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Writing Equations

Writing equations. Once we assign variables, we have to write an equation and solve. Many students can do the algebra once they have an equation, but getting the right equation can be tricky. I’ll review some basics of translating from words to math. So some of this might be familiar already.

The words is or are, the verbs in the text, correspond to the equals signs. That’s very important. It means as we’re translating, when we get to is or are, that’s where the equal sign is in the equation. If we have A is 50 meters, is 50 more than B. So A is, that’s A equals and then this way of saying it, at least in English, 50 more than B, what we’re saying is B plus 50.

So it’s a little bit backwards in the way that we write it mathematically. We write B plus 50 but we say 50 more than B. So A equals B plus 50. If we had A is 50 less than B, well, this one, also, it’s backwards, 50 less than B, it means we start at B and go down by 50, so that’s B minus 50, so A equals, A is, 50 less than B, B minus 50.

Suppose we have something like twice A is 100 less than 3 times B. Well, twice A, that would be 2 times A is, so 2 times A equals, that’s the first part of the sentence. Three times B that’s 3B and then a hundred less than 3B would be 3B minus 100. So that’s how we translate it. Suppose we have, A is 50% of B.

So, A is, A equals 50% of B. Well, as, if we’re finding a percent of something, we simply change that percent to a decimal and multiply. So that would be 0.5 times B. And notice that whole equation, A equals 0.5 times B, or A equals one-half B. We could just multiply both sides by two, 2A equals B, and that would be the same information but without any fractions or decimals.

Now, by contrast, suppose we had A is 50% greater than B, well this is very different. This is not the same as the one before, and you may remember from our discussion way, way back in the percents and ratios and fraction module, 50% greater, that is a percent increase. And the multiplier for a percent increase is one plus the perc, percent as a decimal, so 50% as a decimal is 0.5, 1.5 is the multiplier.

So A is, A equals 1.5 B. We could also write that as three halves B and so forth. Here’s a practice problem. Pause the video and then we’ll talk about this. Okay. Alice and Bruce each bought a refrigerator, and the sum of their purchases was $900, okay?

So we the use the very simple variables A and B for the price they spent. A plus B, the sum equals 900. That’s one equation. If twice of what Alice paid was $75 more than what Bruce paid, what did Alice pay for her refrigerator. So the first equation, the sum of their purchases is 900.

Twice of what Alice paid, 2A was, equals, 2A equals 75 more than Bruce, B plus 75. So that second equation is 2A equals B plus 75. What we want to get what Alice paid, so what I’m gonna do is I’m gonna solve that first equation, the equation on the left, I’m gonna solve that for B, so that when I plug that in to the second equation, I replace B, then I’ll just get an equation with A.

So, I’ll plug that in, then I’ll add A to both sides. I get 3A equals 900 plus 75. I’m just gonna leave it like that, not even add them up, because I have to divide by three. Well if I divide by 3, clearly 900 divided by 3 is 300. And 75 divided by 3 is 25, so it’s just 300 plus 25 or 325.

And that’s what Alice paid for her refrigerator. Here’s another practice problem. Pause the video and then we’ll talk about this. Okay. In a certain company, 50% of Ed’s salary is 20% of Ruth’s salary.

If the difference between their salaries is $90,000, what is Ruth’s salary? Okay, so first of all, 50% of Ed’s salary, so that, we could write that as 0.5 or we could write that as one-half of Ed’s salary is 20%, so that’s 0.2 or one-fifth of Ruth’s salary. So, we can write this in terms of fractions. So, half of Ed is a fifth of Ruth.

So notice, incidentally, what this means is that Ruth much, must have a much larger salary, because we can take only a fifth, a small part of her salary, and it equals half of Ed’s salary. So Ruth is the bigger one, Ruth minus Ed is 90,000. Okay, so as in the last problem we want to solve for Ruth, so it means one of the equations I’m gonna solve for Ed’s salary, and then use that to plug into the other.

And I’m gonna choose to solve the left equation. I’m just gonna multiply by two. I’ll get E equals two-fifths R, and then I can plug that in to the equation on the right. So, I get R minus two-fifths R, and of course, R minus two-fifths R is three-fifths R.

You take away two-fifths of anything, you’re left with three-fifths of it. Three-fifths R equals 90,000. Divide both sides by three, one-fifth is 30,000. Multiply by five, R equals 150,000. And that must be Ruth’s salary. Many times, writing the correct equation depends on the specifics of the problem.

In the upcoming lessons, we will talk about several specific topics in word problems, and in those lessons, we will discuss typical equations for those topics. In summary, we talked about how to translate words into math. Those are very, very important skills to have mastered. Remember that you may have to do a certain amount of algebraic set-up before you can write the equation you will solve.

And as I said at the end, many topics have typical equations, which you will learn in the upcoming lessons on the different word problem topics.

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