The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Solving Another Equation with Two Variables

Date: 1/30/96 at 20:28:1
From: McKellar Clan
Subject: how to do algebra

Dear Dr. Math,

     I am in the 7th grade and our math class is learning algebra.  
I do not understand it and my teacher is no help in trying to 
explain it.  My mother doesn't understand algebra and my father is 
too busy to help me. I hope you can.  I have included a sample 
problem from my book;  6y = 3D = 54.  

I am supposed to solve the equation, but I don't even understand 
it!  PLEASE HELP!!!!  I really need to get a handle on this stuff.

My name is Brandon.

Date: 2/1/96 at 14:44:17
From: Doctor Elise
Subject: Re: how to do algebra


You sound really frustrated.  Algebra makes a lot of people 
confused, especially at the beginning, but there's no magic to it.  
You really can learn it!  I'll try to help by talking about 
algebra in general, and your sample problem in particular.

Algebra is the next step in math after you can pretty much add, 
subtract, multiply, divide, and do anything else you want with 
actual numbers.  In Algebra, instead of using a specific number, 
like '5', we start using a letter, like 'y', to represent a number 
we don't know yet.

At this point, we also start leaving out the "times" sign when we 
write equations if we're multiplying a number "times" a letter.  
What '6y' really means is "six times y, which is a number I don't 
know yet."

The goal of almost every algebra problem is to find out what 
number (or numbers) the letter could equal.  The mathbooks usually 
call it "solving for y", and you've done it when your equation 
finally looks like "y = some number".

Your example is:

 6y = 3D = 54

This looks like a pretty funny equation, doesn't it?  That's
because...Surprise!  It's really three entirely different
problems.  They are:

 6y = 3D
 3D = 54
 6y = 54

Does that make a little more sense?  Let's start with the third 

 6y = 54
In words, that's "six times y equals 54".

The goal is to figure out what number we can plug in for 'y'
that works.  If you just think about it for a minute, you
know the answer already, 6 * 9 = 54, right?  Here's how you
get there using algebra.

The way you solve any algebra problem is by putting all the
letters on one side of the equals sign, and all the numbers
on the other.  The big rule is that you can do anything you
want to the equation as long as you do the same thing to
both sides of the equals sign.  Think about it.  If I start with

2     = 2,     I know this is true.  What if I add 4 to both 
2 + 4 = 2 + 4  I get:
6     = 6      This is also true.  What if I multiply both sides 
               by 3:

6 * 3 = 6 * 3  I get:
18    = 18     Still works.  Okay, what if I multiply both sides
               by some unknown number 'x':

18 * x = 18 * x  This is still always true.
                 Remember, in algebra we write this as:
18x = 18x

Okay, what if I add 4 * x to both sides? Can I do that? Sure!

18x + 4x = 18x + 4x

Works just fine.  Now.  Remember how you used to factor your
plain old vanilla numbers?  As in, 6 = 3 * 2?  And if you
wanted to add 6 + 4 you could write it like

  6   + 4

3 * 2 + 2 * 2

2 * (3 + 2)

2 * 5


See, you get 10 this way, too.

Well, you can do the same thing with this silly 'x' number.

18x + 4x

18 * x + 4 * x  write it out the long way

x * (18 + 4)    pull out the 'x'

x * 22          add the numbers

22x             here's the answer.  Of course, without an "equals"
                sign, we can't solve it any further.

Anyhow, if you have "6y = 54", and what you want is "y = 
something", you just have to divide both sides by 6, right?

6y   =  54
6y / 6 = 54 / 6

and we can write 6y/6 in a bunch of different ways, but it 
basically boils down to the exact same thing you used to do with 
fractions. The same way you can reduce

10/15 by factoring it into

5 * 2 / 5 * 3  and then cross out both 5's, we can write

6y/6 as

6 * y / 6 * 1 and cross out the 6's to get plain old y/1,

which is 'y'.

And, of course, 54/6 = 9, so we have

y = 9


So, for 3D = 54 we do the same approach:

3D /3 = 54/3

D = 18

We can even use 6y = 3D to check our work.  If you substitute 9 
for y and 18 for D, is the equation true?

I hope this helps.  Good luck!

-Doctor Elise,  The Math Forum

Associated Topics:
Middle School Algebra

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.