Solving an Equation
• Finding ALL possible values that make an
equation true.
• You are done solving an equation when you
can look at an equation and “know” what
values make it true.
• If you can’t look at an equation and “know”
what values make it true, try rearranging the
equation to make it easier to look at.
Examples
x2
3x  1
2x  3  7
4( x  1)  8  2 x
Examples
x2
x=2
x=1/3
3x  1
2x=10;
2x  3  7
4( x  1)  8  2 x
x=5
Solving an Equation
• Finding ALL possible values that make an
equation true.
• You are done solving an equation when you
can look at an equation and “know” what
values make it true.
• If you can’t look at an equation and “know”
what values make it true, try rearranging the
equation to make it easier to look at.
Ways to rearrange equations
• Properties of arithmetic
– Associative, Commutative, and Distributive
– Use these to get rid of parentheses and combine like terms
– These change the shape of one side of the equation, but not its value.
4(2 +1) + 3* 2 = 18
4 * 2 + 4 *1+ 3* 2 = 18
4(x +1) + 3x = 8 + 2x
4x + 4 + 3x = 8 + 2x
4 * 2 + 3* 2 + 4 *1 = 18
4x + 3x + 4 = 8 + 2x
7* 2 + 4 = 18
7x + 4 = 8 + 2x
Doing the same thing to both sides
• Adding, Subtracting, Multiplying, Dividing, a number
from both sides of the equation.
– Changes the value of both sides, but not the equality.
12 = 12
- 6 = -6
6=6
4x = 4 + 2x
- 2 x = -2 x
2x = 4
/2=/2
/2=/2
3=3
x=2
WARNING
• Each side is a number. When multiplying (or dividing) multiply
(or divide) the whole number.
GOOD
BAD
GOOD
2+4x = 6
2+4 = 6
2+4 = 6
2 + 4 / 2 = 6 / 2 (2 + 4) / 2 = 6 / 2 (2 + 4x) / 2 = 6 / 2
2 / 2 + 4x / 2 = 3
2+2 =3
6/2=3
1+ 2x = 3
4=3
3=3
x =1
Rearranging an equation to solve
4( x  1)  8  2 x
4x  4  8  2x
 4  4
4x  4  2x
 2 x  2 x
2x  4
x2
Solve for K: 2(K-7) = 4 - 3(K+2)
A)
B)
C)
D)
K=12/5
K=5/24
K=24/5
None of the above
Solve for K: 2(K-7) = 4 - 3(K+2)
2(k - 7) = 4 - 3(k + 2)
2(k - 7) = 4 + -3(k + 2)
2k -14 = 4 + -3k - 3(2)
2k -14 + 3k = 4 + -3k - 6 + 3k
2k -14 + 3k = 4 - 6
5k = -2 +14
5k = 12
k = 12 / 5
A
Solve an inequality
• Method 1: Convert to an equality, solve, then
test.
– Slower
– Works for complex problems later in the course
• Method 2: Solve the inequality like an equality
– Faster
– Easier to make mistakes
– Does not work for complicated problems like
quadratic inequalities.
Method 1
•
•
•
•
•
•
Start with inequality:
8<4-2x
Convert to an equality:
8=4-2x
Solve:
-2=x
Draw a number line <------------|------------------> x
x<-2 -2
x>-2
Your possible answers are x<-2 and x>-2. Pick a number on
each side to test which is correct.
• Test (x<-2) with x=-3: 8<4-2(-3) 8<10 TRUE
• Test (x>-2) with x=0: 8<4-2(0) 8<4
FALSE
• ANSWER: x<-2
Method 2
• Start with inequality:
• Solve by doing the same thing to both sides
•
•
•
8<4-2x
8+2x<4
2x<4-8
2x<-4
x<-2
PROBLEM
• Using method 2, if I do this problem two
different ways, I get two different answers.
8 < 4 - 2x
8 + 2x < 4
8 < 4 - 2x
8 - 4 < -2x
2x < -4
4 < -2x
x < -2
DON’T MATCH!
-2 < x
Which is Correct?
• I know x<-2 is correct, because when I did
method 1, I plugged in to check my work.
8 < 4 - 2x
8 + 2x < 4
8 < 4 - 2x
8 - 4 < -2x
2x < -4
4 < -2x
x < -2
-2 < x
CORRECT
INCORRECT
How did it happen?
• When I turned -2x into +2x, I moved it to the other
side of the <. When I divided by -2, I turned negative
into positive, but didn’t move it to the other side.
Moved x
8 < 4 - 2x
8 + 2x < 4
8 < 4 - 2x
8 - 4 < -2x
2x < -4
4 < -2x
x < -2
-2 < x
Never moved x.
How to fix it
• When I divide by a negative, I can have the same
effect as “moving to the other side” by “flipping the
sign”
Moved x
8 < 4 - 2x
8 + 2x < 4
8 < 4 - 2x
8 - 4 < -2x
2x < -4
4 < -2x
x < -2
-2 > x
Answers Match
Flipped the sign
NEW RULE
• When multiplying or dividing an inequality by
a negative number, flip the sign of the
inequality
-x < 7
x > -7
• Or, just never multiply by a negative.
-x < 7
0 < 7+ x
-7 < x
Solve the following inequality:
-3(x+4) > 53
A)
B)
C)
D)
E)
x > 65/3
x > -65/3
x < 65/3
x < -65/3
None of the above
Solve the following inequality:
-3(x+4) > 53
-3(x + 4) > 53
-3x + -3(4) > 53
-12 > 53+ 3x
-65 > 3x
-65 / 3 > x
D
How to set up word problems
How to set up a word problem
(with equation given)
• A jewelry maker has total revenue for necklaces given by
R(x)=90.75x, and incurs a total cost of C(x)=24.50x+4770,
where x is the number of necklaces made and sold. How many
necklaces must be produced and sold in order to break even?
Step 1: Define the variables
• A jewelry maker has total revenue for necklaces given by
R(x)=90.75x, and incurs a total cost of C(x)=24.50x+4770,
where x is the number of necklaces made and sold. How many
necklaces must be produced and sold in order to break even?
• x=number of necklaces
• R(x)=number of $ earned for x necklaces
• C(x)=number of $ spend to make x necklaces
Step 2: Interpret the question
• A jewelry maker has total revenue for necklaces given by
R(x)=90.75x, and incurs a total cost of C(x)=24.50x+4770,
where x is the number of necklaces made and sold. How many
necklaces must be produced and sold in order to break even?
• x=number of necklaces
• R(x)=number of $ earned for x necklaces
• C(x)=number of $ spend to make x necklaces
• x=?
• R(x)=C(x)
Step 3: Substitute
• A jewelry maker has total revenue for necklaces given by
R(x)=90.75x, and incurs a total cost of C(x)=24.50x+4770,
where x is the number of necklaces made and sold. How many
necklaces must be produced and sold in order to break even?
• x=number of necklaces
• R(x)=number of $ earned for x necklaces
• C(x)=number of $ spend to make x necklaces
• x=?
• R(x)=C(x)
• 90.75x=24.50x+4770
DONE!
• A jewelry maker has total revenue for necklaces given by
R(x)=90.75x, and incurs a total cost of C(x)=24.50x+4770,
where x is the number of necklaces made and sold. How many
necklaces must be produced and sold in order to break even?
• x=number of necklaces
• R(x)=number of $ earned for x necklaces
• C(x)=number of $ spend to make x necklaces
• x=?
• R(x)=C(x)
• 90.75x=24.50x+4770 solve for x.
How to set up a word problem
(without an equation)
• Suppose that in a certain business, there were 30 employees
in the year 1920, and 38 employees in the year 1940.
Assuming linear growth, In what years will the business have
more than 40 employees?
Step 1: Identify your variables
• Suppose that in a certain business, there were 30 employees
in the year 1920, and 38 employees in the year 1940.
Assuming linear growth, In what years will the business have
more than 40 employees?
• P=number of employees
• t=number of years
Step 2: Identify your Points
• Suppose that in a certain business, there were 30 employees
in the year 1920, and 38 employees in the year 1940.
Assuming linear growth, In what years will the business have
more than 40 employees?
• P=number of employees
• t=number of years [Put independent variable first]
• (t,P): (1920,30), (1940,38)
Step 3: Interpret your question
• Suppose that in a certain business, there were 30 employees
in the year 1920, and 38 employees in the year 1940.
Assuming linear growth, In what years will the business have
more than 40 employees?
• P=number of employees
• t=number of years
• (t,P): (1920,30), (1940,38)
• t=?
• P>40?
Step 4: Choose your generic formula
• Suppose that in a certain business, there were 30 employees
in the year 1920, and 38 employees in the year 1940.
Assuming linear growth, In what years will the business have
more than 40 employees?
• P=number of employees
• t=number of years
• (t,P): (1920,30), (1940,38)
• t=?
• P>40?
• (y-b)=m(x-a)
Step 5: Match variables and plug in numbers
• Suppose that in a certain business, there were 30 employees
in the year 1920, and 38 employees in the year 1940.
Assuming linear growth, In what years will the business have
more than 40 employees?
• P=number of employees
• t=number of years [Make independent variable x]
• (t,P): (1920,30), (1940,38)
• t=?
• P>40?
• (y-b)=m(x-a)
• (P-30)=[(38-30)/(1940-1920)](t-1920)
Step 6: Rearrange and substitute
• Suppose that in a certain business, there were 30 employees in the
year 1920, and 38 employees in the year 1940. Assuming linear
growth, In what years will the business have more than 40
employees?
• P=number of employees
• t=number of years [Make independent variable x]
• (t,P): (1920,30), (1940,38)
• t=?
• P>40?
• y=m(x-a)+b
• (P-30)=[8/20](t-1920)
• P=[8/20](t-1920)+30>40
DONE!
• Suppose that in a certain business, there were 30 employees in the
year 1920, and 38 employees in the year 1940. Assuming linear
growth, In what years will the business have more than 40
employees?
• P=number of employees
• t=number of years [Make independent variable x]
• (t,P): (1920,30), (1940,38)
• t=?
• P>40?
• y=m(x-a)+b
• (P-30)=[(38-30)/(1940-1920)](t-1920)
• P=[8/20](t-1920)+30>40, solve for t.
A plumbing store’s monthly profit from the sale of PVC
pipe can be described by P(x)=5.25x-1150 dollars, where
x is the number of feet of PVC pipe sold. Set up an
inequality that would help determine what level of
monthly sales is necessary to incur positive profit.
A)
B)
C)
D)
E)
5.25x >1150
5.25x+1150>0
5.25x+1150=0
Both A & B
All of the above
A plumbing store’s monthly profit from the sale of PVC pipe can be
described by P(x)=5.25x-1150 dollars, where x is the number of feet of
PVC pipe sold. Set up an inequality that would help determine what
level of monthly sales is necessary to incur positive profit.
Define Variables
P(x)=number of dollars of profit
x=number of feet of pipe sold
Interpret Question
x=?
P(x)>0
Substitute
5.25x-1150>0
A) 5.25x>1150