Download Algebra fundamentals

Algebra Fundamentals
The Basics
Imagine the following simple arithmetic equation.
12=3
Although it is a stretch in this case, suppose that one does not know one of the numbers
?2=3
This states that some unknown number plus 2 equals 3. While this example is simple
enough that one could probably determine that the unknown number is 1, again let us
pretend that we do not know this. In that case it is convenient to replace the question
mark with a letter called a variables.
x2=3
Doing math with unknown numbers is all that algebra is about! The rest of one’s study
of algebra is all about technique, strategy and terminology.
Before continuing a piece of terminology must be defined. In algebra we define a term as
a segment of a mathematical statement separated by an addition, subtraction or equal
sign. So in our last equation the x, 2 and 3 are all separate terms.
Example Set 1: Replace the number in bold as a letter representing an unknown number.
Choice of variables can vary. Solutions are included at the end.
1. 3−2=1
2. 14=5
3. 27=9
Adding in an unknown
Let us return to the previous example and explore how we can tweak it and still maintain
a true statement. So first try simply subtracting 2 from the left side.
12=3 (TRUE)
12−2=3
1=3 (FALSE)
We took a true statement and corrupted it into a false statement. However we can still
save the day by subtracting 2 from the other side as well.
12−2=3−2
1=1 (TRUE)
So by subtracting both sides by the same number we maintain the truthfulness of a
statement. But also note that by subtracting by 2 the number two previously there
dropped out. So the lessons learned are as follows.
• When one subtracts from one side of an equation, one must do likewise to the other
side to maintain a true statement.
• When one subtracts a number from an equation its like number drops out of the
equation.
To illustrate the point consider the equation
x2=3
1.
We will subtract two from both sides, maintaining a true statement
x2−2=3−2
Subtracting 2 will eliminate the original two. This gives us
x=1
So our unknown number is 1 (surprise!) A similar logic exists when one has subtraction
x−2=3
x−22=32
x=5
Let’s check this statement by substituting 5 into the original equation.
5−2=3 (TRUE)
Strategy lesson:
To get rid of a term being added one subtracts the same number. Likewise if a term is
subtracted then adds. Always add or subtract to same number to both sides.
Example Set 2: Identify (aka solve) the unknown number using addition or subtraction.
Solutions are at the end.
x5=12
4.
y1=2
5.
3
z=5
6.
7. a−5=3
8. h−7=10
s3=2
9.
Multiplying
Now suppose that the first equation above is modified to use multiplication.
1×2=2
We can do the same trick as above by supposing that we do not know one of the
numbers.
x×2=2
One can imagine the confusion that could arise if one did not know that is multiplication
symbol was not an x! To avoid confusion we express multiplication using a ∙ between
known numbers or leave nothing otherwise.
2 x =2
A house divided
So is there a trick for multiplication similar to the above? Go back to the original
equation and divide both sides by 2
1×2=2 (TRUE)
1×2÷2=2÷2
1=1 (TRUE)
So it appears that, like addition, dividing eliminates a product and maintains a true
statement if it is done to both sides. So now repeat using variables. Only now utilize the
notation that a fraction is really a division
2.
2 x =2
2x 2
=
2 2
x=1
The reverse is also true. If one has division that needs to be undone then use
multiplication.
x
=4
3
3x
=3⋅4
3
x=12
Since 12 divided by 3 is 4 then this is a valid solution.
Strategy lesson:
To eliminate a portion of a term use either multiplication or division. If the term
involves multiplication then use division and vice versa.
Example set 3: Solve the unknown number using multiplication or division. Solutions
are at the end.
10. 4 y=20
11. 7 x=56
12. 9 w=7
t
=25
13.
5
p
=20
14.
3
2e
=4
15.
5
Mixing things up
More general mathematical problems involve a combination of addition, subtraction,
multiplication and division. While our lessons learned are still of use we need to consider
our strategy for their use in solving for unknown numbers. First consider the following
true arithmetic statement that we can apply subtraction to
3×23=9 (TRUE)
3×23−3=9−3
6=6 (TRUE)
There is nothing new here as we’ve seen all this before. However we could have taken
this in a different direction and ended up with
3×2=9−3
Note that the 3 term on the left is effectively removed. We can repeat the process
(almost) using a variable
3 x3=9
3 x3−3=9−3
3.
3 x=9−3
3 x=6
3x 6
=
3 3
x=2
The moral to the story is that we can use subtraction (or addition as occasion requires) to
remove a single term to the other side. We can use this knowledge to remove any
unwanted term we wish from a single side of an equation. The price to be paid is that we
must apply the same process to the other side.
What about division? Let us repeat the same as above
3×23=9
3×23÷3=9÷3
61=3
7=3 (FALSE)
Here we must have incorrectly applied the division since we began with a true statement
and ended up with a false one. So let us apply a slightly different logic
3×23=9
3×2÷33÷3=9÷3
21=3
3=3 (TRUE)
This approach is correct, but it must be remembered that we applied the division to all
terms, not just to both sides. Again applying algebra to a previous conclusion
3 x=6
3x 6
=
3 3
x=2
The result is that just a part of the term (here the 3) has been surgically removed, this is in
contrast to subtraction in which the entire term gets removed. So now to summarize our
conclusions:
•
•
Apply addition or subtraction to both sides to remove a term from its original side.
Multiplication or division must be applied to all terms on both sides of the equal sign.
These can be used to remove just a part of a term.
This leads to our grand strategy for solving these types of math problems.
 The final goal is to totally isolate the variable to solve for, such as “x=…” or “y=…”
 Use addition or subtraction to move terms across the equal sign.
 Use multiplication or division to remove a portion of a term so as to free up the
variable. One usually isolates the term with the variable using addition or subtraction
and then multiplies or divides as a last step.
Example Set 4: Solve for the unknown variable. Solutions are below.
16. 2 d 6=4
4.
17. 6 t−5=13
18. 5 w−2=w
19. x−7=5−2 x
y
10=7
20.
7
Solutions
The original questions are given with the steps and solutions indented. The final
solutions are boxed.
Set 1:
1. 3−2=1
x−2=1
2. 14=5
1z =5
3. 27=9
q7=9
Set 2:
x5=12
4.
x5−5=12−5
x=7
y1=2
5.
y1−1=2−1
y=1
6. 3 z=5
z 3=5
z 3−3=5−3
z =2
7. a−5=3
a−55=35
a=8
8. h−7=10
h−77=107
h=17
s3=2
9.
s3−3=2−3
s=−1
Set 3:
10. 4 y=20
4 y 20
=
4
4
y=5
11. 7 x=56
Replace the 3 with an x
Replace the 4 with a z
Replace the 2 with a q
Subtract 5 from both sides
The 5s cancel
Subtract 1 from both sides
The 1s cancel
Swapping the ordering of addition (optional)
Subtracting 3 from both sides
The 2s cancel
Add 5 to both sides
The 5s cancel
Add 5 to both sides
The 7s cancel
Subtract 3 from both sides
The 3s cancel
Divide all terms by 4
The 4s cancel and 20/4 = 5
5.
7 x 56
=
7
7
x=8
Divide all terms by 7
The 7s cancel and 56/7 = 8
12. 9 w=7
w=
13.
7
9
t
=25
5
5 t
⋅ =25⋅5
1 5
t=125
14.
Multiply all terms by 5
The 5s cancel and 25×5 = 125
p
=20
3
3 p
⋅ =20⋅3
1 3
p=60
15.
Divide all terms by 9, the answer is irreducible
Multiply all terms by 3
The 3s cancel and 20×3 = 60
2e
=4
5
5 2e 4 5
⋅ = ⋅
2 5 1 2
20
e=
2
e=10
Set 4:
16. 2 d 6=4
2 d 6−6=4−6
2 d =−2
2 d −2
=
2
2
d =−1
17. 6 t−5=13
6 t−55=135
6 t=18
6 t 18
=
6
6
t=3
18. 5 w−2=w
5 w−22=w2
5 w=w2
5 w−w=w2−w
4 w=2
4w 2
=
4 4
Multiply both sides by 5/2, remember that 4 = 4/1
The fraction on the left cancels
Reduce the remaining fraction with 20/2 = 10
Subtract 6 from both sides
The 6s cancel
Divide all terms by 2
The 2s cancel
Add 5 to both sides
The 5s cancel
Divide by 6
The 6s cancel, and reduce 18/6 to 3
Add 2 to cancel the “-2” term on the left
Cancel the ”-2” term on the left
Subtract w to cancel the w term on the right
Cancel the w term on the right side
Divide both sides by 4
6.
1
2
x−7=5−2 x
x−77=5−2 x7
x=12−2 x
x2 x=12−2 x2 x
3 x=12
3 x 12
=
3
3
x=4
y
10=7
7
y
10−10=7−10
7
y
=−3
7
7 y
⋅ =−3⋅7
1 7
y=−21
w=
19.
20.
Divide by 4 and reduce 2/4 to ½
Add to both sides to remove it from the left
Cancel the 7s on the left and add 5+7 on the right
Add 2x to both sides to bring the xs to one side
Combine like terms and cancel the 2x on the right
Divide by 3
The 3s cancel and 12/3=4
Subtract 10 from both sides
The 10s cancel and 7-10 = -3
Multiply by 7 (or 7/1) to isolate the y
The 7s cancel
Prepared by Kevin Gibson
www.mc.maricopa.edu/~kevinlg/supplements/
7.