10.4 applications of linear equations

Slide - 1Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
2
Equations,
Inequalities, and
Applications
10

1. Learn the six steps for solving applied
problems.
2. Solve problems involving unknown
numbers.
3. Solve problems involving sums of
quantities.
4. Solve problems involving consecutive
integers.
5. Solve problems involving complementary
and supplementary angles.
Objectives
10.4 An Introduction to Applications of
Linear Equations

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.
Solving Applied Problems

Example
The product of 3, and a number decreased by 2, is 42. What is the
number?
Solve Problems Involving Unknown Numbers
Read the problem carefully. We are asked to find a number.Step 1
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.
Step 2

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.
Example (cont) The product of 3, and a number decreased by
2, is 42. What is the number?
Write an equation.Step 3
The product of 3, a number decreased by 2,and is 42.
3 • x – = 422 )(
Notice the placement of the commas!

Example (cont)
The product of 3, and a number decreased by 2, is 42. What is
the number?
3 ( x – 2 ) = 42
3x – 6 = 42
3x – 6 + 6 = 42 + 6
3x = 48
3x 48
3 3
=
Distribute.
Add 6.
Combine terms.
Divide by 3.
x = 16

Example (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.

Example
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?
Solve Problems Involving Sums of Quantities
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 1
Assign a variable.
Let x = the number of pencils in the box.
Then x + 16 = the number of pens in the box.
Step 2

Example (cont)
Solving Problems Involving Sums of Quantities
The total is the number of pens the number of pencilsplus
68 = (x + 16) + x
Recall: x = # of pencils, x + 16 = # of pens

Example (cont)
68 = 2x + 16
68 – 16 = 2x + 16 – 16
52 = 2x
52 2x
2 2
=
26 = x
68 = (x + 16) + x
Combine terms.
Subtract 16.
Combine terms.
Divide by 2.
or x = 26

Example (cont)
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.

Example 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?
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 1
Assign a variable.
Let x = the number of milliliters of acid required.
Then 12x = the number of milliliters of water required.
Step 2

Example (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
Write an equation. A diagram is sometimes helpful.Step 3
Acid
x
Water
12x
Beaker
= 286
x 12x =+ 286
Recall: x = ml. of acid, 12x = ml. of water.

Step 4 Solve.
13x = 286
13x 286
13 13
=
x = 22
x + 12x = 286
Combine terms.
Divide by 13.

Step 5 State the answer. The beaker requires 22 ml of acid
and 12(22) = 264 ml of water.
Recall: x = ml of acid, 12x = ml of water
Step 6 Check. Since 22 + 264 = 286, and 264 is 12 times
22, the answer checks.

Example 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?
Read the problem carefully. Three lengths must be found.Step 1
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.
Step 2

Example (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 x x – 14
Longest Middle-sized Shortest is Total length
3x x x – 14 = 96+ +

Step 4 Solve.
96 = 5x – 14
96 + 14 = 5x – 14 + 14
110 = 5x
110 5x
5 5
=
22 = x
96 = 3x + x + x – 14
Combine terms.
Add 14.
Combine terms.
Divide by 5.
or x = 22

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.
Recall: x = length of middle-sized piece, 3x = length of longest
piece, and x – 14 = length of shortest piece.
Step 6 Check. The sum of the lengths is 96 in. All
conditions of the problem are satisfied.

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.
Solving Problems with Consecutive Integers

Example The sum of two consecutive checkbook check
numbers is 893. Find the numbers.
Solve Problems with Consecutive Integers
Read the problem. We are to find the check numbers.Step 1
Assign a variable.
Let x = the lesser check number.
Then x + 1 = the greater check number.
Step 2

Example (cont) The sum of two consecutive checkbook check
Recall: Let x = the lesser check number.
Then x + 1 = the greater check number.
Write an equation. The sum of the check numbers is 893, soStep 3
x + ( x + 1 ) = 893
Solve.Step 4 2x + 1 = 893
2x = 892
x = 446
Subtract 1.
Combine terms.
Divide by 2.

Example (cont) The sum of two consecutive checkbook check
State the answer. The lesser check number is 446 and the
greater check number is 447.
Step 5
Check. The sum of 446 and 447 is 893. The answer is correct.Step 6

4 times the smaller is added to 3 times the larger, the result is 125
Example If four times the smaller of 2 consecutive odd integers
is added to three times the larger, the result is 125. Find the integers.
Read the problem. We are to find two consecutive odd
integers.
Step 1
Assign a variable.
Let x = the lesser consecutive odd integer.
Then x + 2 = the greater consecutive odd integer.
Step 2
4 • x + 3 • ( x + 2 ) = 125

Step 4
Example If four times the smaller of 2 consecutive odd integers is
added to three times the larger, the result is 125. Find the integers.
Solve.
State the answer. The lesser integer is 17 and the greater
integer is 17 + 2 = 19.
Step 5
Check. The sum of 4(17) and 3(19) is 125. The answer is
correct.
Step 6
4x + 3 ( x + 2 ) = 125
4x + 3x + 6 = 125
7x + 6 = 125
7x + 6 – 6 = 125 – 6
7x = 119
x = 17
Combine terms.
Distribute.
Subtract 6.
Combine terms.
Divide by 7.

3 4
1
2
3 4Angles and
are supplementary.
They form a straight
angle.
Straight angle1 2Angles and
are complementary.
They form a right
angle.
Solve Problems with Supplementary and
Complementary Angles
180 (degrees)

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.
x
180 – x
x
90 – x
Complement of x:
x
Supplement of x:Measure of x:
x x

Example Find the measure of an angle whose supplement is 20°
more than three times its complement.
Read the problem. We are to find the measure of an angle,
given information about its complement and supplement.
Step 1
Assign a variable.
Let x = the degree measure of the angle.
Then 90 – x = the degree measure of its complement;
180 – x = the degree measure of its supplement.
Step 2
Supplement is 20 more than three times its complement.
180 – x = 20 + 3 • (90 – x)

Example (cont.) Find the measure of an angle whose supplement
is 20° more than three times its complement.
Solve.Step 4 180 – x = 20 + 3 (90 – x)
180 – x = 20 + 270 – 3x
180 – x = 290 – 3x
180 – x + 3x = 290 – 3x + 3x
2x 110
2 2
=
Distribute.
Combine terms.
Add 3x.
Divide by 2.
x = 55
180 + 2x = 290
180 + 2x – 180 = 290 – 180
2x = 110
Combine terms.
Subtract 180.
Combine terms.

Example (cont) Find the measure of an angle whose supplement is
20° more than three times its complement.
State the answer. The measure of the angle is 55°.Step 5
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.
Step 6

10.4 applications of linear equations

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