Example

Answer: Find the least common multiple (LCM) of 4 and 8. Remember, to find the LCM, identify the greatest number of times each factor appears in each factorization. Here, 2 appears 3 times, so [latex]2\cdot2\cdot2[/latex], or 8, will be the LCM.

[latex]\begin{array}{l}4=2\cdot2\\8=2\cdot2\cdot2\cdot2\\\text{LCM}=2\cdot2\cdot2\\\text{LCM}=8\end{array}[/latex]

[latex]\begin{array}{r}8\cdot \frac{x+5}{8}=\frac{7}{4}\cdot 8\,\,\,\,\,\,\,\\\\\frac{8(x+5)}{8}=\frac{7(8)}{4}\,\,\,\,\,\,\\\\\frac{8}{8}\cdot (x+5)=\frac{7(4\cdot 2)}{4}\\\\\frac{8}{8}\cdot (x+5)=7\cdot 2\cdot \frac{4}{4}\\\\1\cdot (x+5)=14\cdot 1\,\,\,\end{array}[/latex]

[latex]\begin{array}{r}x+5=14\\x=9\,\,\,\end{array}[/latex]

[latex]\begin{array}{r}\frac{x+5}{8}=\frac{7}{4}\\\\\frac{9+5}{8}=\frac{7}{4}\\\\\frac{14}{8}=\frac{7}{4}\\\\\frac{7}{4}=\frac{7}{4}\end{array}[/latex]

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Example

Answer: Clear the denominators by multiplying each side by the common denominator. The common denominator is [latex]3\left(x+1\right)[/latex] since [latex]3\text{ and }x+1[/latex] don't have any common factors.

[latex]\begin{array}{c}3\left(x+1\right)\left(\frac{8}{x+1}\right)=3\left(x+1\right)\left(\frac{4}{3}\right)\end{array}[/latex]

Simplify common factors.

[latex]\begin{array}{c}3\cancel{\left(x+1\right)}\left(\frac{8}{\cancel{x+1}}\right)=\cancel{3}\left(x+1\right)\left(\frac{4}{\cancel{3}}\right)\\24=4\left(x+1\right)\\24=4x+4\end{array}[/latex]

Now this looks like a linear equation, and we can use the addition and multiplication properties of equality to solve it.

[latex]\begin{array}{c}24=4x+4\\\underline{-4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-4}\\20=4x\,\,\,\,\,\,\,\,\\\\x=5\,\,\,\,\,\,\,\,\,\end{array}[/latex]

[latex]\begin{array}{r}\,\,\,\,\,\frac{8}{\left(x+1\right)}=\frac{4}{3}\\\\\frac{8}{\left(5+1\right)}=\frac{4}{3}\\\\\frac{8}{6}=\frac{4}{3}\end{array}[/latex]

Reduce the fraction [latex]\frac{8}{6}[/latex] by simplifying the common factor of 2:

[latex]\large\frac{\cancel{2}\cdot4}{\cancel{2}\cdot3}\normalsize=\large\frac{4}{3}[/latex]

Answer

Example

Answer: Both fractions in the equation have a denominator of 3. Multiply both sides of the equation (not just the fractions!) by 3 to eliminate the denominators.

[latex] 3\left( \frac{x}{3}+1 \right)=3\left( \frac{4}{3} \right)[/latex]

[latex]\begin{array}{r}3\left( \frac{x}{3} \right)+3\left( 1 \right)=3\left( \frac{4}{3} \right)\\\\\cancel{3}\left( \frac{x}{\cancel{3}} \right)+3\left( 1 \right)=\cancel{3}\left( \frac{4}{\cancel{3}} \right)\\\\ x+3=4\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-3}\,\,\,\,\,\underline{-3}\\\\x=1\end{array}[/latex]

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Example

Answer: Determine any values for x that would make the denominator 0.

[latex] \frac{2x-5}{x-5}=\frac{15}{x-5}[/latex]

[latex]\begin{array}{r}2x-5=15\\2x=20\\x=10\end{array}[/latex]

[latex]\begin{array}{r}\frac{2x-5}{x-5}=\frac{15}{x-5}\,\,\\\\\frac{2(10)-5}{10-5}=\frac{15}{10-5}\\\\\frac{20-5}{10-5}=\frac{15}{10-5}\\\\\frac{15}{5}=\frac{15}{5}\,\,\,\,\,\,\,\,\,\end{array}[/latex]

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Example

Answer: Determine any values for m that would make the denominator 0. [latex]−4[/latex] is an excluded value because it makes [latex]m+4[/latex] equal to 0. Since the denominator of each expression in the equation is the same, the numerators must be equal. Set the numerators equal to one another and solve for m.

[latex]\begin{array}{l}16=m^{2}\\\,\,\,0={{m}^{2}}-16\\\,\,\,0=\left( m+4 \right)\left( m-4 \right)\end{array}[/latex]

[latex]\begin{array}{c}0=m+4\,\,\,\,\,\,\text{or}\,\,\,\,\,\,0=m-4\\m=-4\,\,\,\,\,\,\text{or}\,\,\,\,\,\,m=4\\m=4,-4\end{array}[/latex]

[latex]\begin{array}{c}\frac{16}{m+4}=\frac{{{m}^{2}}}{m+4}\\\\\frac{16}{-4+4}=\frac{{{(-4)}^{2}}}{-4+4}\\\\\frac{16}{0}=\frac{16}{0}\end{array}[/latex]

[latex]\begin{array}{c}\frac{16}{4+4}=\frac{{{(4)}^{2}}}{4+4}\\\\\frac{16}{8}=\frac{16}{8}\end{array}[/latex]

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Example

Answer: Start with the formula for density. [latex-display] D=\frac{m}{v}[/latex-display] Multiply both side of the equation by v to isolate m. [latex-display] v\cdot D=\frac{m}{v}\cdot v[/latex-display] Simplify and rewrite the equation, solving for m. [latex-display]\begin{array}{l}v\cdot D=m\cdot \frac{v}{v}\\v\cdot D=m\cdot 1\\v\cdot D=m\end{array}[/latex-display] 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-display]\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-display]

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Example

Answer: Start with the formula for the volume of a cylinder. [latex-display] V=\pi{{r}^{2}}h[/latex-display] Divide both sides by [latex] \pi {{r}^{2}}[/latex] to isolate h. [latex-display] \frac{V}{\pi {{r}^{2}}}=\frac{\pi {{r}^{2}}h}{\pi {{r}^{2}}}[/latex-display] Simplify. You find the height, h, is equal to [latex] \frac{V}{\pi {{r}^{2}}}[/latex]. [latex-display] \frac{V}{\pi {{r}^{2}}}=h[/latex-display]

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Example

Answer: 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-display] \frac{25\,\,\text{bulbs}}{1\,\,\text{hour}}+\frac{15\,\,\text{bulbs}}{1\,\,\text{hour}}=\frac{40\,\,\text{bulbs}}{1\,\,\text{hour}}[/latex-display] 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 don't know is how much time it will take to do the required work at the designated rate. [latex-display] \frac{40}{1}=\frac{150}{t}[/latex-display] Solve the equation by multiplying both sides by the common denominator, then isolating t. [latex-display]\begin{array}{c}\frac{40}{1}\cdot 1t=\frac{150}{t}\cdot 1t\\\\40t=150\\\\t=\frac{150}{40}=\frac{15}{4}\\\\t=3\frac{3}{4}\text{hours}\end{array}[/latex-display]

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Example

Answer: Choose variables to represent the unknowns. Since it takes John 3 times as long as Joe to paint the house, his time is represented as 3x. Let x = time it takes Joe to complete the job 3x = time it takes John to complete the job The work is painting 1 house or 1. 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 24 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 24, you will get 1, which is the number of houses they can paint in 24 hours. [latex-display] \begin{array}{l}1=\left( \frac{1}{x}+\frac{1}{3x} \right)24\\\\1=\frac{24}{x}+\frac{24}{3x}\end{array}[/latex-display] Now solve the equation for x. (Remember that x represents the number of hours it will take Joe to finish the job.) [latex-display]\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-display] Check the solutions in the original equation. [latex-display]\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-display] The solution checks. Since [latex]x=32[/latex], it takes Joe 32 hours to paint the house by himself. John’s time is 3x, so it would take him 96 hours to do the same amount of work.

Answer

Example

Answer: Write the formula for a direct variation relationship.

[latex]y=kx[/latex]

[latex]300=k\left(10\right)[/latex]

[latex]\begin{array}{l}\frac{300}{10}=\frac{10k}{10}\\\\\,\,\,\,30=k\end{array}[/latex]

Answer

Example

Answer: Write the formula for an inverse variation relationship.

[latex] y=\frac{k}{x}[/latex]

[latex] 25=\frac{k}{5}[/latex]

[latex] \begin{array}{c}5\cdot 25=\frac{k}{5}\cdot 5\\\\125=\frac{5k}{5}\\\\125=k\,\,\,\end{array}[/latex]

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Example

Answer: You are told that this is a joint variation relationship, and that the area of a triangle (A) varies jointly with the lengths of the base (b) and height (h).

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,\,\,y=kxz\\Area=k(base)(height)\end{array}[/latex]

[latex]30=k\left(10\right)\left(6\right)\\30=60k\\\\\frac{30}{60}=\frac{60k}{60}\\\\\frac{1}{2}=k[/latex]

[latex]\begin{array}{l}Area=k(base)(height)\\\\Area=(15)(20)(\frac{1}{2})\\\\Area=\frac{300}{2}\\\\Area=150\,\,\text{square inches}\end{array}[/latex]

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