MATH
MADEEASY
I am a student at UGHS, and
the purpose of this web site is to make the understanding of a few types
of Algebra problems simple, and easy to understand. Often Algebra
is perceived as being difficult, and frustrating.
Hopefully after my
explanation, you will see that is NOT the case. We will explore 4
different types of problems, with those being, GRAPHING, SUBSTITUTION,
ELIMINATION, and a WORD PROBLEM.
The 1st type of problem we will
explore is a simple Substitution Problem
You are given the following:
y = 5 - 4x
3x - 2y = 12
First, you must realize the
'point' of a substitution problem is to get the values for the variables.
(in this case, 'x' and 'y') You will use one equation to get a value
for either of the variables, and then plug that in to get the exact value
of the exponent. That's a little difficult to understand at this
point, but it will become very simple after a little explanation.
y= 5 - 4x
3x - 2y = 12
|
You already
have a 'value' for y. This is known because the first equation begins
with 'y='
|
|
3x - 2(5
- 4x) = 12
|
You plug
that value for y into the other equation where the variable y is given.
In this case you will replace the 2y with the value for y.
|
3x - 2(5
- 4x) = 12
3x - 10 (2*-4x)
= 12
3x -10 + 8x
= 12
|
You now
use simply the distributive property to distribute the -2 amongst all that
lies within the parenthesis
|
3x
-10 + 8x = 12<
11x-10=12
|
Now you
combine like terms. (the x's with the x's and the numbers with the numbers)
|
11x - 10
= 12
+10 +10
--------------------
11x = 22
|
You must
now try to isolate the variable. (x) You do this by adding 10 to
each side of the equation. (or add negative 10)
|
11x
= 22
11
11
-------------------
x= 2
|
You now
go another step to get the x isolated by dividing each side by 11
|
|
y= 5 - 4(2)
|
Yay! 
Yay!
You now have
an exact value for x. Now all you have to do is plug that value in either
equation.
|
y= 5 - 4(2)
y= 5 -
8
y= -3
|
All that
needs to be done is simple arithmetic. (multiply the -4 with the 2, then
subtract that value from 5)
|
|
|
Congratulations!
The values for
x and y are now clearly evident. Just put them into ordered pair
format.
|
7x + 2y = -1
3x - 4y = 19
|
You must first choose
if you want to eliminate the x or the y. Since the y's are easily
multiplied together, it is wisest to choose to eliminate the y
|
|
2(7x + 2y =- 1)
|
To get the y's to
eliminate you must make the sum of the y's from both equations equal 0.
This is easily done by multiplying the 1st equation by 2
|
|
14x + 4y = -2
|
Now the 2 must be
distributed amongst the rest of the equation. (Multiply 2 with 7x,
2 with 2y and 2 with 1)
|
14x + 4y
= -2
3x - 4y =
19
------------------
17x =
17
|
You are now left with
14x + 4y = -2 and the other original equation of 3x - 4y = 19. You
now combine the like terms with the equations like it were a simple addition
problem.
|
17x = 17
17
17
--------------
x =
1
|
All that is left is
a simple arithmetic problem. Just divide the 17x by 17 to get the
value of x
|
|
3(1) - 4y = 19
|
Just plug the value
for x (1) into either equation to get y.
|
3(1) - 4y = 19
3 - 4y = 19
|
Distribute the 3 to
the 1
|
3 - 4y = 19
-3
-3
---------------
-4y = 16
|
Subtract 3 from both
sides to get the y isolated
|
-4y = 16
-4
-4
------------
y = -4
|
Divide 16 by -4 to
get the y completely isolated.
|
|
|
Congratulations,
the values for x and y
are found!
|
Y
OU COULD EASILY DO PROBLEMS ALL DAY THAT HAVE THE ACTUAL PROBLEM GIVEN
TO YOU IN A SIMPLE FORMAT.
THESE
SKILLS CAN BE USED TO SOLVE EVERYDAY PROBLEMS. W
E WILL WORK ONE OUT NOW.
Calvin
was having a hard day trying to sell lemonade for 15 dollars, so he decided
to mark his prices down a bit. Calvin sold 30 glasses of lemonade
total. He sells regular lemonade, and pink lemonade. Regular lemonade
sells for $9.95 and pink lemonade sells for $10.50. If their total
amount of money received for lemonade sales $310.60, how many of each lemonade
were sold?
r = regular
lemonade
p= pink
lemonade. |
You must
first develop 2 equations from the given information. Lets allow
"r'' to be the variable for number of regular lemonades sold, and "p" for
amount of pink lemonade sold.. |
|
r
+ p = 30
|
You
must then gather the given information, and put it into a format where
it would be easy to determine the values for the variables. The most
obvious equation is that pink lemonade sales plus regular lemonade sales
equals 30.
|
r(9.95)
+ p(10.50) = 310.50
|
What
else do you know?
You know
the prices for each type of lemonade. Well how can you work that into an
equation? Since you also know how much money total was made, you
COULD say that the price of each lemonade times the amount sold of each
is equal to $310.50. Just put that into equation form.
|
10(r9.95
+ p10.50 = 310.50)
r995 +
p1050 + 31050
|
Decimals
aren't impossible to work with, but it's simplier to use whole numbers.
So just multiply the equation by 100.
|
r995
+ p1050= 31050
r + p
= 30
|
You
now have two equations to work with. Good job!
|
|
Now
the question is, do we use linear combination, or substitution. For
the time being, lets use elimination.
|
r
+ p = 30
- p=
- p
r = 30
- p
|
The
second equation (r + p = 30) is easier to put into a format that will allow
you to perform substitution. If you subtract p from both sides you
are left with, "r = 30 - p"
|
995(30
- p) + 1050p = 31060
29850
-995p + 1050p = 31060
29850
+ 55p = 31060
|
Use
this value of r by plugging it into "r995 + p1050 = 31050"
|
29850
+ 55p = 31060
-29850
-29850
55p =1210
|
Now
just subtract 29850 from both side
|
55p
= 1210
55
= 55
p = 22
|
Divide
1200 by 55 to get the value for p
|
r
+ 22 = 30
-
22 22
r = 8
|
Plug
this value for x into either equation
|
p
= 22
r = 8
|
~~CoNgRaTuLaTiOnS~~
You now
have both p and r
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