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Quiz 20
1.) Solve $$\dfrac{3}{7} - \dfrac{1}{2} = \dfrac{x}{4}.$$ Solution: Write the left side in terms of its least common denominator: $$\dfrac{6}{14} - \dfrac{7}{14} = \dfrac{x}{4}.$$ Perform the subtraction on the left to get $$\dfrac{-1}{14} = \dfrac{x}{4}.$$ Multiply both sides by $4$ to get $$\dfrac{-4}{14} = -\dfrac{2}{7}= x.$$

2.) Solve $$\dfrac{3}{x-2} + \dfrac{4}{x+1} = \dfrac{8}{x^2-x-2}.$$ Solution: First notice that $x^2-x-2=(x-2)(x+1)$. The easiest way to proceed is to now multiply both sides of the equation by $(x-2)(x+1)$ to get $$\left[ (x-2)(x+1) \right] \left[ \dfrac{3}{x-2} + \dfrac{4}{x+1} \right] = \left[ (x-2)(x+1) \right] \left[ \dfrac{8}{(x-2)(x+1)} \right].$$ Now on the left distribute and cancel and on the right just cancel to get $$3(x+1)+4(x-2)=8.$$ Now distribute the numbers in the multiplcations on the left to get $$3x+3+4x-8=8.$$ Now combine like terms to get $$7x-5=8.$$ Add $5$ and divide by $7$ to get $$x = \dfrac{13}{7}.$$

3.) Bob mows his lawn in $4$ hours. It takes Mary $3$ hours to mow the same lawn. How long would it take both to mow the lawn if they worked together? Solution: We are told Bob mows the lawn at a rate of $4 \dfrac{\mathrm{hours}}{\rm{lawn}}$ and Mary mows the lawn at the rate $3 \dfrac{\rm{hours}}{\rm{lawn}}$. It does not make sense to add these quantities, because when working together, it should take less time. The key is to express these rates in another way (i.e. in the form $\dfrac{\rm{lawns}}{\rm{hour}}$). To do this take the reciprocal: $$4 \dfrac{\rm{hours}}{\rm{lawn}} \longleftrightarrow \dfrac{1}{4} \dfrac{\rm{lawns}}{\rm{hour}},$$ and $$3 \dfrac{\rm{hours}}{\rm{lawn}} \longleftrightarrow \dfrac{1}{3} \dfrac{\rm{lawns}}{\rm{hour}}.$$ It does make sense to add these quantities because, when working together, the fraction of the lawn that gets mowed by Bob and Mary should be greater than the fraction of the lawn that they mow individually. Therefore, together, their rate is $$\dfrac{1}{4} \dfrac{\rm{lawns}}{\rm{hour}} + \dfrac{1}{3} \dfrac{\rm{lawns}}{\rm{hour}} = \left( \dfrac{1}{4} + \dfrac{1}{3} \right) \dfrac{\rm{lawns}}{\rm{hour}}=\dfrac{7}{12} \dfrac{\rm{lawns}}{\rm{hour}}.$$

Therefore the rate corresponding to the time it takes them to mow the lawn working together is $\dfrac{12}{7} \dfrac{\rm{hours}}{\rm{lawn}}$.

4.) The following triangles are similar triangles. Find $x$:

Solution: By the similar triangles property, we get the equation $$\dfrac{5}{10} = \dfrac{x}{14}.$$ Therefore multiply by $14$ to get $$\dfrac{5 \cdot 14}{10} = x,$$ or $$\dfrac{14}{2}=7=x.$$

5.) The speed of train A is $10 \dfrac{\mathrm{miles}}{\mathrm{hour}}$ slower than the speed of train B. Train A travels $100$ miles in the same time that it takes train $B$ to travel $200$ miles. What is the speed of each train?
Solution: Let $s_A$ denotes the speed of train $A$ and $s_B$ denotes the speed of train $B$. Since the times are equal, we may deduce the following equation: $$\dfrac{s_A}{100} = \dfrac{s_B}{200}.$$ This is an equation with two variables. To make it an equation of one variable, take the information that says the speed of $A$ is $10 \dfrac{\rm{miles}}{\rm{hour}}$ slower than train $B$, i.e. $$s_A = s_B-10.$$ Plug this in for $s_A$ in our equation to get $$\dfrac{s_B-10}{100} = \dfrac{s_B}{200}.$$ Multiply both sides by $200$ to get $$\dfrac{200}{100} (s_B-10) = s_b,$$ or $$2(s_B-10)=s_B,$$ or $$2s_B - 20 = s_B.$$ Subtract $s_B$ from both sides to get $$s_B-20=0.$$ Add $20$ to conclude $$s_B=20,$$ (slow train!) Use the equation relating $s_A$ and $s_B$ to deduce that $$s_A = 20-10 = 10.$$