The Math Forum

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

Solve for x; Find 5 Ordered Pairs

Date: 11/02/2002 at 13:09:50
From: Rachel Brock
Subject: Two questions

I have two questions. The first one reads: 

   Solve for x. 3x/4 - 4 = 7 + x/2

The second one reads: 

   Determine 5 ordered pairs that will satisfy the equation 
    4y = 8 - 2x.

I am really stuck and I appreciate the way you explain things instead 
of making me feel stupid. You are wonderful and I am so glad my sister 
recommended you.


Date: 11/02/2002 at 14:35:13
From: Doctor Ian
Subject: Re: Two questions

Hi Rachel,

Let's look at your second question first. An ordered pair is a pair
of values for x and y that satisfy some equation. Suppose we have an
equation like 

  6y = 9 - 3x

Now, we could _choose_ to give x a value like x=1. If we do that, what 
happens to the equation? 

  6y = 9 - 3*1

  6y = 9 - 3

  6y = 6

Now, for _that_ choice of x, the value of y is determined: it can only
be y=1. So the values x=1 and y=1 are one possible solution to the
equation, and we can write that as an ordered pair: (x=1,y=1), or just

But what about other choices? What if we give x the value x=2? Now we 

  6y = 9 - 3*2

  6y = 9 - 6

  6y = 3

Now the corresponding value of y must be y=1/2, so (2,1/2) is another
ordered pair that satisfies the equation. 

It's always a good idea to check your work by substituting _both_
values into the equation, to see if you end up with something true:

  6(1/2) = 9 - 3(2)

       3 = 9 - 6

       3 = 3                   (Okay)

Note that if you substitute two incompatible values, you'll end up 
with something that doesn't make sense. For example, let's try the
ordered pair (5,4):

    6(4) = 9 - 3(5)

      24 = 9 - 15

      24 = -6                  (Oops!)

When you get something like this, you need to go back and find out
what you did wrong. 

When you have two variables but only one equation, you can have an
infinite number of ordered pairs like this. Note that you don't have 
to choose an x and figure out a y. You can choose a y and figure out
an x if that seems easier.  

More importantly, you can 'solve for' either variable in terms of the
other, which trades a little work up front for an easier time later. 
In this case, we can divide everything on both sides of the equation
by 6 to get 

  (6y)/6 = (9 - 3x)/6

       y = (9 - 3x)/6

Now we can just plug in a value for x, and the corresponding value of
y pops out:

  y = (9 - 3*3)/6 = 0/6 = 0

  y = (9 - 3*4)/6 = -3/6 = -1/2

and so on.  It's like turning a crank.  

Which sort of leads us back to your first question.  To 'solve for' a
variable, you look for things that you can do to both sides of an
equation to make it look like

  some variable = some expression not involving the variable

What kinds of things can you do?  Basically, anything except divide by
zero!  In practice, the three main things you do are add the same
thing to both sides of an equation; multiply both sides of an equation
by the same thing; and use the distributive property to combine or
split things up. 

For example, if you have something like 

  3x = 4y - 2x

you can add 2x to both sides of the equation to get

  3x + 2x = 4y - 2x + 2x

   (3+2)x = 4y

       5x = 4y

which is close to solving for x, but there is one more step: multiply
both side of the equation by (1/5) to leave x by itself on one side:

  (1/5)5x = (1/5)4y

        x = 4y/5

Note that we could have solved for y instead, by making the last step 

   (1/4)5x = (1/4)4y

    (5/4)x = y

The earlier steps are still necessary:

  1. Get all the terms involving the variable you want to solve
     for on one side of the equation.

  2. Simplify that side of the equation to get something that 
     looks like 

       this*variable + that = ....

  3. Add (-that) to both sides of the equation.

  4. Multiply both sides of the equation by (1/this). 

If this seems new or strange to you, you might want to look at this:

   Basic Tips on Solving for X 

Does this help? 

- Doctor Ian, The Math Forum 
Associated Topics:
High School Linear Equations

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.