Transcript Ch2-Sec 2.4

Copyright © 2010 Pearson Education, Inc. All rights reserved.
2.4 – Slide 1
Chapter 2
Equations, Inequalities, and
Applications
Copyright © 2010 Pearson Education, Inc. All rights reserved.
2.4 – Slide 2
2.4
An Introduction to Applications
of Linear Equations
Copyright © 2010 Pearson Education, Inc. All rights reserved.
2.4 – Slide 3
2.4 An Introduction to Applications of
Linear Equations
Objectives
1.
2.
3.
4.
5.
Learn the six steps for solving applied problems.
Solve problems involving unknown numbers.
Solve problems involving sums of quantities.
Solve problems involving supplementary and
complementary angles.
Solve problems involving consecutive integers.
Copyright © 2010 Pearson Education, Inc. All rights reserved.
2.4 – Slide 4
2.4 An Introduction to Applications of
Linear Equations
Solving Applied Problems
Solving an Applied Problem
Step 1 Read the problem, several times if necessary, until you
understand what is given and what is to be found.
Step 2 Assign a variable to represent the unknown value, using
diagrams or tables as needed. Write down what the variable
represents. Express any other unknown values in terms of
the variable.
Step 3 Write an equation using the variable expression(s) .
Step 4 Solve the equation.
Step 5 State your answer. Does it seem reasonable?
Step 6 Check the answer in the words of the original problem.
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2.4 – Slide 5
2.4 An Introduction to Applications of
Linear Equations
Solving Problems Involving Unknown Numbers
Example 1
The product of 3, and a number decreased by 2, is 42. What is the
number?
Step 1 Read the problem carefully. We are asked to find a number.
Step 2 Assign a variable to represent the unknown quantity. In this
problem, we are asked to find a number, so we write
Let x = the number.
There are no other unknown quantities to find.
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2.4 – Slide 6
2.4 An Introduction to Applications of
Linear Equations
Solving Problems Involving Unknown Numbers
Example 1 (cont.) The product of 3, and a number decreased
by 2, is 42. What is the number?
Notice the placement of the commas!
Step 3 Write an equation.
The product of 3, and
3•
a number decreased by 2,
(
x
–
2
)
is
42.
=
42
The equation 3x – 2 = 42 corresponds to the statement “The product
of 3 and a number, decreased by 2, is 42.” Be careful when reading
these types of problems. The placement of the commas is important.
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2.4 – Slide 7
2.4 An Introduction to Applications of
Linear Equations
Solving Problems Involving Unknown Numbers
Example 1 (cont.) The product of 3, and a number decreased
by 2, is 42. What is the number?
Step 4
Solve the equation.
3 ( x – 2 ) = 42
3x – 6 = 42
3x – 6 + 6 = 42 + 6
Distribute.
Add 6.
3x = 48
Combine terms.
3x
48
=
3
3
Divide by 3.
x = 16
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2.4 – Slide 8
2.4 An Introduction to Applications of
Linear Equations
Solving Problems Involving Unknown Numbers
Example 1 (cont.) The product of 3, and a number decreased
by 2, is 42. What is the number?
Step 5
State the answer. The number is 16.
Step 6
Check. When 16 is decreased by 2, we get 16 – 2 = 14.
If 3 is multiplied by 14, we get 42, as required. The
answer, 16, is correct.
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2.4 – Slide 9
2.4 An Introduction to Applications of
Linear Equations
Solving Problems Involving Sums of Quantities
Example 2
A box contains a combined total of 68 pens and pencils. If the box
contains 16 more pens than pencils, how many of each are in the box?
Step 1 Read the problem. We are given information about the total
number of pens and pencils and asked to find the number of
each in the box.
Step 2 Assign a variable.
Let
x = the number pencils in the box.
Then x + 16 = the number pens in the box.
Copyright © 2010 Pearson Education, Inc. All rights reserved.
2.4 – Slide 10
2.4 An Introduction to Applications of
Linear Equations
Solving Problems Involving Sums of Quantities
Example 2 (cont.)
A box contains a combined total of 68 pens and pencils. If the box
contains 16 more pens than pencils, how many of each are in the box?
Recall: x = # of pencils, x + 16 = # of pens
Step 3 Write an equation.
The total
is
the number of pens
plus
the number of pencils
68
=
( x + 16 )
+
x
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2.4 – Slide 11
2.4 An Introduction to Applications of
Linear Equations
Solving Problems Involving Sums of Quantities
Example 2 (cont.)
A box contains a combined total of 68 pens and pencils. If the box
contains 16 more pens than pencils, how many of each are in the box?
Step 4
Solve the equation.
68 = ( x + 16 ) + x
68 = 2x + 16
68 – 16 = 2x + 16 – 16
Combine terms.
Subtract 16.
52 = 2x
Combine terms.
52
2x
=
2
2
26 = x
Divide by 2.
or
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x = 26
2.4 – Slide 12
2.4 An Introduction to Applications of
Linear Equations
Solving Problems Involving Sums of Quantities
Example 2 (cont.)
A box contains a combined total of 68 pens and pencils. If the box
contains 16 more pens than pencils, how many of each are in the box?
Recall: x = # of pencils, x + 16 = # of pens
Step 5
State the answer. The variable x represents the
number of pencils, so there are 26 pencils. Then the
number of pens is x + 16 = 26 + 16 = 42.
Step 6
Check. Since there are 26 pencils and 42 pens, the
combined total number of pencils and pens is
26 + 42 = 68. Because 42 – 26 = 16, there are 16
more pens than pencils. This information agrees
with what is given in the problem, so the answer
checks.
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2.4 – Slide 13
2.4 An Introduction to Applications of
Linear Equations
Solving Problems Involving Sums of Quantities
Example 3 A mixture of acid and water must be combined to fill a
286 ml beaker. If the mixture must contain 12 ml of water for each
milliliter of acid, how many milliliters of water and how many
milliliters of acid does it require to fill the beaker?
Step 1 Read the problem carefully. We must find how many
milliliters of water and how many milliliters of acid are
needed to fill the beaker.
Step 2 Assign a variable.
Let
x = the number of milliliters of acid required.
Then 12x = the number of milliliters of water required.
Copyright © 2010 Pearson Education, Inc. All rights reserved.
2.4 – Slide 14
2.4 An Introduction to Applications of
Linear Equations
Solving Problems Involving Sums of Quantities
Example 3 (cont.) A mixture of acid and water must be combined
to fill a 286 ml beaker. If the mixture must contain 12 ml of water for
each milliliter of acid, how many milliliters of water and how many
milliliters of acid does it require to fill the beaker?
Recall: x = ml. of acid, 12x = ml. of water.
Step 3 Write an equation. A diagram is sometimes helpful.
Beaker
Acid
Water
= 286
x
12x
x
+
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12x
= 286
2.4 – Slide 15
2.4 An Introduction to Applications of
Linear Equations
Solving Problems Involving Sums of Quantities
Example 3 (cont.) A mixture of acid and water must be combined
to fill a 286 ml beaker. If the mixture must contain 12 ml of water for
each milliliter of acid, how many milliliters of water and how many
milliliters of acid does it require to fill the beaker?
Step 4
Solve.
x + 12x = 286
13x = 286
13x = 286
13
13
Combine terms.
Divide by 13.
x = 22
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2.4 – Slide 16
2.4 An Introduction to Applications of
Linear Equations
Solving Problems Involving Sums of Quantities
Example 3 (cont.) A mixture of acid and water must be combined
to fill a 286 ml beaker. If the mixture must contain 12 ml of water for
each milliliter of acid, how many milliliters of water and how many
milliliters of acid does it require to fill the beaker?
Recall: x = ml of acid, 12x = ml of water
Step 5
State the answer. The beaker requires 22 ml of acid
and 12(22) = 264 ml of water.
Step 6
Check. Since 22 + 264 = 286, and 264 is 12 times
22, the answer checks.
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2.4 – Slide 17
2.4 An Introduction to Applications of
Linear Equations
Solving Problems Involving Sums of Quantities
Example 4 Gerald has a wire 96 inches long that he must cut into
three pieces. The longest piece must be three times the length of the
middle-sized piece and the shortest piece must be 14 inches shorter
than the middle-sized piece. How long must each piece be?
Step 1 Read the problem carefully. Three lengths must be found.
Step 2 Assign a variable.
x = the length of the middle-sized piece,
3x = the length of the longest piece, and
x – 14 = the length of the shortest piece.
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2.4 – Slide 18
2.4 An Introduction to Applications of
Linear Equations
Solving Problems Involving Sums of Quantities
Example 4 (cont.) Gerald has a wire 96 inches long that he must
cut into three pieces. The longest piece must be three times the length
of the middle-sized piece and the shortest piece must be 14 inches
shorter than the middle-sized piece. How long must each piece be?
Step 3 Write an equation.
96 inches
3x
Longest
Middle-sized
Shortest
3x
+
x
+ x – 14
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x – 14
x
is
=
Total length
96
2.4 – Slide 19
2.4 An Introduction to Applications of
Linear Equations
Solving Problems Involving Sums of Quantities
Example 4 (cont.) Gerald has a wire 96 inches long that he must
cut into three pieces. The longest piece must be three times the length
of the middle-sized piece and the shortest piece must be 14 inches
shorter than the middle-sized piece. How long must each piece be?
Step 4 Solve.
96 = 3x + x + x – 14
96
96 + 14
110
110
5
22
= 5x – 14
Combine terms.
Add 14.
= 5x – 14 + 14
= 5x
Combine terms.
5x
Divide by 5.
=
5
= x
or
x = 22
Copyright © 2010 Pearson Education, Inc. All rights reserved.
2.4 – Slide 20
2.4 An Introduction to Applications of
Linear Equations
Solving Problems Involving Sums of Quantities
Example 4 (cont.) Gerald has a wire 96 inches long that he must
cut into three pieces. The longest piece must be three times the length
of the middle-sized piece and the shortest piece must be 14 inches
shorter than the middle-sized piece. How long must each piece be?
Recall: x = length of middle-sized piece, 3x = length of longest
piece, and x – 14 = length of shortest piece.
Step 5
State the answer. The middle-sized piece is 22 in.
long, the longest piece is 3(22) = 66 in. long, and the
shortest piece is 22 – 14 = 8 in. long.
Step 6
Check. The sum of the lengths is 96 in. All
conditions of the problem are satisfied.
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2.4 – Slide 21
2.4 An Introduction to Applications of
Linear Equations
Solving with Supplementary and Complementary Angles
1
2
Angles 1 and 2
are complementary.
They form a right
angle, indicated by .
3
4
Angles 3 and 4
are supplementary.
They form a straight
angle.
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180 (degrees)
Straight angle
2.4 – Slide 22
2.4 An Introduction to Applications of
Linear Equations
Solving with Supplementary and Complementary Angles
Measure of x:
Complement of x:
Supplement of x:
90 – x
180 – x
x
x
x
Problem-Solving Hint
If x represents the degree measure of an angle, then
90 – x represents the degree measure of its complement, and
180 – x represents the degree measure of its supplement.
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2.4 – Slide 23
2.4 An Introduction to Applications of
Linear Equations
Solving with Supplementary and Complementary Angles
Example 5 Find the measure of an angle whose supplement is 20°
more than three times its complement.
Step 1 Read the problem. We are to find the measure of an angle,
given information about its complement and supplement.
Step 2 Assign a variable.
Let
Then
x = the degree measure of the angle.
90 – x = the degree measure of its complement;
180 – x = the degree measure of its supplement.
Step 3 Write an equation.
Supplement is 20 more than
three times its complement.
180 – x
=
20
+
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3•
( 90 – x )
2.4 – Slide 24
2.4 An Introduction to Applications of
Linear Equations
Solving with Supplementary and Complementary Angles
Example 5 (cont.) Find the measure of an angle whose
supplement is 20° more than three times its complement.
Step 4 Solve. 180 – x = 20 + 3 ( 90 – x )
180 – x = 20 + 270 – 3x
Distribute.
Combine terms.
180 – x = 290 – 3x
180 – x + 3x = 290 – 3x + 3x
Add 3x.
Combine terms.
180 + 2x = 290
Subtract 180.
180 + 2x – 180 = 290 – 180
2x = 110
Combine terms.
2x
110
Divide by 2.
=
2
2
x = 55
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2.4 – Slide 25
2.4 An Introduction to Applications of
Linear Equations
Solving with Supplementary and Complementary Angles
Example 5 (cont.) Find the measure of an angle whose
supplement is 20° more than three times its complement.
Step 5 State the answer. The measure of the angle is 55°.
Step 6 Check. The complement of 55° is 35° and the
supplement of 55° is 125°. Also, 125° is equal to 20°
more than three times 35° (that is, 125 = 20 + 3(35) is
true). Therefore, the answer is correct.
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2.4 – Slide 26
2.4 An Introduction to Applications of
Linear Equations
Solving Problems with Consecutive Integers
Problem-Solving Hint
When solving consecutive integer problems, if x = the lesser integer,
then for any
two consecutive integers, use
x,
x + 1;
two consecutive even integers, use
x,
x + 2;
two consecutive odd integers, use
x,
x + 2.
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2.4 – Slide 27
2.4 An Introduction to Applications of
Linear Equations
Solving Problems with Consecutive Integers
Example 6 The sum of two consecutive checkbook check
numbers is 893. Find the numbers.
Step 1 Read the problem. We are to find the check numbers.
Step 2 Assign a variable.
Let
x = the lesser check number.
Then x + 1 = the greater check number.
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2.4 – Slide 28
2.4 An Introduction to Applications of
Linear Equations
Solving Problems with Consecutive Integers
Example 6 (cont.) The sum of two consecutive checkbook check
numbers is 893. Find the numbers.
Recall:
Let
Then
x = the lesser check number.
x + 1 = the greater check number.
Step 3 Write an equation. The sum of the check numbers is 893, so
Step 4 Solve.
x + (x + 1)
2x + 1
2x
x
Copyright © 2010 Pearson Education, Inc. All rights reserved.
=
=
=
=
893
893
892
446
Combine terms.
Subtract 1.
Divide by 2.
2.4 – Slide 29
2.4 An Introduction to Applications of
Linear Equations
Solving Problems with Consecutive Integers
Example 6 (cont.) The sum of two consecutive checkbook check
numbers is 893. Find the numbers.
Step 5 State the answer. The lesser check number is 446 and the
greater check number is 447.
Step 6 Check. The sum of 446 and 447 is 893. The answer is correct.
Copyright © 2010 Pearson Education, Inc. All rights reserved.
2.4 – Slide 30
2.4 An Introduction to Applications of
Linear Equations
Solving Problems with Consecutive Integers
Example 7 If four times the smaller of 2 consecutive odd integers
is added to three times the larger, the result is 125. Find the integers.
Step 1 Read the problem. We are to find two consecutive odd
integers.
Step 2 Assign a variable.
Let
x = the lesser integer.
Then x + 2 = the greater integer.
Step 3 Write an equation.
4 times the smaller is added to 3 times the larger, the result is 125
4•
x
+
Copyright © 2010 Pearson Education, Inc. All rights reserved.
3•
(x+2)
=
125
2.4 – Slide 31
2.4 An Introduction to Applications of
Linear Equations
Solving Problems with Consecutive Integers
Example 7 If four times the smaller of 2 consecutive odd integers
is added to three times the larger, the result is 125. Find the integers.
Step 4 Solve. 4x + 3 ( x + 2 ) = 125
4x + 3x + 6 = 125
Distribute.
Combine terms.
7x + 6 = 125
7x + 6 – 6 = 125 – 6
Subtract 6.
7x = 119
Combine terms.
x = 17
Divide by 7.
Step 5 State the answer. The lesser integer is 17 and the greater
integer is 17 + 2 = 19.
Step 6 Check. The sum of 4(17) and 3(19) is 125. The answer is
correct.
Copyright © 2010 Pearson Education, Inc. All rights reserved.
2.4 – Slide 32