This section covers:
 Introducing Rational Expressions
 Multiplying and Dividing, and Simplifying Rationals
 Finding the Common Denominator
 Adding and Subtracting Rationals
 Restricted Domains of Rational Functions
 Solving Rational Equations
 Applications of Rationals
 More Practice
Note that we talk about how to graph rationals using their asymptotes in the Graphing Rational Functions, including Asymptotes section.
Rational Functions are just those with polynomials (expression with two or more algebraic terms) in the numerator and denominator, so they are the ratio of two polynomials.
Since factoring is so important in algebra, you may want to revisit it first. Remember that we first learned factoring here in the Solving Quadratics by Factoring and Completing the Square section, and more Advanced Factoring can be found here.
Introducing Rational Expressions
Again, think of a rational expression as a ratio of two polynomials.
Here are some examples of expressions that are and aren’t rational expressions:
Expression 
Rational? 
Explanation 
\(\displaystyle \frac{{{{x}^{2}}1}}{{x+3}}\) 
Yes 
Ratio of two polynomials 
\(\displaystyle \frac{1}{x}\) 
Yes 
Ratio of two polynomials. “1” is a constant monomial. 
\(\displaystyle \frac{{{{x}^{3}}8}}{1}\) or \({{x}^{3}}8\) 
Yes 
Note that this polynomial is technically a rational expression, but we typically think of them as having a variable in the denominator. 
\(\displaystyle \frac{{\sqrt{{2x}}+1}}{{{{x}^{3}}}}\) 
No 
The numerator isn’t a polynomial, because of the radical. 
\(\displaystyle \frac{{\frac{1}{x}}}{{x+4}}\) 
Not technically, but can be simplified to a rational. 
This expression isn’t a rational since the numerator (\(\displaystyle \frac{1}{x}\)) isn’t a polynomial, but with some algebra, we can turn it into one: \(\displaystyle \frac{{\frac{1}{x}}}{{x+4}}=\frac{1}{x}\left( {\frac{1}{{x+4}}} \right)=\frac{1}{{x\left( {x+4} \right)}}\).
(Flip the denominator and multiply). 
Multiplying and Dividing, and Simplifying Rationals
Frequently, rationals can be simplified by factoring the numerator, denominator, or both, and crossing out factors. They can be multiplied and divided like regular fractions.
Here are some examples. Note that these look really difficult, but we’re just using a lot of steps of things we already know. That’s the fun of math! Also, note in the last example, we are dividing rationals, so we flip the second and multiply.
Remember that when you cross out factors, you can cross out from the top and bottom of the same fraction, or top and bottom from different factors that you are multiplying. You can never cross out two things on top, or two things on bottom.
Also remember that at any point in the problem, when variables are in the denominator, we’ll have domain restrictions, since denominators can’t be \(0\).
Multiplying and Dividing, and Simplifying Rationals 
\(\displaystyle \require{cancel} \frac{{\left( {3{{x}^{2}}5x2} \right)}}{{{{x}^{2}}4}}=\frac{{\left( {3x+1} \right)\cancel{{\left( {x2} \right)}}}}{{\left( {x+2} \right)\cancel{{\left( {x2} \right)}}}}=\frac{{3x+1}}{{x+2}};\,\,\,x\ne 2,\,\,2\) 
\(\displaystyle \require{cancel} \begin{align}\frac{{4{{x}^{2}}+12xy+2x+6y}}{{2{{x}^{2}}+6xy+6x+18y}}\,&=\,\frac{{2\left( {2{{x}^{2}}+6xy+x+3y} \right)}}{{2\left( {{{x}^{2}}+3xy+3x+9y} \right)}}\,=\,\frac{{2\left[ {2x\left( {x+3y} \right)+1\left( {x+3y} \right)} \right]}}{{2\left[ {x\left( {x+3y} \right)+3\left( {x+3y} \right)} \right]}}\,\,\,\\&=\,\frac{{\cancel{2}\left( {2x+1} \right)\cancel{{\left( {x+3y} \right)}}}}{{\cancel{2}\left( {x+3} \right)\cancel{{\left( {x+3y} \right)}}}}\,=\,\frac{{2x+1}}{{x+3}};\,\,\,\,x\ne 3,\,\,3y\end{align}\) 
\(\displaystyle \require{cancel} \begin{align}\frac{{\frac{{7{{y}^{2}}}}{{{{y}^{2}}16}}}}{{\frac{{14{{y}^{2}}+49y}}{{3{{y}^{2}}10y8}}}}\,&=\,\frac{{\frac{{7{{y}^{2}}}}{{\left( {y4} \right)\left( {y+4} \right)}}}}{{\frac{{7y\left( {2y+7} \right)}}{{\left( {y4} \right)\left( {3y+2} \right)}}}}\,=\,\frac{{\cancel{7}{{{\cancel{{{{y}^{2}}}}}}^{y}}}}{{\cancel{{\left( {y4} \right)}}\left( {y+4} \right)}}\cdot \frac{{\cancel{{\left( {y4} \right)}}\left( {3y+2} \right)}}{{\cancel{7}\cancel{y}\left( {2y+7} \right)}}\,\,\,\\&=\,\frac{{y\left( {3y+2} \right)}}{{\left( {y+4} \right)\left( {2y+7} \right)}};\,\,\,\,\,\,y\ne 4,\,\frac{7}{2},\,\,\frac{2}{3},\,\,0,\,\,4\end{align}\) 
Finding the Common Denominator
When we add or subtract two or more rationals, we need to find the least common denominator (LCD), just like when we add or subtract regular fractions. If the denominators are the same, we can just add the numerators across, leaving the denominators as they are. We then must be sure we can’t do any further factoring:
\(\require{cancel} \displaystyle \frac{2}{{3x}}+\frac{4}{{3x}}\,=\,\frac{{(2+4)}}{{3x}}\,=\,\frac{{{{{\cancel{6}}}^{2}}}}{{{{{\cancel{3}}}^{1}}x}}\,=\,\frac{2}{x};\,\,\,\,x\ne 0\).
Just like with regular fractions, we want to use the factors in the denominators in every fraction, but not repeat them across denominators. When nothing is common, just multiply the factors.
Let’s find the least common denominators for the following denominators (ignore the numerators for now).
Rational Expressions 
LCD 

Rational Expressions 
LCD 
\(\displaystyle \frac{{3x}}{{\left( {2x+1} \right)\left( {x+4} \right)}}\,;\,\,\,\,\frac{1}{{8{{x}^{2}}}}\) 
\(8{{x}^{2}}\left( {2x+1} \right)\left( {x+4} \right)\)
None of the factors are the same, so multiply factors.

\(\displaystyle \frac{2}{{\left( {x4} \right)\left( {x3} \right)}}\,;\,\,\,\frac{1}{{8{{{\left( {x3} \right)}}^{2}}}}\) 
\(8\left( {x4} \right){{\left( {x3} \right)}^{2}}\)
Don’t have to use the \(\left( {x3} \right)\) in the first fraction, since it’s in the second – but need the whole \({{\left( {x3} \right)}^{2}}\)! 

\(\displaystyle \frac{{4x}}{{3{{x}^{2}}\left( {x+2} \right)}}\,;\,\,\,\,\frac{1}{{9\left( {{{x}^{2}}4} \right)}}\)
Factor denominators: \(\displaystyle \frac{{4x}}{{3{{x}^{2}}\left( {x+2} \right)}}\,;\,\,\,\,\frac{1}{{9\left( {x+2} \right)\left( {x2} \right)}}\) 
\(\begin{array}{l}9{{x}^{2}}\left( {x+2} \right)\left( {x2} \right)\\\,\,\,=9{{x}^{2}}\left( {{{x}^{2}}4} \right)\end{array}\)
Don’t have to use the 3 and the \(\left( {x+2} \right)\) in the first fraction, since it’s in the second. 
\(\displaystyle \frac{1}{{6{{x}^{4}}3{{x}^{3}}63{{x}^{2}}}}\,;\,\,\,\frac{x}{{36{{x}^{2}}126x}}\)
Factor denominators: \(\displaystyle \frac{1}{{3{{x}^{2}}\left( {2x7} \right)\left( {x+3} \right)}}\,;\,\,\,\frac{x}{{18x\left( {2x7} \right)}}\) 
\(18{{x}^{2}}\left( {2x7} \right)\left( {x+3} \right)\)
Don’t have to use the 3 and the \(\left( {2x7} \right)\) in the first fraction, since it’s in the second. Don’t have to use the \(x\) in the second since it’s in the first.

Adding and Subtracting Rationals
Now let’s add and subtract the following rational expressions.
Note that one way to look at finding the LCD is to multiply the top by what’s missing in the bottom. For example, in the first example, the LCD is \(\left( {x+3} \right)\left( {x+4} \right)\), and we need to multiply the first fraction’s numerator by \(\left( {x+4} \right)\), since that’s missing in the denominator.
Adding and Subtracting Rationals 
Notes 
\(\displaystyle \begin{align}\frac{{x+2}}{{x3}}+\frac{{x1}}{{x+4}}\,&=\,\frac{{x+2}}{{x3}}\left( {\frac{{x+4}}{{x+4}}} \right)\,\,+\,\,\frac{{x1}}{{x+4}}\left( {\frac{{x3}}{{x3}}} \right)\,\,\\&=\frac{{\left( {x+2} \right)\left( {x+4} \right)}}{{\left( {x3} \right)\left( {x+4} \right)}}+\frac{{\left( {x1} \right)\left( {x3} \right)}}{{\left( {x3} \right)\left( {x+4} \right)}}\,\\&=\,\frac{{{{x}^{2}}+6x+8+{{x}^{2}}4x+3}}{{\left( {x3} \right)\left( {x+4} \right)}}=\,\frac{{2{{x}^{2}}+2x+11}}{{\left( {x3} \right)\left( {x+4} \right)}}\,\\x&\ne 4,\,\,3\,\end{align}\) 
Since there are no common factors, it’s almost like “cross multiplying” to get the numerators (do you see it?).
When you get the final answer, make sure you can’t factor the top and further reduce. 
\(\begin{align}\frac{{2{{y}^{2}}16}}{{{{y}^{2}}4}}\frac{{y+4}}{{y+2}}\,\,&=\,\,\frac{{2\left( {{{y}^{2}}8} \right)}}{{\left( {y2} \right)\left( {y+2} \right)}}\frac{{y+4}}{{y+2}}\left( {\frac{{y2}}{{y2}}} \right)\,\\&=\frac{{2{{y}^{2}}16\left( {{{y}^{2}}+2y8} \right)}}{{\left( {y2} \right)\left( {y+2} \right)}}\,\,=\,\,\frac{{{{y}^{2}}2y8}}{{\left( {y2} \right)\left( {y+2} \right)}}\\\,&=\frac{{\left( {y4} \right)\cancel{{\left( {y+2} \right)}}}}{{\left( {y2} \right)\cancel{{\left( {y+2} \right)}}}}\,\,=\,\,\frac{{y4}}{{y2}}\,\,\\y&\ne 2,\,\,2\end{align}\) 
Multiply the top by what you don’t have in the bottom.
We had to multiply the second term by \(\displaystyle \left( {\frac{{y2}}{{y2}}} \right)\) since we didn’t have \(\left( {y2} \right)\) on the bottom.
Watch out for negatives; don’t forget to push negatives through parentheses! And don’t forget to simplify! 
\(\displaystyle \begin{align}\frac{{\frac{2}{{x3}}}}{{\frac{3}{{x+1}}+\frac{1}{{x3}}}}\,&=\,\frac{{\frac{2}{{x3}}}}{{\frac{{3\left( {x3} \right)+1\left( {x+1} \right)}}{{\left( {x+1} \right)\left( {x3} \right)}}}}=\,\frac{{\frac{2}{{x3}}}}{{\frac{{4x8}}{{\left( {x+1} \right)\left( {x3} \right)}}}}\,\\&=\frac{{{}^{1}\cancel{2}}}{{\cancel{{x3}}}}\cdot \frac{{\left( {x+1} \right)\cancel{{\left( {x3} \right)}}}}{{{{{\cancel{4}}}_{2}}\left( {x2} \right)}}\,=\,\frac{{x+1}}{{2\left( {x2} \right)}}\,\\x&\ne 1,\,\,2,\,3\end{align}\) 
Now let’s combine what we know about adding/subtracting and multiplying/dividing rationals.
Find the common denominator on the bottom first, combine terms, and then flip and multiply to the top.
Then simplify. 
Restricted Domains of Rational Functions
As we’ve noticed, since rational functions have variables in denominators, we must make sure that the denominators won’t end up as “0” at any point of solving the problem.
So the domain of \(\displaystyle \frac{{x+1}}{{2x(x2)(x+3)}}\) is \(\{x:x\ne 3,\,\,0,\,\,2\}\). This means if we ever get a solution to an equation that contains rational expressions and has variables in the denominator (which they probably will!), we must make sure that none of our answers would make any denominator in that equation “0”. These “answers” that we can’t use are called extraneous solutions. We’ll see this in the first example below.
Solving Rational Equations
When we solve rational equations, we can multiply both sides of the equations by the least common denominator (LCD) and not even worry about working with fractions! The denominators will cancel out and we just solve the equation using the numerators. Remember that with quadratics, we need to get everything to one side with 0 on the other side and either factor or use the Quadratic Formula.
Again, think of multiplying the top by what’s missing in the bottom from the LCD.
Notice that sometimes you’ll have to solve literal equations, which just means that you have to solve an equation for a variable, but you’ll have other variables in the answer. The last example shows this.
Solving Rational Equations 
Notes 
\(\displaystyle \frac{3}{{x+3}}\frac{1}{x}\,=\,\frac{{9}}{{x\left( {x+3} \right)}}\)
\(\displaystyle \begin{array}{c}\left( {x\left( {x+3} \right)} \right)\left( {\frac{3}{{x+3}}\frac{1}{x}} \right)\,=\,\left( {\frac{{9}}{{x(x+3)}}} \right)\left( {x\left( {x+3} \right)} \right)\\3x(x+3)\,=\,9\\\,\,2x3\,=\,9\\\,\,\,\,2x\,=\,6;\,\,\,\,\,\,\,x\,=\,3\end{array}\) Does not work! No solution, or Ø 
Multiply both sides by the least common (LCD) denominator, which is \(\displaystyle x\left( {x+3} \right)\).
Note that you multiply the numerators with what you don’t have in the denominator. For example, \(\displaystyle \frac{3}{{x+3}}\) doesn’t have \(x\), so you multiply the top 3 by \(x\) to get \(3x\).
Check your answers to make sure no denominators are 0. Our answer doesn’t work (–3 is an extraneous solution), so there is no solution. 
\(\displaystyle \frac{{5{{x}^{2}}+8x4}}{{3{{x}^{2}}+11x+6}}\frac{1}{{x+3}}\,=\,\frac{x}{{3x+2}}\)
\(\displaystyle \begin{array}{c}\left( {\left( {3x+2} \right)\left( {x+3} \right)} \right)\left( {\frac{{5{{x}^{2}}+8x4}}{{\left( {3x+2} \right)\left( {x+3} \right)}}\frac{1}{{x+3}}} \right)\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\,\frac{x}{{3x+2}}\left( {\left( {3x+2} \right)\left( {x+3} \right)} \right)\\5{{x}^{2}}+8x4\left( {3x+2} \right)\left( 1 \right)\,=\,x\left( {x+3} \right)\\5{{x}^{2}}+8x43x2\,=\,{{x}^{2}}+3x\\4{{x}^{2}}+2x6\,=\,0\\2{{x}^{2}}+x3\,=\,\frac{0}{2}\,=0\\\,\left( {2x+3} \right)\left( {x1} \right)=0;\,\,\,\,\,\,\,\,\,\,x=\frac{3}{2}\,\,\,\,;x=1\,\,\,\,\,\end{array}\) 
Notice we had to factor the denominator first, so we could get our LCD. We would typically factor the numerator too; we get \(\left( {5x2} \right)\left( {x+2} \right)\), but it doesn’t really help since we can’t cancel anything out.
Sometimes we end up with a quadratic and we have to factor (“unfoil”) or maybe even use the quadratic formula, if factoring doesn’t work.
In this example, we could factor. Then set each factor to 0 to solve for \(x\). Both work!
It’s still a good to check each answer. This isn’t easy; you may want to use the STO> function in your calculator to store the solutions, and then type in the sides of the equations using \(X,{\mathrm T},\theta ,n\). 
\(\displaystyle \frac{1}{{1x}}\,=1\frac{x}{{x1}}\)
\(\displaystyle \begin{array}{c}\frac{{1}}{{x1}}=1\frac{x}{{x1}}\\\left( {x1} \right)\left( {\frac{{1}}{{x1}}} \right)\,=\,\left( {1\frac{x}{{x1}}} \right)\left( {x1} \right)\\1=\left( {x1} \right)x\\1=1;\,\,\,\,\,\,\,\mathbb{R}\end{array}\) But, \(x\ne 1\), since this would make the denominator 0. Answer is all reals, except \(x\ne 1\), or \(\left( {\infty ,1} \right)\cup \left( {1,\infty } \right)\). 
Multiply the first term by –1 to make the denominators the same: \(\displaystyle \frac{1}{{1x}}=\frac{{1}}{{\left( {1x} \right)}}=\frac{1}{{x1}}\). The LCD is \(\left( {x1} \right)\).
Multiply the numerators by “what’s missing” in the denominator; we end up with \(1=1\), which is always true.
The answer would be all real numbers, except we can’t have a 1 in the denominator, so we have to “skip over” the value of 1. 
Solve for \(f\): \(\displaystyle \,\,\frac{1}{f}+\frac{1}{p}\,=\,\frac{1}{q}\,\,\,\,\)
\(\displaystyle \begin{array}{c}fpq\left( {\frac{1}{f}+\frac{1}{p}} \right)\,=\,\left( {\frac{1}{q}} \right)fpq\\\,\,pq+fq\,=\,fp\\\,\,fpfq\,=\,pq\\f(pq)\,=\,pq;\,\,\,\,\,\,\,f\,=\,\frac{{pq}}{{pq}}\end{array}\) 
When solving for literal equations (equations where your answer will have other variables in it), just pretend the variables are numbers, and perform the same steps.
Here we are solving for \(f\), so, after multiplying both sides by the LCD, we need to get everything with an \(f\) in it to one side.
Then you can factor the \(f\) out, and solve for it. 
Applications of Rationals
There are certain types of word problems that typically use rational expressions. These tend to deal with rates, since rates are typically fractions (such as distance over time). We also see problems dealing with plain fractions or percentages in fraction form.
Fraction Problems
The denominator of a fraction is 2 less than twice the numerator. If 7 is added to both the numerator and denominator, the resulting fraction is \(\displaystyle \frac{4}{5}\). What is the original fraction?
Solution:
This problem is a little tricky, since we don’t want to set the variable to what the problem is asking – the original fraction. We want to set \(n=\) the numerator, since we’ll have to get both the numerator and the denominator. Since the denominator is in terms of the numerator, it’s easier to make the variable the numerator. So, from the first sentence of the problem, the original fraction is \(\displaystyle \frac{n}{{2n2}}\).
Now to solve, we just have to add 7 to the numerator and denominator, and set to \(\displaystyle \frac{4}{5}:\,\,\,\,\,\,\,\,\frac{{n+7}}{{\left( {2n2} \right)+7}}=\frac{4}{5}\).
Now let’s do the math:
Math 
Notes 
\(\displaystyle \begin{align}\frac{{n+7}}{{\left( {2n2} \right)+7}}\,&=\,\frac{4}{5}\\\,\frac{{n+7}}{{2n+5}}\,&=\,\frac{4}{5}\\\,\left( 5 \right)\left( {n+7} \right)\,&=\,\left( 4 \right)\left( {2n+5} \right)\\\,\,5n+35&=8n+20\\\,3n&=15\\\,n&=5\end{align}\)
Original fraction is \(\displaystyle \frac{n}{{2n2}}\,=\,\frac{5}{8}.\) 
Simplify the fraction with variables in it, and we can just cross multiply, since there’s only one term on each side. Solve to get \(n=5\).
When we get \(n\), we can build the fraction easily, by using the first sentence of the word problem.
Let’s check our answer: The denominator of \(\displaystyle \frac{5}{8}\) is 2 less than twice the numerator. If 7 is added to both the numerator and denominator, the resulting fraction is \(\displaystyle \frac{{5+7}}{{8+7}}=\frac{{12}}{{15}}=\frac{4}{5}.\) So the original fraction is \(\displaystyle \frac{5}{8}\).

OK, it was totally coincidental that the answer was the same fraction as in the picture above. Weird! 😆
Problem:
Bethany has scored 10 free throws out of 18 tries. She would really like to bring her free throw average up to at least 68%. How many consecutive free throws should she score in order to bring up her average to 68%?
Solution:
Let’s let \(x=\) the number of free throws that Bethany should score (in a row) in order to bring up her average. Again, we can use fractions, and this time they will represent the fraction of free throws that she scores. We’ll start out with her current fraction (rate) of consecutive free throws, and then we’ll add the number she needs to score to both the numerator (number she scores) and denominator (total number of throws): \(\displaystyle \frac{{10+x}}{{18+x}}\,=\,\frac{{68}}{{100}}\).
Let’s do the math:
Math 
Notes 
\(\displaystyle \begin{align}\frac{{10+x}}{{18+x}}\,\,&=\,\,\frac{{68}}{{100}}\\\\\left( {100} \right)\left( {10+x} \right)\,\,&=\,\,\left( {68} \right)\left( {18+x} \right)\\1000+100x&=1224+68x\\32x&=224\\x&=7\end{align}\) 
Again, we can just cross multiply, since there’s only one term on each side. Solve to get \(x=7\).
Let’s check our answer: If she throws 7 more free throws, she’ll have 17 scores in a row out of 25 tries. \(\frac{{17}}{{25}}\,=\,68%\)
So Bethany needs 7 more consecutive free throws to bring her free throw percentage up to 68%.

Distance/Rate/Time Problems
With rational rate problems, we must always remember: \(\text{Distance}=\text{Rate }\times \,\text{Time}\). It seems most of the problems deal with comparing times or adding times.
Problem:
Shalini can run 3 miles per hour faster than her sister Meena can walk. If Shalini ran 12 miles in the same time it took Meena to walk 8 miles, what is the speed of each sister in this case?
Solution:
This is a “\(\text{Distance}=\text{Rate }\times \,\text{Time}\)” problem, and let’s go ahead and use a table to organize this information like we did in the Algebra Word Problems and Systems of Linear Equations and Word Problems sections. Let’s let \(x=\) Meena’s speed, since Shalini runs faster and it’s easier to add than subtract:
Now let’s do the fun part – the math:
Math 
Notes 
\(\displaystyle \begin{align}\frac{{12}}{{x+3}}&=\,\frac{8}{x}\\12x&=8\left( {x+3} \right)\\\,12x&=8x+24\\\,\,4x&=24\\\,\,\,x&=6\end{align}\) 
Since we know that Shalini ran 12 miles in the same time that Meena walked 8 miles, we can set the times equal to each other.
Now we can just cross multiply since we only have one term on each side.
So Meena’s speed is 6 miles per hour, and Shalini’s speed is \(6+3=\) 9 miles per hour – pretty fast!

Problem:
Here’s a problem where we use the rates of a boat going upstream (rate in still water rate minus the rate of the water current) and downstream (rate in still water plus the rate of the water current) to add times. (Think when you’re going downstream, it’s like you’re going down a hill, so it’s faster.)
The time it takes for a canoe to go 3 miles upstream and back 3 miles downstream is 4 hours. The current in the lake is 1 mile per hour. Find the average speed (rate) of the canoe in still water.
Solution:
We’ll let \(x=\) the speed of the boat in still water. So we know that the rate of the canoe going upstream (again, think that “up” is more difficult so we go slower) is “\(x1\)”, and the rate going downstream is “\(x+1\)”. (Think about going “down” a hill; we go faster).
We use that the fact that \(\displaystyle \text{Time}=\frac{{\text{Distance}}}{{\text{Rate}}}\) to add the time upstream to the time downstream; this equals 4 hours.
Math 
Notes 
\(\require{cancel} \displaystyle \begin{align}\left( {x1} \right)\left( {x+1} \right)\left( {\frac{3}{{x1}}+\frac{3}{{x+1}}} \right)\,&=\,4\left( {{{x}^{2}}1} \right)\\3\left( {x+1} \right)+3\left( {x1} \right)\,&=\,4{{x}^{2}}4\\3x\cancel{{+3}}+3x\cancel{{3}}\,&=\,4{{x}^{2}}4\\4{{x}^{2}}6x4\,&=\,0\\\,\,2{{x}^{2}}3x2\,&=\,0\\\,\left( {2x+1} \right)\left( {x2} \right)\,&=\,0\end{align}\)
\(\cancel{{x=\frac{1}{2}}}\,\,\,\,\,\text{or}\,\,\,\,\,x=2\) 
You have to add up the time upstream and time downstream, and set this equal to 4 hours.
Remember again that \(\displaystyle \text{time}=\frac{{\text{distance}}}{{\text{rate}}}\).
Then we solve for \(x\); we multiply both sides by the LCD, which is \(\left( {x1} \right)\left( {x+1} \right)\), or \({{x}^{2}}1\).
We end up with a quadratic that we can factor, so we get two possible solutions.
Since we can’t have a negative rate in this problem, the average rate of the canoe in still water is 2 miles per hour.

Note: If we were given the rate of the canoe in still water and had to find the rate of the current, we’d do the problem in a similar way. We’d have to add and subtract variables (the rate of the current) from numbers (the rates of the canoe in still water) in the denominators.
Work Problems:
Work problems typically have to do with different people or things working together and alone, at different rates.
I find that usually the easiest way to work these problems is to remember:
\(\displaystyle \frac{{\operatorname{time} \,\text{together}}}{{\text{time alone}}}\,\,\,+\,\,…\,\,+\,\,\,\frac{{\text{time together}}}{{\text{time alone}}}\,\,\,=\,\,\,1\) (or whatever part of the job or jobs is done; if they do half the job, this equals \(\displaystyle \frac{1}{2}\); if they do a job twice, this equals 2).
“Proof”: For work problems, \(\text{Rate }\times \,\,\text{Time = }1\text{ Job}\). Add up individuals’ portions of a job with this formula, using the time working with others (time together):
\(\displaystyle \begin{array}{c}\left( {\text{individual rate }\times \text{ time }} \right)\text{ + }…\text{ +}\left( {\text{individual rate }\times \text{ time }} \right)=1\\\left( {\frac{1}{{\text{individual time to do 1 job}}}\text{ }\times \text{ time}} \right)\text{ + }…\text{ +}\left( {\frac{1}{{\text{individual time to do 1 job}}}\text{ }\times \text{ time}} \right)=1\\\left( {\frac{{\text{time working with others}}}{{\text{individual time to do 1 job}}}} \right)\text{ + }…\text{ +}\left( {\frac{{\text{time working with others}}}{{\text{individual time to do 1 job}}}} \right)=1\end{array}\)
Also, as explained after the first example below, often you see this formula as \(\displaystyle \frac{\text{1}}{{\text{time alone}}}\,\,+…+\,\frac{\text{1}}{{\text{time alone}}}\,=\,\frac{\text{1}}{{\text{time together}}}\).
Problem:
Erica can paint her room in 5 hours. If she has her friend Rachel help her, they can paint the room together in 3 hours. How long would it take for Rachel to paint the room alone, if Erica wanted to go play tennis for the afternoon?
Solution:
We’ll let \(R=\) amount of time (in hours) Rachel can paint the room by herself. We know that the time the girls paint the room together is 3 hours, and the time Erica paints the room by herself is 5 hours.
Math 
Notes 
\(\displaystyle \begin{align}\frac{3}{5}+\frac{3}{R}&=1\\\left( {5R} \right)\left( {\frac{3}{5}+\frac{3}{R}} \right)&=1\left( {5R} \right)\\3R+15&=5R\\2R&=15\\R&=7.5\end{align}\) 
Let’s use the formula \(\displaystyle \frac{{\text{time together}}}{{\text{time alone}}}\,\,\text{+}\,\,\frac{{\text{time together}}}{{\text{time alone}}}\,=\,1\), since they need to get the whole job done.
Erica’s time to paint is 5 hours, Rachel’s is \(R\), since we’re trying to find it, and their time together is 3 hours.
Multiply both sides the LCD, which is \(5R\).
So it would take Rachel 7.5 hours to paint Erica’s room. Let’s hope Rachel comes through to help, so Erica can play tennis!

Note: The formula above can also be derived by using the concept that you can figure out how much of the job the girls do per hour (or whatever the time unit is), both together and alone. Then you can add the individual “rates” to get the “rate” of their painting together.
We are actually adding the Work they complete (together and alone) using formula \(\text{Rate }\times \text{ }\text{Time }=\text{ Work}\), where the Time is 1 hour (or whatever the unit is).
In this example, Erica’s rate per hour is \(\displaystyle \frac{1}{5}\) (she can do \(\displaystyle \frac{1}{5}\) of the job in 1 hour); Rachel’s rate per hour is \(\displaystyle \frac{1}{R}\); we can add their rates to get the rate of their painting together: \(\displaystyle \frac{1}{5}+\frac{1}{R}=\frac{1}{3}\). If we multiply all the terms by 3, we come up with the equation above! (Sometimes you see this as \(\displaystyle \frac{\text{1}}{{\text{time alone}}}\,\,+…+\,\frac{\text{1}}{{\text{time alone}}}\,=\,\frac{\text{1}}{{\text{time together}}}\).)
Problem:
It takes Jill 3 hours to make sandwiches for a charity organization (as many as they need). Allie starts one hour later and (if she were working alone) it would take her 4 hours to make the sandwiches needed. How long will it take both of them working together to complete the job?
Solution:
This one’s a little trickier since “Allie starts one hour later”. Since it takes Jill 3 hours to make sandwiches by herself, after one hour (when Allie starts), she’s already one third of the way through the job (one third of 3 hours is 1 hour). The amount of job left then is two thirds.
We’ll let \(T=\) amount of time (in hours) the girls can make the sandwiches together. We know it takes Jill 3 hours to make them alone, and Allie 4 hours to make them alone. So we can use the \(\displaystyle \frac{{\text{time together}}}{{\text{time alone}}}\,\,+\,\,\frac{{\text{time together}}}{{\text{time alone}}}\,\,\,=\,\,\,1\text{ job}\) formula, but in this case, we’ll use \(\displaystyle \frac{2}{3}\) job instead of 1:
Math 
Notes 
\(\displaystyle \begin{align}\frac{T}{3}+\frac{T}{4}\,&=\,\frac{2}{3}\\\left( {12} \right)\left( {\frac{T}{3}+\frac{T}{4}} \right)\,&=\,\frac{2}{3}\left( {12} \right)\\4T+3T\,&=\,8\\7T&=\,8\\T&=\frac{8}{7}=1\frac{1}{7}\end{align}\) 
We’ll use \(\displaystyle \frac{{\text{time together}}}{{\text{time alone}}}\,\,\text{+}\,\,\frac{{\text{time together}}}{{\text{time alone}}}\,\,=\,\,\frac{2}{3}\), since they only need to get \(\displaystyle \frac{2}{3}\) of the job done.
Jill’s time to make sandwiches is 3 hours, Allie’s is 4 hours, and we’re trying to find their time together, \(T\).
Multiply both sides the LCD, which is 12. So it would take the girls \(\displaystyle 1\frac{1}{7}\) or about 1.14 hours to make the sandwiches. It really helps to have them work together, and it’s more fun that way, too!

Problem:
Two hoses are used to fill Maddie’s neighborhood swimming pool. One hose alone can fill the pool in 10 hours; the second hose can fill it in 12 hours. The pool’s drain can empty the pool in 8 hours. If the two hoses are working, and the drain is open (by mistake), how long will it take to fill the swimming pool?
Solution:
This is the same type of “work” problem where we have “alone” times, but want to find the “together” time. The difference with this problem is that we have 3 different things working to fill the pool, one of which is working against filling the pool (the drain), which we need to make negative.
Let’s let \(x=\) the time it takes for the 2 hoses (positive work) and the drain (negative work) all together to fill the pool:
Math 
Notes 
\(\displaystyle \begin{align}\frac{x}{{10}}+\frac{x}{{12}}\frac{x}{8}\,&=\,1\\\left( {120} \right)\left( {\frac{x}{{10}}+\frac{x}{{12}}\frac{x}{8}} \right)\,&=\,1\left( {120} \right)\\\,12x+10x15x\,&=\,120\\\,7x\,&=\,120\\x\,&=\,\frac{{120}}{7}=17\frac{1}{7}\,\,\text{hrs}\end{align}\) 
We’ll use \(\displaystyle \frac{{\text{time together}}}{{\text{time alone}}}\,\,\text{+}\,\,\frac{{\text{time together}}}{{\text{time alone}}}\,\,\text{}\,\,\frac{{\text{time together}}}{{\text{time alone}}}=\,\,1\), since we have 2 hoses coming into the pool, and 1 drain where the water is going out.
Multiply both sides the LCD, which is 120.
So it would take the 2 hoses coming in, working with the drain with water going out \(\displaystyle \frac{{120}}{7}\) or \(\displaystyle 17\frac{1}{7}\) hours to fill the pool. It’s taking a lot longer having that drain open!

Problem:
8 women and 12 girls can paint a large mural in 10 hours. 6 women and 8 girls can paint it in 14 hours. Find the time by 1 woman alone, and 1 girl alone to paint the mural.
Solution:
This is actually a systems problem and also a work problem at the same time. We did this problem without using rationals here in the Systems of Linear Equations and Word Problems section (and be careful, since the variables we assigned were different).
Let’s use the approach that you can figure out what per hour rates are (together and alone), and add the individual “rates” to get the “rate” of their painting together.
(Since a work rate is \(\displaystyle \frac{1}{{\text{time}}}\),this is actually the \(\displaystyle \frac{\text{1}}{{\text{time alone}}}\,\,+…+\,\frac{\text{1}}{{\text{time alone}}}\,=\,\frac{\text{1}}{{\text{time together}}}\) method, but with multiple people).
Let \(w=\) the time (in hours) it takes for one woman to paint the mural alone, and \(g=\) the time it takes for one of the girls to paint the mural alone. Set up the systems, and solve:
Math 
Notes 
\(\displaystyle \begin{align}\frac{8}{w}+\frac{{12}}{g}\,&=\,\frac{1}{{10}}\\\frac{6}{w}+\frac{8}{g}\,&=\,\frac{1}{{14}}\\\left( {10gw} \right)\left( {\frac{8}{w}+\frac{{12}}{g}} \right)\,&=\,\frac{1}{{10}}\left( {10gw} \right)\\\left( {14gw} \right)\left( {\frac{6}{w}+\frac{8}{g}} \right)\,&=\,\frac{1}{{14}}\left( {14gw} \right)\\80g+120w=gw;\,\,\,\,\,&84g+112w=gw\\80g+120w&=84g+112w\\8w=4g;\,\,\,&\,\,\,\,g=2w\end{align}\)
Substitute in first equation:
\(\displaystyle \begin{align}\frac{8}{w}+\frac{{12}}{{2w}}\,&=\,\frac{1}{{10}}\\\,\left( {10w} \right)\frac{8}{w}+\frac{{12}}{{2w}}\,&=\,\frac{1}{{10}}\,\left( {10w} \right)\\\,80+60&=w;\\\\w&=140\\g=2w&=280\end{align}\) 
Each woman can paint \(\displaystyle \frac{1}{w}\) murals in one hour, and each girl can paint \(\displaystyle \frac{1}{g}\) murals in one hour. From the first part of the problem, this means that 8 women can paint \(\displaystyle 8\left( {\frac{1}{w}} \right)\) murals in an hour, and 12 girls can paint \(\displaystyle 12\left( {\frac{1}{g}} \right)\) murals in an hour. Together the 8 women and 12 girls can paint a mural in 10 hours (\(\displaystyle \frac{1}{{10}}\) of a mural in an hour). (We’re adding amounts of work per hour; their total work in an hour is \(\displaystyle \frac{1}{{10}}\).)
From the second part of the problem, 6 women can paint \(\displaystyle 6\left( {\frac{1}{w}} \right)\) murals in an hour, and 8 girls can paint \(\displaystyle 8\left( {\frac{1}{g}} \right)\) murals in an hour. Together the 6 women and 8 girls can paint a mural in 14 hours (\(\displaystyle \frac{1}{{14}}\) of a mural in an hour).
Add what they can paint individually together in 1 hour to get how much they can paint together in 1 hour, and, with the 2 parts of the problem, we have a system.
Multiply both sides of the equations by their respective LCDs. We try not to work with the \(gw\), so we eliminate it by setting the two equations together (we got lucky here!). Now we can solve to get \(g=2w\). We then use substitution with either one of the equations; we used the first one.
So it would take one of the women 140 hours, and one of the girls 280 hours to paint the mural by herself. Whew – that was a tough one!!

Cost Problems
A third type of problem you might get while studying rational has to do with costs per person (or unit cost), which is again a rate.
Problem:
A group of girls decided to go on a trip to the beach and the organizer said that the bus would cost $360 to rent. The organizer also told them that if they got 3 more girls to go on the trip, each girl could pay $6 less (which they ended up doing). How many girls ended up going on the trip?
Solution:
Let n be the original number of girls going on the trip. The cost per girl for the original trip (like a rate) is \(\displaystyle \frac{{360}}{n}\) (use real numbers to see this – if 10 girls were going, it would be $36 per girl).
If they added 3 more girls, each girl would have to pay \(\displaystyle \frac{{360}}{{n+3}}\) since the total cost is still $360, but the number of girls is \(\displaystyle n+3\).
So now with \(\displaystyle n+3\) girls, the new cost is 6 dollars less, or the original cost minus 6, so we can set up an equation like this: \(\displaystyle \frac{{360}}{{n+3}}=\frac{{360}}{n}6\).
Let’s do the math:
Math 
Notes 
\(\displaystyle \frac{{360}}{{n+3}}\,=\,\frac{{360}}{n}6\) \(\require{cancel} \begin{array}{c}\left( {n\left( {\cancel{{n+3}}} \right)} \right)\left( {\frac{{360}}{{\cancel{{n+3}}}}} \right)=\left( {\frac{{360}}{{\cancel{n}}}6} \right)\left( {n\left( {n+3} \right)} \right)\\\,360n=360\left( {n+3} \right)6\left( {n\left( {n+3} \right)} \right)\\\cancel{{360n}}=\cancel{{360n}}+10806{{n}^{2}}18n\\\,\,6{{n}^{2}}+18n1080=0\,\,\,\,\,\\\,\,6\left( {{{n}^{2}}+3n180} \right)=0\\\,\,\,\,{{n}^{2}}+3n180=0\\\,\,\left( {n+15} \right)\left( {n12} \right)=0\\\,\,\,\,\,\,\,\cancel{{n=15}};\,\,\,\,\,\,\,n=12\,\,\,\,\,\,\,\,\\n+3=15\,\,\,\text{girls}\end{array}\) 
We need to multiply both sides by the LCD, which is \(\left( {n\left( {n+3} \right)} \right)\). We have to be careful to “push through” the LCD \(\left( {n\left( {n+3} \right)} \right)\) on the right side to both \(\displaystyle \frac{{360}}{n}\) and 6, with a minus sign in between.
Since we have a quadratic, we have to move everything to one side and set it to 0. When we simplify and factor, we need to “throw away” the negative number; this won’t work for the number of girls.
When we get the answer, we have to be careful and add 3 to that number. We solved for \(n\), which is the original number of girls, but the problem calls for the new number of girls, which is \(n+3\).
So the original number of girls planning on going on the trip is 12, but the new number is 15.

Understand these problems, and practice, practice, practice!
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