Download Evaluating and simplifying variable expressions, translating verbal

Chapter 2: Variable Expressions
Expressions (contain no “=” sign) :
An expression is one or numbers or variables having some mathematical operations done on
them.
Numerical Expressions:
3+5
3(4)
6/2
5-1
4
Expressions can be evaluated or simplified: 3 + 5 can be simplified to 8
This just means, “Whatever you see, do”
Algebraic Expressions:
x+5
3x ç in Algebra, it is implied that 3 x means 3 times x. The “proper” way to write the
product of a number and a variable is to always write the number to the left of the variable.
x times 5 = 5x
When numbers are multiplied by variables, they are given a special name, “ coefficient”.
5 is the coefficient of 5x.
The quantities being added in an algebraic expression are called the terms.
If a term is a variable or a combination of variables multiplied by numbers, it is called a
variable term. Numbers that are just added in the expression are called constant terms.
Example:
3 terms
4x2 - 2x – 3
variable terms
constant term
Hey! 2x and 3 aren’t being added, they are being subtracted! If we write this expression as
4x2 + -2x + -3, then we have all term being added, and the constant term is -3 and the
variable terms are 4x2 and -2x.
Algebraic expressions can be simplified by using the associative and
distributive properties.
-4m(-5n) can be simplified by rearranging the terms (we can do this when
the only operation is multiplication. This is the associative property.) so
that all the constants are grouped together and all the variables are grouped
together(alphabetically))
-4m(-5n)
= (-4)(-5)mn
= 20mn
2(-4z)(6y)=2(-4)(6)yz = -48yz
3(s+7) can be simplified by using the distributive property
=3s + 3(7)
= 3x + 21
-6(-3x - 6y + 8) can also be simplified. Change any subtraction to adding a
negative.
-6(-3x + -6y + 8)
Now distribute the -6 and make sure to glue the negative sign on that -6
wherever you distribute it!
-6(-3x) + -6(-6y) + -6(8)
= 18x + 36y + -48
= 18x + 36y - 48.
Example 7: Simplify
4(x - y) – 2(-3x + 6y)
=4x – 4y + 6x - 12y use distributive property to get rid if parentheses
= 4x + 6x - 4y -12y regroup terms to put like terms together
= 10x - 16y
combine like terms
You do this one: Simplify 7(x - 2y) – 3(-x – 2y)
LIKE TERMS
Like terms are terms with exactly the same variables raised to the exactly the
same powers. Any constants in an expression are considered like terms. Terms
that are not like terms are called unlike terms.
Like Terms
Unlike Terms
2x, 3x, -4x
2x, 2x2
Same variables, each
with a power of 1.
Different powers
3, 5, -1
3, 3x, 3x2
Constants
Different powers
5x2, -x2
5x2, 5y2
Same variables and
same powers
Different variables
Only Like Terms can be combined!
Combining like terms is to add or subtract like terms. Combine like
terms containing variables by combining their coefficients and keeping
the same variables with the same exponents.
Example: 3x2 – 8x2 = (3 - 8)x2 = -5x2
Example: 5x + 2x = (5 + 2)x = 7x
Example: 3x2 + 5x + 2 + x 2 – x - 3
=3x2 + x2 + 5x-x + 2-3
= 4x2 + 4x – 1
Algebraic expressions can only be simplified if they have “like terms,”
which are terms with the same variable and same exponent.
x+5+2
Only 5 + 2 can be simplified.
x + 5 + 2 becomes x + 7
3x + x + 2
can be simplified to 4x + 2
x2 + x
cannot be simplified because they don’t have the same
exponents and are therefore not like terms.
Algebraic expressions can only be EVALUATED if you are given the value for the
variables.
−x −15
6
Example: Evaluate
for x = 3. Substitute (3) for x in the expression.
− (3) − 15 − 3 − 15 − 3 + ( −15) − 18
=
=
=
= −3
6
6
6
6
Example 2 p.68
If a=2 and b=-3, evaluate ab - b2
Don’t forget PEMDAS!
(2)(-3) - (-3)2 Do exponents first
= (2)(-3) - 9 Multiplication left to right
= -6 – 9
Subtraction
= -15
Example 3 p. 68 Evaluate when a=3 and b=-4
a 2 − b2
a−b
(3) 2 − ( −4 ) 2
=
[(3) − (−4)]
[9 − 16 ]
=
[7]
−7
=
7
= −1
[
]
For fractions, assume that the numerator and denominator are
separate groupings (put [ ] around each one) and do what’s
inside each grouping first.
Do exponents.
Do subtraction in the numerator grouping.
Now that the numerator and denominator groupings have been
simplified, you can divide.
You do this one : Evaluate for a = 5 and b = -3
a 2 + b2
a+b
Translating Verbal Expressions in to Mathematical Expressions
Verbal Expressions Examples
Math Translation
Addition
added to
more than
the sum of
increased by
the total of
6 added to y
8 more than x
the sum of x and z
t increased by 9
the total of 5 and y
6+y
8+x
x+z
t+9
5+y
Subtraction
minus
less than
subtracted from
decreased by
the difference
between
x minus 2
x-2
7 less than t
t-7
5 subtracted from 8
8-5
m decreased by 3
m-3
the difference between y and 4 y-4
Multiplication times
of
the product of
10 times 2
one half of 6
the product of 4 and 3
10 X 2
(1/2) X 6
4X3
multiplied by
y multiplied by 11
11y
divided by
the quotient of
x divided by 12
the quotient of y and z
x/12
y/z
the ratio of
the ratio of t to 9
t/9
Power
the square of
the cube of
squared
the square of x
the cube of z
y squared
x2
3
z
y2
Equivalency
equals
is
is the same as
1+2 equals 3
2 is half of 4
½ is the same as 2/4
1+2 = 3
2 = (½)X4
yields
represents
3+1 yields 4
y represents x+1
3+1 = 4
y=x+1
greater than
less than
greater than or equal
to
at least
no less than
less than or equal to
at most
no more than
-3 is greater than -5
-3 > -5
-5 is less then -3
-5 < -3
x is greater than or equal to 5 x = 5
Division
1
2
=
2
4
Comparison
x is at least 80
x is no less than 70
x is less then or equal to -6
y is at most 23
y is no more than 21
x
x
x
y
y
= 80
= 70
= -6
= 23
= 21
Solving Application Problems
Problem-Solving Strategy:
•
Analyze the problem. What are you trying to find? Label variables to the unknown quantities.
What’s the given info?
•
Work out a plan before starting. Draw a sketch if possible. Look for indicator words (e.g.
gained, lost, times, per) to know which operations (+,-, x,÷) to use.
•
Estimate a reasonable answer.
•
Solve the problem.
•
Check your work. If the answer is not reasonable, start over.
Example 6 on p. 89:
The length of a swimming pool is 20ft longer than the width. Express the length of the pool in
terms of the width.
Step 1) What are we trying to find? The length of the pool. Let l = length.
What’s the given info? The length is 20ft longer than the width. What’s the width?
We don’t know. So set w = width.
Step 2) Look for indicator words. “is” means =
Draw a sketch.
, “longer than” means +
l=20+w
The length is 20 ft longer than the width.
l
w
= 20 + w
We have now expressed the length of the pool in terms of the width. Since we don’t know what
the width actually is, this is as far as we can go.
The answer is 20 + w
Is this reasonable for the length of the pool (this is what we were asked to express)?
Put in any number of w and see if the length 20 + w is reasonable. Yes.
YOU DO THIS ONE:
An older computer takes twice as long to process a set of data as does a new model. Express the
amount of time it takes the older computer to process the data in terms of the amount of
time it takes the newer model.
Let c = time older computer takes to process data.
n = time newer computer takes to process data.
Write c as expression with n in it.
p. 95 #121
A coin bank contains thirty-five coins in nickels and dimes. Use the same
variable to express the number of nickels and the number of dimes in the coin
bank.
Step 1) What are we being asked to find? The number of nickels and dimes
expressed with the same variable. We’ll let d = number of dimes. Right now
let’s use another variable, n = number of nickels, and then we’ll make an
equation to express nickels in terms of dimes.
Step 2) What’s the given info. Total coins are 35. That is,
The number of dimes + the number of nickels = 35
d
+
n
= 35
Solving for n, we get
n
= 35 – d
Now we can say the number of dimes = d
and the number of nickels = 35 – d
Step 3) Is this what we were asked to find? Yes.
You do #120
A halyard 12ft long is cut into two pieces. Use the same variable to express
the lengths of the two pieces.
Example:
p. 96 #127 A wire whose length is given as x inches is bent into a square.
Express the length of a side of the square in terms of x.
Draw a picture. We are asked to express the length of a side of the square in
terms of x. Let’s let s = length of a side of the square. What’s given? The
wire (of length x) is bent into a square.
x
What do we know about squares? The perimeter
(or length going around the square) is s + s + s +s, and the total length going
around the square is x (given). s + s+ s+s = x
1s + 1s +1s + 1s = x
4s =x
s = x/4