Using “Word Equations” to Help Write Algebraic Equations

Algebraic techniques can often help us solve word problems. The process typically involves assigning one or more variables, writing an equation (or two or three), and then solving that equation. It sounds easy enough, but creating an equation from a word problem can sometimes prove difficult. In these instances, I like to use a “word equation” as an intermediate step. To better understand what I mean, please consider the following:

In 5 years, Al will be 3 times as old as Mike.

If we let A = Al’s present age, and let M = Mike’s present age, what equation can we write?

If we’re unsure where to begin, we can start with a word equation that encapsulates the given information.

How about: Al’s future age = 3 times Mike’s future age

Even better: (Al’s age in 5 years) = 3(Mike’s age in 5 years)

At this point, we can transition to variables to get: (A + 5) = 3(M + 5)

This next one is a little harder.

If the retail cost of a certain toy were reduced by 20%, Barb could buy 12 more toys for $48, than she could at the original price. 

If we let x = the original price (in dollars) for one toy, what equation can we write?  

Here’s one option: (# of toys at original price) + 12 = (# of toys at reduced price) 

Even better: (# of toys purchased for $48 at original price) + 12 = (# of toys purchased for $48 at reduced price) 

If the original price (x) is reduced by 20%, the reduced price = 0.8x. So, we get: (# of toys purchased for $48 at x dollars apiece) + 12 = (#of toys purchased for $48 at 0.8x dollars apiece) 

Note: # of toyspurchased = (total $ spent)/(price per toy) 

We’re now ready to write an algebraic equation: 48/x + 12 = 48/0.8x

Now let’s use word equations to solve the following question:

Fiona and Gail departed A-town at the same time. Fiona traveled at 60 miles per hour, and Gail traveled at 45 miles per hour. If Fiona arrived in B-ville two hours earlier than Gail did, what is the distance from A-town to B-ville?

We’ll examine two different approaches that both begin with a word equation.

Approach #1: Since Fiona and Gail both traveled the same distance, we can write:

distance Fiona traveled = distance Gail traveled

Note: distance = (speed)(time)

We’re given their speeds, but not their travel times. So, let’s let f = Fiona’s travel time (in hours). Since Gail’s trip took 2 hours longer, we can say that f + 2 = Gail’s travel time (in hours)

We can now rewrite our word equation as: (60)(f) = (45)(f + 2)

When we solve this equation, we get f = 6. So, at a speed of 60 mph, Fiona took 6 hours to complete the trip. This means the distance from A-town to B-ville is 360 miles.

Approach #2: Since Gail’s trip took 2 hours longer than Fiona’s, we can write:

Gail’s travel time (in hours) = Fiona’s travel time (in hours) + 2

Note:  time = distance/speed

We’re given their speeds, but not the distance. So, let’s let d = the distance from A-town to B-ville.

We can now rewrite our word equation as: d/45 = d/60 + 2

When we solve this equation, we get d = 360. So, once again, the distance from A-town to B-ville is 360 miles.

As you can see, a word equation or two can help make it easier to transition from words to variables. If you’d like some practice writing word equations, you can try the some questions I’ve solved using this technique as part of my approach:

beatthegmat.com/equation-t107935.html

beatthegmat.com/average-problem-t270924.html

beatthegmat.com/boat-upstream-t276296.html

beatthegmat.com/contradictory-ps-problem-t274974.html

beatthegmat.com/relative-speed-on-circular-path-t178172.html

beatthegmat.com/relative-rates-problem-t116531.html

beatthegmat.com/rate-distance-and-time-problem-t121963.html

beatthegmat.com/3-grades-of-milk-t277111.html

beatthegmat.com/a-reduction-in-the-price-of-petrol-by-10-enables-a-motorist-t265044.html

beatthegmat.com/stuck-with-mixture-problem-help-please-t263906.html

beatthegmat.com/insurance-t278612.html

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