Transcript x - Wiley

C3
Simple Equations
and Word
Problems
Prepared by:
Richard Mitchell
Humber College
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
3.1-Solving First-Degree Equations
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
3.1-STRATEGY-Page 86
Eliminate Fractions:
Multiply both sides of the equation by the lowest common denominator.
Remove Parenthesis:
Brackets are multiplied away.
Collect x Terms:
Move all x terms to one side and all other terms to other side.
Combine Like Terms:
Always simplify.
Remove Coefficient of x:
Divide both sides by coefficient.
Check Answer:
Be sure to substitute the answer back into the original equation.
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
3.1-EXAMPLE 4-Page 84
Solve the equation for x:
3x = 8 + 2x
CHECK:
3(8) = 8 + 2(8)
3x – 2x = 8 + 2x – 2x
24 = 8 + 16
x=8+0
24 = 24 (checks)
x=8
ANS:
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
x=8
3.1-EXAMPLE 7-Page 85
Solve the equation for x:
3(3x + 1) – 6 = 5(x – 2) + 15
CHECK:
3[3(2)+1] - 6 = 5[(2)-2] + 15
9x + 3 – 6 = 5x – 10 + 15
3[6 + 1] – 6 = 5[2 – 2] + 15
9x – 3 = 5x + 5
3[7] – 6 = 5[0] + 15
9x – 5x = 5 + 3
21 – 6 = 0 + 15
4x = 8
15 = 15
(checks)
x=2
ANS:
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
x=2
3.1-EXAMPLE 8-Page 86
Solve the equation for x:
CHECK:
x
25
3
(21)
25
3
x

3    2   3  (5)
3

7–2=5
 x
3     3  (2)  3  (5)
 3 1
5=5
1
(checks)
x – 6 = 15
ANS:
x = 15 + 6
x = 21
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
x = 21
3.1-EXAMPLE 9-Page 86
Solve the equation for x:
CHECK:
(18)
(18)
3
2
3
x 3 x
 
2 1 3
 x 3
 x
6    6 
 2 1
3
-9 + 3 = -6
 x
 3 2  x 
6   6   6 
 21
1
 31 
-6 = -6
3
(checks)
3x + 18 = 2x
ANS:
3x – 2x = -18
x = -18
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
x = -18
3.1-EXAMPLE 10-Page 87
Solve the equation for x:
3x  5 2( x  1) 4


2
3
1
 3x  5 
 2( x  1) 4 
6
 
  6
1
 2 
 3
 3x  5  2  2( x  1) 
4
6
  6
  6 
 21 
 31 
1
3
9x – 15 = 4x - 4 + 24
9x – 4x = 24 - 4 + 15
5x = 35
x=7
3(3x - 5) = 4(x - 1) + 6(4)
ANS:
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
x = 7 (con’t)
3.1-EXAMPLE 10-Page 87
Solve the equation for x:
3x  5 2( x  1) 4


2
3
1
16 12
 4
2 3
CHECK:
8 44
3(7)  5 2(7  1)

4
2
3
8  8 (checks)
21  5 2(6)

4
2
3
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
ANS:
x=7
3.1-EXAMPLE 11-Page 87
Solve the following for x:
3(x + 2)(2 – x) = 3(x – 2)(x – 3) + 2x(4 - 3x)
3[2x – x2 + 4 – 2x] = 3[x2 – 3x – 2x + 6] + 8x – 6x2.
3[– x2 + 4] = 3[x2 - 5x + 6] + 8x – 6x2
-3x2 + 12 = 3x2 – 15x + 18 + 8x – 6x2
12 = -15x + 18 + 8x
15x – 8x = 18 – 12
ANS :
7x = 6
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
6
x
7
(con't)
3.1-EXAMPLE 11-Page 88
Solve the following for x:
3(x + 2)(2 – x) = 3(x – 2)(x – 3) + 2x(4 - 3x)
CHECK:
3(6/7 + 2)(2 – 6/7) = 3(6/7 – 2)(6/7 – 3) + 2(6/7)[4 – 3(6/7)]
3(20/7)(8/7) = 3(-8/7)(-15/7) + 12/7[10/7]
9 39/49 = 7 17/49 + 2 22/49
9 39/49 = 9 39/49
(checks)
ANS: x 
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
6
7
3.1-EXAMPLE 15-Page 89
Solve the equation for x:
3x + b = 5
3x = 5 - b
5b
x
3
CHECK:
3(x)+b=5
5b
3
  5b
 31 
1
5b  5b
(checks)
5b
ANS: x 
3
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
3.2-Solving Word Problems
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
3.2-STRATEGY-Pages 90 to 94
Verbal Statements
Always Check Your ANSWER
Picture the Problem:
Use a detailed sketch to help visualize the problem.
Understand the Words:
Look up meanings of unfamiliar words (eg. Difference, quotient, consecutive etc).
Additionthe Unknown:
Identify
Sum, total, and,
plus,
increased
by,found
more (eg.find,
than, added
to, riseshow much etc).
Determine
exactly
what
is to be
calculate,
Subtraction
Define Other Unknowns:
Difference, minus, less, remains, is subtracted from, decreases by, diminished by
Define
additional unknowns in terms of the original unknown if possible.
Multiplication
Estimate
the Answer:
Times, product,
of, multiplied by, triple, twice
Estimate or guess the answer to see if the result is reasonable.
Division
Divided,
dividend,
remainder,
quotient, split into, equal amounts
Write
and
Solve the
Equation:
Use an equation or formula to relate the unknown quantity to the given quantities.
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
3.2-EXAMPLE 26-Page 94
If twice a certain number is increased by 9, the result is
equal to 2 less than triple the number. Find the number.
2 x  9  3x  2
Let x = the number
9  2  3x  2 x
Solve for x
x  11
Check Your ANSWER
Twice the number 11 is equal to 22. Twice this number increased by nine is equal
to 31. Triple the number 11 is equal to 33. Two less than triple this number is also
equal to 31. Thus, our number 11 checks.
ANS: x  11
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
3.2-EXAMPLE 28-Page 95
In a group of 102 employees, there are three times as many employees on the
day shift as on the night shift, and two more on the swing shift than on the
night shift. How many are on each shift ?
Let x = number of employees on the night shift
Check Your ANSWER
Let 3x = number of employees on the day shift
If x is 20, then there are 20 employees
on the night shift. Three times as
many would equal 60 on the day shift.
Two more than the night shift would
Let (x + 2) = number of employees on the swing shift
equal 22 on the swing shift.
Equation
x  3x  ( x  2)  102
5 x  100
x  20
ANS: NS=20, DS=60 and SS=22
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
3.3-Financial Problems
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
3.3-EXAMPLE 30-Page 97
A person invests part of his $10 000 savings in a bank at 6%, and part in a
certificate of deposit at 8%, both simple interest. He gets a total of $750 per
year in interest from the two investments. How much is invested at each rate ?
AMOUNT
of
Investment
x
RATE
of
Interest
=
INTEREST
Account
A
x
0.06
0.06 (x)
Account
B
(10 000 – x)
0.08
0.08 (10 000 – x)
TOTALS
10 000
-
750
Equation
0.06( x)  0.08(10000  x)  750
0.06 x  800  0.08 x  750
0.02 x  50
ANS: $2 500 at 6%
$7 500 at 8%
x  $2 500
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
3.4-Mixture Problems
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
3.4-EXAMPLE 34-Page 99
How much steel containing 5.25% nickle must be combined with another steel
containing 2.84% nickle to make 3.25 tonnes of steel containing 4.15% nickle ?
AMOUNT
of
Steel
x
PERCENT
=
of
Nickle
AMOUNT of NICKLE
Mixture
A
x
0.0525
0.0525 (x)
Mixture
B
(3.25 – x)
0.0284
0.0284 (3.25 – x)
TOTALS
3.25
0.0415
0.0415 (3.25)
Equation
0.0525( x)  0.0284(3.25  x)  0.0415(3.25)
0.0525 x  0.0923  0.0284 x  0.1349
0.0241x  0.0426
x  1.77 tonnes
ANS: 1.77 t of 5.25%
1.48 t of 2.84%
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
3.5-Statics Problems
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
3.5-EXAMPLE 36-Page 102
A horizontal uniform beam of negligible weight is 6.35 m long and is supported
by columns at either end. A concentrated load of 525 N is applied to the beam.
At what distance from one end must this load be located so that the vertical
reaction at that same end is 315 N, and what will be the reaction at the other end ?
Taking moments
ESTIMATE:
If the
about
525p,Nwe
load
setwere
the moments
at the mid-span,
that tendthe
to two
turnreactions
the bar inwould
a
have equal values of ½(525) or
262.5 N. Since
clockwise
(CW)the
direction
left reaction
equal(315
to theN)moments
is greaterthat
than
tend
that,
to we
turndeduce
the barthat
in athe load is to the left of the mid-span
and that the reaction
counter-clockwise
(CCW)
at the direction.
right will be
(Byless
Eq.A15)
than 262.5 N.
R  315  523
R  210 N
525 x  210(6.35)
210(6.35)
x
525
ANS: R= 210 N
x=2.54 m
x  2.54 m
Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.
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Calter & Calter, Technical Mathematics with Calculus, Canadian Edition
©2008 John Wiley & Sons Canada, Ltd.