[latex] D=\frac{m}{v}[/latex]

Multiply both side of the equation by v to isolate m.

[latex] v\cdot D=\frac{m}{v}\cdot v[/latex]

Simplify and rewrite the equation, solving for m.

[latex]\begin{array}{l}v\cdot D=m\cdot \frac{v}{v}\\v\cdot D=m\cdot 1\\v\cdot D=m\end{array}[/latex]

To solve the equation [latex] D=\frac{m}{v}[/latex] in terms of v, you will need do the same steps to this point and then divide both sides by D.

[latex]\begin{array}{r}\frac{v\cdot D}{D}=\frac{m}{D}\\\\\frac{D}{D}\cdot v=\frac{m}{D}\\\\1\cdot v=\frac{m}{D}\\\\v=\frac{m}{D}\end{array}[/latex]

Therefore, [latex] m=D\cdot v[/latex] and [latex] v=\frac{m}{D}[/latex].

[latex] V=\pi{{r}^{2}}h[/latex]

Divide both sides by [latex] \pi {{r}^{2}}[/latex] to isolate h.

[latex] \frac{V}{\pi {{r}^{2}}}=\frac{\pi {{r}^{2}}h}{\pi {{r}^{2}}}[/latex]

Simplify. You find the height, h, is equal to [latex] \frac{V}{\pi {{r}^{2}}}[/latex].

[latex] \frac{V}{\pi {{r}^{2}}}=h[/latex]

Therefore, [latex] h=\frac{V}{\pi {{r}^{2}}}[/latex].

Think about how many bulbs each person can plant in one hour. This is their planting rate.

Myra: [latex] \frac{50\,\,\text{bulbs}}{2\,\,\text{hours}}[/latex] or [latex] \frac{25\,\,\text{bulbs}}{1\,\,\text{hour}}[/latex]

Francis: [latex] \frac{45\,\,\text{bulbs}}{3\,\,\text{hours}}[/latex] or [latex] \frac{15\,\,\text{bulbs}}{1\,\,\text{hour}}[/latex]

Combine their hourly rates to determine the rate they work together.

Myra and Francis together:

[latex] \frac{25\,\,\text{bulbs}}{1\,\,\text{hour}}+\frac{15\,\,\text{bulbs}}{1\,\,\text{hour}}=\frac{40\,\,\text{bulbs}}{1\,\,\text{hour}}[/latex]

Use one of the work formulas to write a rational equation, for example, [latex] r=\frac{W}{t}[/latex]. You know r, the combined work rate, and you know W, the amount of work that must be done. What you do not know is how much time it will take to do the required work at the designated rate.

[latex] \frac{40}{1}=\frac{150}{t}[/latex]

Solve the equation by multiplying both sides by the common denominator and then isolating t.

[latex]\begin{array}{c}1t\cdot\frac{40}{1} =\frac{150}{t}\cdot 1t\\\\40t=150\\\\t=\frac{150}{40}=\frac{15}{4}\\\\t=3\frac{3}{4}\text{hours}\end{array}[/latex]

It should take [latex]3[/latex] hours [latex]45[/latex] minutes for Myra and Francis to plant [latex]150[/latex] bulbs together.

Choose variables to represent the unknowns. Since it takes John [latex]3[/latex] times as long as Joe to paint the house, his time is represented as [latex]3x[/latex].

Let [latex]x[/latex] = time it takes Joe to complete the job

[latex]3x[/latex] = time it takes John to complete the job

The work is painting [latex]1[/latex] house or [latex]1[/latex]. Write an expression to represent each person’s rate using the formula [latex] r=\frac{W}{t}[/latex]_.

Joe’s rate: [latex] \frac{1}{x}[/latex]

John’s rate: [latex] \frac{1}{3x}[/latex]

Their combined rate is the sum of their individual rates. Use this rate to write a new equation using the formula [latex]W=rt[/latex].

combined rate: [latex] \frac{1}{x}+\frac{1}{3x}[/latex]

The problem states that it takes them [latex]24[/latex] hours together to paint a house, so if you multiply their combined hourly rate [latex] \left( \frac{1}{x}+\frac{1}{3x} \right)[/latex] by [latex]24[/latex], you will get [latex]1[/latex], which is the number of houses they can paint in [latex]24[/latex] hours.

[latex] \begin{array}{l}1=\left( \frac{1}{x}+\frac{1}{3x} \right)24\\\\1=\frac{24}{x}+\frac{24}{3x}\end{array}[/latex]

Now solve the equation for x. (Remember that x represents the number of hours it will take Joe to finish the job.)

[latex]\begin{array}{l}\,\,\,1=\frac{3}{3}\cdot \frac{24}{x}+\frac{24}{3x}\\\\\,\,\,1=\frac{3\cdot 24}{3x}+\frac{24}{3x}\\\\\,\,\,1=\frac{72}{3x}+\frac{24}{3x}\\\\\,\,\,1=\frac{72+24}{3x}\\\\\,\,\,1=\frac{96}{3x}\\\\3x=96\\\\\,\,\,x=32\end{array}[/latex]

Check the solution in the original equation.

[latex]\begin{array}{l}1=\left( \frac{1}{x}+\frac{1}{3x} \right)24\\\\1=\left[ \frac{\text{1}}{\text{32}}+\frac{1}{3\text{(32})} \right]24\\\\1=\frac{24}{\text{32}}+\frac{24}{3\text{(32})}\\\\1=\frac{24}{\text{32}}+\frac{24}{96}\\\\1=\frac{3}{3}\cdot \frac{24}{\text{32}}+\frac{24}{96}\\\\1=\frac{72}{96}+\frac{24}{96}\end{array}[/latex]

The solution checks. Since [latex]x=32[/latex], it takes Joe [latex]32[/latex] hours to paint the house by himself. John’s time is [latex]3x[/latex], so it would take him [latex]96[/latex] hours to do the same amount of work.