displaystyle z = 114 \div 0.04$

To divide, move the decimal point in both numbers two units to the right;

$\displaystyle z = 11400 \div 4$

$\displaystyle z= 2,850$

Solve the following equation for h, when l is equal to 5

$\displaystyle 3h+12=\frac{l^3-25}{5}$

Let's begin by substituting in 5 for l

$\displaystyle 3h+12=\frac{5^3-25}{5}$

Simplify

$\displaystyle 3h+12=\frac{125-25}{5}=\frac{100}{5}=20$

$\displaystyle 3h+12=20$

Next, move the twelve over:

$\displaystyle 3h=20-12=8$

$\displaystyle 3h=8$

Finally, we get:

$\displaystyle h=\frac{8}{3}$