Problem solving using Linear Equations

Problem solving using Linear Equations

From sale prices to trip distances, many real life problems can be solved using linear equations. In this lesson, we'll practice translating word problems into linear equations, then solving the problems.

Real World Math

A train leaves Chicago at 7 a.m., traveling at 70 mph to New York, which is 800 miles away. Another train leaves New York at the same time, traveling on a parallel track to Chicago at 85 mph. When will they meet? The question is: why do we care so much about trains?

Well, I like trains, but I still feel a little nervous when I read a math problem that starts with a train. If I'm going to have to translate a real world scenario to an algebraic equation, can't it be something I might actually encounter in my life? I mean, I've ridden trains between Chicago and New York, but I've never plotted when my train will pass another.

In this lesson, we'll not only practice solving problems that can be translated into linear equations, we'll also focus on problems you may encounter in your life - problems not involving trains passing each other.

Linear Equations

As a reminder, a linear equation is just an algebraic expression that represents a line. These equations typically have one variable and look like 3x = 9 or y + 4 = 10. In these equations, we're trying to figure out the variable, which involves getting it alone on one side of the equals sign.

Simple Problems

There are simple problems that involve linear equations. For example, the sum of 35 and a number is 72. What is the number?

The thing we don't know is our variable. Let's use x here. We know x + 35 = 72, so that's our equation. If we solve for x by subtracting 35 from both sides, we get x = 37. Now we know our number.

They can be a bit more complex, like this: 15 less than four times a number is 57. What is the number? Again, let's use x for the number. Four times that number is 4x. 15 less than that is 4x -15. So, our equation is 4x - 15 = 57. To solve for x, we add 15 to both sides. Then, we divide by 4, and x = 18.

Practice Problems

Let's take that knowledge and look at some real life situations. Let's start with money. We all like money, right?

Let's say you're a little short on cash and need a loan. Your cousin agrees to loan you money, and you agree that you'll repay him in full plus 4% interest. We're going to ignore the questionable judgment he displays in loaning money to family. If he loans you $500, how much interest will you need to pay?

Our variable here is the amount of interest, so let's call that x. The interest will be the amount of the loan, $500, multiplied by the interest rate, 4%. To multiply with a percent, we convert it to a decimal. So, x = 500*.04. What is 500 * .04? 20. So, you'll owe him $20, plus the original $500.

You decide that you want to be better about saving money. Your cell phone company is promoting a text message plan that costs $10 each month plus five cents per text. You currently pay $20 each month for an unlimited plan, but you want to save a few dollars. If you want to try the new plan and spend only $15 each month, how many texts can you send?

Ok, let's use t for texts. It costs five cents per text, so that's .05t. Plus, there's that $10 fee. So, you want .05t + 10 to equal 15. First, subtract 10 from both sides. Then, divide by .05. So, you could send 100 texts each month. You've been averaging way more than that, so maybe this isn't a great plan.

But, then you get a new job, and suddenly you have some extra cash. You decide that you want to save up for new bike. You find one you like that costs $400. If you can save $35 each week, how many weeks will it take you to get the bike?

So, we want to know weeks, or w. You're saving $35 each week, so that's 35w. If we want to save $400, then our equation is 35w = 400. This one's simple. Just divide 400 by 35 and you get 11.4. So, it'll take you just over 11 weeks to get that bike.

After you get the bike, you decide to have some fun. Let's say you and two friends go bowling. At the end of the night, your bill is $42. If you played 3 games and paid $3 each for shoes, how much did you pay per game?

Let's call the cost per game g. You played three games, so that's 3g. You also paid $3 each for shoes, and there were three of you, so that's 3*3, or 9. That means 3g + 9 = 42. That's our equation.

First, subtract 9 from both sides. 3g = 33. Now, divide by 3. g = 11. So, each game costs $11.

Ok, before we go, why don't we try that train problem, you know, just to show that we can. How did it go? A train leaves Chicago at 7 a.m. traveling to New York, which is 800 miles away, at 75 mph. Another train leaves New York at the same time, traveling on a parallel track to Chicago at 85 mph. We want to know when they'll meet.

We're trying to find how much time it will take, or t. The first train is traveling at a rate of 75 mph, so the distance it covers in t time is 75t. The second train is going 85 mph for t time, or 85t. We want to know when 75t + 85t = 800. In other words, when do the two distances add up to the total distance, 800 miles.

We add 75t and 85t to get 160t. Now, we divide both sides by 160, and we get t = 5. So, they'll meet 5 hours into their respective trips.

Hopefully, the passengers will have finished their linear equation word problems and look up in time to wave. I mean, two trains passing each other at 75 and 80 miles per hour won't see each other very long.

Lesson Summary

In summary, we learned how to translate word problems in linear equations, or algebraic expressions that represent lines. We looked at simple examples, where the problem describes a number in terms of details about it, like the sum of twice a number and 52 is 174.

Then, we looked at problems that involve real-life scenarios, from loaning money to bowling. We focused on defining the variable, or the unknown quantity, in terms of what is known, then solving for the variable.

Oh, and we solved the dreaded algebra train problem. Nice work!