Transcript UNIT 10

Unit 15
COMPLEX EQUATIONS
SOLVING EQUATIONS
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Remove parentheses
Combine like terms on each side of equation
Get all unknown terms on one side and all
known terms on the other using addition and
subtraction principle
Combine like terms
Apply multiplication and division principles
of equality
Apply power and root principles of quality
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EQUATIONS CONSISTING OF
COMBINED OPERATIONS
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Complex equations require the use of two
or more principles of equality for their
solutions
Solve for x: 7x + 4 = 39
7 x  4 39

Subtract 4 from both sides
4
4
7 x 35

7
7
Divide by 7
x = 5 Ans
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EQUATIONS CONSISTING OF
COMBINED OPERATIONS
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Solve for y: 5y – 3(2y – 6) = 12
5y – 3(2y – 6) = 12
5y – 6y + 18 = 12
–1y + 18 = 12
Remove parentheses
Combine like terms
– 18 – 18
Subtract 18 from both sides
 1 y 12

Divide by  1 on both sides
1 1
y = –12 Ans
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EQUATIONS CONSISTING OF
COMBINED OPERATIONS
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Solve for
2
2
p
p:  37.5  52.5
40
p
 37.5  52.5
40
 37.5  37.5
Add 37.5 to both sides
2
p
 90
40
 p2 

 
(40)  (90)(40) Multiply both sides by 40
40 
p2 = 3600
Square root both sides
p = 60 Ans
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SUBSTITUTING INTO FORMULAS
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The volume of a sphere is given as 4/3
times  times the radius cubed. Set up the
formula and determine the volume given
that the radius is 2 inches
4 3
V  r
– Set up the formula:
3
4
– Substitute 2 inches for r: V   ( 2" ) 3
3
– Solve for the volume:
V = 33.51 in3 Ans
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REARRANGING FORMULAS
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Sometimes it is necessary to rearrange a
formula to solve for another value.
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The formula must be rearranged so that the
unknown term is on one side of the equation and
all other values are on the other side.
The formula is rearranged by using the same
procedure that is used for solving equations
consisting of combined operations.
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REARRANGING FORMULAS
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Rearrange the formula A = 1/2bh to solve for
b:
1
A  bh
2
Multiply both sides by 2
2 A bh

h
h
Divide both sides by h
2A
b
h
Ans
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REARRANGING FORMULAS
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Rearrange P = I2R to solve for I:
P I 2R

R
R
Divide both sides by R
P
 I2
R
Square root both sides
I 
P
Ans
R
Whenever taking a square root there are two solutions that are
usually denoted by ± and should be in front of all roots but then are
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determined viable or not.
PRACTICE PROBLEMS
1.
7x – 3 = 11
2.
15 – (y – 7) = 50
2
r
3.    11  70
3
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PRACTICE PROBLEMS (Cont)
4.
y(3 + y) + 14 = y2 – 7
5.
x3 – 3x = 3(72 – x)
6.
Given that the volume of a right circular
cylinder is  times radius squared times
height. Determine the volume given that the
radius is three centimeters and the height is
seven centimeters.
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PRACTICE PROBLEMS (Cont)
7.
Find the Celsius temperature using the
formula C = 5/9(F – 32º) when the
Fahrenheit temperature is 72 degrees
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8. Solve F  C  32 for C
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1
9. Solve X C 
for C
2fC
10. Solve r  x 2  y 2 for x
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PROBLEM ANSWER KEY
1.
2.
3.
4.
5.
6.
7.
2
–28
27
–7
6
197.92 cm3
22.22º
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8. C  F  32
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1
9. C 
2fX C
10. x  r 2  y 2
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