# Linear Equations

```

1:  3X  +   6Y   =    9
X  -   6Y   =   11
---------------------------------
4X               =    20
-------                  --------
4                     4

X = 5

NOW WE KNOW, X=5,     (  X=5 , Y = ?  )

plug X back into either equation to figure out what Y is,

3X   +    6y  =  9
3(5) +    6Y  =  9

15   +  6y   =       9
-15           =      -15
--------------------------
6Y   =  -6

y = -1      SO SOLUTION POINT IS  (  5,  -1 )

--------------------------------------------------------------------

Problem 4 :   2X  +  3Y   =  6
-5X   + 2Y   =  4

Opps, when you add the Xs or Ys...guess what? They don't add to 0, so they don't cancel each other out.

So now what?

We have to decide, do we get rid of the Xs first or the Ys? Your choice! Both paths work towards the right solutions.

As for me, I'll pick the Y first. Let's get rid of Y first, then solve for X,
and when you calculate X, you can plug it back into either equation(both work),
and finally calculate Y.  Giving you the ( x, y ) solution point.

so take the 2 in front of the Y in the second equation and multiply the first equations by it

then take the 3 in front of the Y in the first equation and multiply the second equation by it, messy...but it works.

2(  2X  +  3Y  =  6 )
3( -5x  +  2Y  =  4 )

4X  +   6Y  =  12
-15X  +   6Y  =  12
-----------------------------

opps, the Ys add to 12Y, not zero, so what do I do to make sure they add to 0 ?

I change the sign of one of the equations, I'll do it to the second equation

4X   +    6Y  =  12
+15X   -    6Y  = -12
------------------------------
19X            =  0

19X  =  0
-----   ----
19      19

X = 0   so we know  ( x=0, y=? ) about the solution point

since you know X now, plug it back into one of the original equations(doesn't matter which you use, same answer), to calculate y,

x=0     2X  +  3Y   =  6

2(0) + 3Y   =  6

3Y   =  6
----    ----
3       3

y = 2

so our solution point is  ( 0 , 2 )

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Word Problems
____________________

12:

Find two numbers whose sum is 33 and whose difference is 13.

Ok, how do I start?

Well, I have two numbers right?

So I pick X for one and Y for the other.

Now to solve for 2 variables ( x  y ) I need two different equations.

find two numbers whose sum is  33

so one equation is  X +  Y  = 33

the next equation comes from  'whose difference is 13'

so  X - Y  = 13

now I have the two equations I need to solve for x and y

X  +  Y  =  33
X  -  Y  =  13
---------------------
2X  +  0  =  46

2X  =  46
----   ----
2       2

X =  23

ok, I have x now, so all I have to do is plug it back into an equation to solve for Y

I can pick either equation ( x +  y = 33 ) or ( x - y = 13 ), makes no difference, they will both give me the same Y

so I pick  x + y = 33  x=23

23 + y =  33
-23       -23
----------------
y =  10

so the solution point is  ( 23, 10 )

12:

Number of hot dogs      Number of sodas         Total Cost
_____________________   __________________      ____________
4                        4                    \$20
4                        6                    \$24

let's select good variable names, so H for hot dogs and S for sodas

let's create our two equations

4H  +   4S   =  20
4H  +   6S   =  24
------------------------

which variable do we get rid of first....hmmm

look at 4H and 4H in both equations, if one of those 4H was negative, then they would add to zero, and that is what we want

so change one equation to negative version by changing all the signs

take  4H  +  4S  = 20(top equation) to  its opposite by changing all the signs in the equation

so 4H + 4S  = 20  to    -4H - 4S =  -20

-4H - 4S = -20
4H + 6S =  24
--------------------
2S =  4

2S = 4
---- ---
2    2

S = 2  now we know soda costs  \$2, plug S back into one of the original equations to find out what a Hot Dog costs

4H  +  6S    =  24

4H  +  6(2)  =  24

4H  +  12    =  24
-  12      -12
---------------------
4H  +  0     =  12

4H =  12
----  ----
4      4

H = 3   so a Hot Dog costs \$3

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