How to solve Algebra word problems
What vocabulary should we use to solve ratio and proportion word problems?
In this lesson you will learn how to set up and solve ratio and proportion word problems.
For word problems, the best thing to do is to look at a few examples but first let’s review a few vocabulary terms.
Ratio: A ratio is a comparison of two items and it is often written as a fraction.
Proportion: A proportion is an equality between two ratios.
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Solving algebra word problems using ratios and proportions
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Exploring different examples of algebra word problems
Example
A pet store has ???12??? dogs, ???14??? fish and ???32??? cats. What is the ratio of cats to total animals?
We know there are ???32??? cats and we’re looking for the ratio
???\frac{\text{cats}}{\text{total}}???
We can find the total number of animals by adding all the groups together.
???12+14+32 = 58???
So we get
???\frac{32}{58}???
We need to reduce the ratio to its lowest terms
???\frac{16(2)}{29(2)}???
???\frac{16}{29}???
There are ???16??? cats for every ???29??? animals.
Let’s look at another type of word problem.
A ratio is a comparison of two items and it is often written as a fraction.
Example
Two numbers have a ratio of ???1??? to ???9???, and a sum of ???80???. What are the two numbers?
Let’s call the two unknown numbers ???x??? and ???y???. Then we’ll set up a proportion for ???x??? and ???y??? to relate them to their ratio of ???1??? to ???9???, and solve for a variable.
???\frac{x}{y} = \frac{1}{9}???
Cross multiply.
???9x = 1y???
???y = 9x???
Next set up an equation for ???x??? and ???y??? using the sum.
???x+y=80???
Let’s solve this for ???y???, since we already know ???y=9x???.
???y = 80 - x???
Now let’s use our system of equations and the substitution method. We know:
???y=80-x???
???y=9x???
Since we have two values equal to ???y???, we can set those values equal to each other.
???80-x=9x???
???80=9x+x???
???80=10x???
???\frac{80}{10}=\frac{10x}{10}???
???8=x???
???x=8???
Now we can use ???y=9x??? and the fact that ???x=8??? to solve for ???y???.
???y=9(8)???
???y=72???
Let’s check our work.
It’s true that
???\frac{8}{72}=\frac{1}{9}???
and that
???8+72=80???
Let’s do one more like that one.
Example
Two numbers have a ratio of ???7??? to ???13???, and a sum of ???300???. What are the two numbers?
Let’s call the two unknown numbers ???x??? and ???y???. Set up a proportion for ???x??? and ???y??? to relate them to their ratio of ???7??? to ???13???, and solve for a variable.
???\frac{x}{y} = \frac{7}{13}???
Cross multiply.
???13x=7y???
Solve for ???x???.
???\frac{13x}{13}=\frac{7y}{13}???
???x=\frac{7}{13}y???
Next set up an equation for ???x??? and ???y??? using the sum.
???x+y=300???
Let’s solve for ???x??? since we already know ???x=(7/13)y???.
???x=300-y???
Now let’s use our system of equations and the substitution method. We know:
???x=300-y???
???x=\frac{7}{13}y???
Since we have two values equal to ???x???, we can set those values equal to each other.
???\frac{7}{13}y=300-y???
???\frac{7}{13}y + y = 300???
???\frac{7}{13}y + \frac{13}{13}y=300???
???\frac{20}{13}y=300???
???\frac{13}{20} \cdot \frac{20}{13}y=\frac{13}{20} \cdot 300???
???y=13 \cdot 15???
???y=195???
Now solve for ???x???.
???x=300-195???
???x=105???
Use the original equations to check your work.
It’s true that
???\frac{105}{195} = \frac{7}{13}???
and that
???105+195=300???
