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Solving an Algebra Equation

Date: 3/20/96 at 3:1:35
From: Anonymous
Subject: linear equations

 Dear Dr Math:

I hope you will be able to help me with this problem, I'm 
really stuck.

              10-(k+5) = 3(k+2)

I would also appreciate any explanation. 

Date: 3/20/96 at 15:1:58
From: Doctor Elise
Subject: Re: linear equations


It's a little confusing when you have the same variable on both
sides of the equals sign, isn't it?  We know that when the problem
is solved, it will look like "k = something".  So we have to try to
get all the "k's" on one side of the equation, and all the 
"somethings" on the other, and then we'll be done!

The first step is to do all the multiplication and addition and
subtraction on each side of the equals sign by itself.  That way
we get rid of the parentheses.

10 - (k+5) = 3(k+2)

On the left side, we have 10 minus (k plus 5).  The parentheses
mean we have to subtract BOTH k and 5 from 10. Okay.  So we can 
rewrite it without the parentheses and the left side becomes:

10 - k - 5  

And because addition and subtraction without parentheses can be done 
in any order (which is called the commutative property of addition -
the one that says a + b = b + a) we can rewrite it as 10 - 5 - k and get:

5 - k  

So that's the left side.  Let's look at the right side:

3(k+2)   Okay, we could say this as 3 times the sum of k and 2.  The
         distributive property of multiplication means that we can
         "distribute" the 3 over both the k and the 2, meaning that
         3 times  (k+2) is the same as 3 times k plus 3
         times 2.  (I'll use '*' for "times".)

3*k + 3*2

We usually write 3*k as just 3k, so this becomes:

3k + 6

Now we've simplified the left and the right sides of the original
equation, so by now we have:

5 - k = 3k + 6

Doesn't that look better?  It still doesn't look like "k = something"

Okay.  The key to moving terms around in a polynomial is that
you can do anything to one side of the equation as long as you do
the same thing to the other side. I want to add k to each side
so that I don't have a 'k' on the left, I can do it this way:

5 - k + k = 3k + 6 + k    And now I can simplify both sides by
                          adding the terms with 'k's in them:

5 = 4k + 6                Well, this is looking better!  Let's
                          subtract 6 from each side so that we
                          have only 'k's on the right:

5 - 6 = 4k + 6 - 6

-1 = 4k                   Great!  This is the same as:

4k = -1                   Now we need to get rid of that pesky 4.
                          Let's divide each side by 4:

4k/4 = -1/4

k = -1/4                   Yay!  An answer!  There were an awful lot
                           of numbers flying around to get here, so
                           I'm going to check my work by "plugging in"
                           this value of 'k' into the original equation
                           10 - (k+5) = 3(k+2):

10 - (-1/4 + 5) = 3(-1/4 + 2)

10 - (4 3/4) = 3( 1 3/4)

5 1/4 = 3 * 1 + 3 * 3/4

5 1/4 = 3 + 9/4

5 1/4 = 3 + 2 1/4

5 1/4 = 5 1/4

Okay, it checks out!  Good luck with your next polynomial!

 - Dr. Elise  - The Math Forum

Associated Topics:
Middle School Equations

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