USING RECTANGLES TO MULTIPLY
5.1.1 through 5.1.3
Two ways to find the area of a rectangle are: as a product of the (height) ! (base) or as the
sum of the areas of individual pieces of the rectangle. For a given rectangle these two
areas must be the same, so Area as a product = Area as a sum. Algebra tiles and, later,
generic rectangles, provide area models to help multiply expressions in a visual, concrete
manner. See the Math Notes box on page 218.
Example 1 Using Tiles
The algebra tile pieces x 2 + 6x + 8 are arranged into
a rectangle inside the cornerpiece as shown at right.
The area of the rectangle can be written as the
product of its base and height or as the sum of its
parts.
(x
+ 4) (x
+3
2)!!!!!!!=!!!!!!!x12 4
+2
6x4+38
123
12
base
x
x
height
x
area as a product
cornerpiece
area as a sum
xxxx
x
area
height
x2
base
Example 2 Using Generic Rectangles
A generic rectangle allows us to organize the problem in the same way as the first example
without needing to draw the individual tiles. It does not have to be drawn accurately or to scale.
Multiply ( x ! 3) (2x + 1) .
-3 !6x
-3
!
x
2x
30
+1
x
!3
2x 2
x
2x
+1
! ( x ! 3) ( 2x + 1) = 2x 2 ! 5x ! 3
area as a product
area as a sum
Algebra Connections Parent Guide
Problems
Write a statement showing: area as a product equals area as a sum.
1.
2.
3.
4.
5.
6.
x
x 2 3x
–5 !5x !15
x +3
6
3y 3xy 12y
18y !12x
3y
–2 !2x !8
!2x
x
+4
Multiply.
7.
( 3x + 2 ) ( 2x + 7 )
8.
10.
( 2y ! 1) ( 4y + 7 )
11.
13.
( 3x ! 1) ( x + 2 )
16.
19.
( 2x ! 1) ( 3x + 1)
9.
( 2x ) ( x ! 1)
( y ! 4 )( y + 4 )
12.
( y ) ( x ! 1)
14.
( 2y ! 5 ) ( y + 4 )
15.
( 3y ) ( x ! y )
( 3x ! 5 ) ( 3x + 5 )
17.
( 4x + 1)2
18.
( x + y)( x + 2 )
( 2y ! 3)2
20.
( x ! 1) ( x + y + 1)
21.
( x + 2 )( x + y ! 2 )
Answers
1.
( x + 1) ( x + 3) = x 2 + 4x + 3
2.
( x + 2 ) ( 2x + 1) = 2x 2 + 5x + 2
3.
( x + 2 ) ( 2x + 3) = 2x 2 + 7x + 6
4.
( x ! 5 ) ( x + 3) = x 2 ! 2x ! 15
5.
6 ( 3y ! 2x ) = 18y ! 12x
6.
( x + 4 ) ( 3y ! 2 ) = 3xy ! 2x + 12y ! 8
7.
6x 2 + 25x + 14
8.
10.
8y 2 + 10y ! 7
11.
y 2 ! 16
12.
xy ! y
13.
3x 2 + 5x ! 2
14.
2y 2 + 3y ! 20
15.
3xy ! 3y 2
16.
9x 2 ! 25
17.
16x 2 + 8x + 1
18.
x 2 + 2x + xy + 2y
19.
4y 2 ! 12y + 9
20.
x 2 + xy ! y ! 1
21.
x 2 + xy + 2y ! 4
6x 2 ! x ! 1
Chapter 5: Multiplication and Proportions
9.
2x 2 ! 2x
31
SOLVING EQUATIONS WITH MULTIPLICATION
5.1.4
To solve an equation with multiplication, first use the Distributive Property or a generic
rectangle to rewrite the equation without parentheses, then solve in the usual way. See the
Math Notes box on page 198.
Example 1
Solve 6 ( x + 2 ) = 3( 5x + 1)
Use the Distributive Property
Subtract 6x
6x + 12 = 15x + 3
12 = 9x + 3
Subtract 3
9 = 9x
Divide by 9
1= x
Example 2
Solve x ( 2x ! 4 ) = ( 2x + 1) ( x + 5 )
Rewrite using generic rectangles
2x 2 ! 4x = 2x 2 + 11x + 5
!4x = 11x + 5
Subtract 2x 2
Subtract 11x
Divide by !15
32
!15x = 5
x=
5
!15
= ! 13
Algebra Connections Parent Guide
Problems
Solve each equation.
1.
3( c + 4 ) = 5c + 14
2.
x ! 4 = 5 ( x + 2)
3.
7(x + 7) = 49 ! x
4.
8(x ! 2) = 2 ( 2 ! x )
5.
5x ! 4 ( x ! 3) = 8
6.
4y ! 2 ( 6 ! y ) = 6
7.
2x + 2(2x ! 4) = 244
8.
x ( 2x ! 4 ) = ( 2x + 1) ( x ! 2 )
9.
( x ! 1) ( x + 7 ) = ( x + 1) ( x ! 3)
10.
( x + 3) ( x + 4 ) = ( x + 1) ( x + 2 )
11.
2x ! 5 ( x + 4 ) = !2 ( x + 3)
12.
( x + 2 ) ( x + 3) = x 2 + 5x + 6
13.
( x ! 3) ( x + 5 ) = x 2 ! 7x ! 15
14.
( x + 2 ) ( x ! 2 ) = ( x + 3) ( x ! 3)
15.
1
2
x ( x + 2 ) = ( 12 x + 2 ) ( x ! 3)
Answers
1.
!1
2.
!3.5
3.
0
4.
2
5.
!4
6.
3
7.
42
8.
2
9.
0.5
10.
!2.5
11.
!14
12.
all numbers
13.
0
14.
no solution
15.
!12
Chapter 5: Multiplication and Proportions
33
SOLVING MULTI-VARIABLE EQUATIONS
5.1.5
Solving equations with more than one variable uses the same process as solving an
equation with one variable. The only difference is that instead of the answer always
being a number, it may be an expression that includes numbers and variables. The usual
steps may include: removing parentheses, simplifying by combining like terms, removing
the same thing from both sides of the equation, moving the desired variables to one side
of the equation and the rest of the variables to the other side, and possibly division or
multiplication.
Example 1
Example 2
3x ! 2y = 6
!2y = !3x + 6
Solve for y
Subtract 3x
Divide by !2
y=
Simplify
y=
Solve for y
7 + 2(x + y) = 11
2(x + y) = 4
2x + 2y = 4
2y = !2x + 4
Subtract 7
!3x+6
!2
3x!3
2
Distribute the 2
Subtract 2x
y=
Divide by 2
y = !x + 2
Simplify
Example 3
Example 4
y = 3x ! 4
y + 4 = 3x
y+4
=x
3
Solve for x
Add 4
Divide by 3
!2 x+ 4
2
I = prt
Solve for t
I
=t
pr
Divide by pr
Problems
Solve each equation for the specified variable.
1.
y in 5x + 3y = 15
2.
x in 5x + 3y = 15
3.
w in 2l + 2w = P
4.
m in 4n = 3m ! 1
5.
a in 2a + b = c
6.
a in b ! 2a = c
7.
p in 6 ! 2(q ! 3p) = 4 p
8.
x in y =
9.
r in 4(r ! 3s) = r ! 5s
1
4
x +1
Answers (Other equivalent forms are possible.)
1.
y = ! 53 x + 5
2.
x = ! 53 y + 3
3.
w = !l +
4.
m=
4n+1
3
5.
a=
c!b
2
6.
a=
7.
p=q!3
8.
x = 4y ! 4
9.
r=
34
P
2
c!b !or! b!c
!2
2
7s
3
Algebra Connections Parent Guide
SETTING UP PROPORTIONS
5.2.1 and 5.2.2
A convenient way to set up proportions is to arrange the information in a table. Once
the information is placed into a labeled table, the proportion can be written directly from
the table. See the Math Notes box on page 211.
Example 1
A tree casts a 43-foot shadow. At the same
time, a 4 12 foot tall boy casts a 10-foot
shadow. Determine the height of the tree.
height (ft)
shadow (ft)
x
43
4.5
10
x
43
=
4.5 10
Eliminate the denominators.
! x $
! 43 $
45 #
= 45 # &
&
" 4.5 %
" 10 %
10x = 193.5
x = 19.35
Divide by 10.
Example 2
Kim noticed that 100 vitamins cost $1.89.
At this rate, how much should 350 vitamins
cost?
Vitamins (#)
Cost ($)
100
1.89
350
x
100 1.89
=
350
x
Eliminate the denominators.
Divide by 100.
Chapter 5: Multiplication and Proportions
! 100 $
! 1.89 $
350x #
= 350x #
&
" 350 %
" x &%
100x = 661.5
x = $6.62
35
Problems
Solve each problem by writing a proportion and solving it.
1.
Joe came to bat 464 times in 131 games. At this rate, how many times should he expect to
bat in a full season of 162 games?
2.
Mario’s car needs 12 gallons to go 320 miles. At the same rate, how far can he travel with
10 gallons of gas?
3.
If 50 empty soda cans weigh 3 12 pounds, how much would 70 empty soda cans weigh?
4.
There is a $34 tax on an $800 motor scooter. How much tax would there be on a $1000
motor scooter?
5.
A dozen donuts cost $3.50. How much should three donuts cost?
6.
In 35 minutes, Suki’s car goes 25 miles. If she continues at the same rate, how long will it
take her to drive 90 miles?
7.
In a city of three million people, 3,472 were surveyed. Of those surveyed, 28 of them
watched the last Lonely Alien movie. If the survey represents the city’s TV viewing habits,
about how many people in the city watched the movie?
8.
Julio runs
9.
It is now 7:51 p.m. The movie that you have waited three weeks to see starts at 8:15 p.m.
While standing in line for the movie, you count 146 people ahead of you. Nine people buy
their tickets in 70 seconds. Will you be able to buy your ticket before the movie starts?
3
10
mile in 1 12 minutes. At that rate how long will it take him to run a mile?
10.
A biscuit recipe uses 12 teaspoon of baking powder for
powder is needed for three cups of flour?
3
4
11.
The germination rate for zinnia seeds is 78%. This means that 78 out of every 100 seeds
will sprout and grow. If Jim wants 60 plants for his yard, how many seeds should he plant?
12.
L.J.’s car has a gas tank which holds 19 gallons of fuel. If the gas tank was full and he
used eight gallons to drive 200 miles, does the car have enough gas to go another 250
miles?
cup of flour. How much baking
Answers
1.
4.
7.
10.
36
about 574 times
$42.50
about 24,194 people
2 teaspoons
2.
5.
8.
11.
about 267 miles
$0.88
5 minutes
about 77 seeds
3.
6.
9.
12.
4.9 pounds
126 minutes
yes, in about 19 min.
yes, 275 miles
Algebra Connections Parent Guide
SOLVING PROPORTIONS
5.2.2
To solve proportions (equal ratio equations) remove the denominators by multiplying
both fractions (ratios) by the common denominator. If you cannot determine the common
denominator, multiply the denominators and then multiply both fractions (ratios) by the
result. Then solve in the usual manner.
Another approach is to begin by multiplying to remove the denominator of the fraction
(ratio) with the variable. In the first example below, multiply by 12; in the second
example, multiply by 3. When there is a variable in both fractions, multiply by the
product of both denominators.
Example 1
Solve for x
Multiply by 60
Simplify
Divide by 5
Example 2
x
7
=
12 10
! x$
! 7$
60 # & = 60 # &
" 12 %
" 10 %
5x = 42
42
5
x = 8.4
x=
Solve for p
Multiply by 15
11 p + 2
=
5
3
! 11 $
! p + 2$
15 # & = 15 #
" 5%
" 3 &%
Simplify
33 = 5 ( p + 2 )
Distribute the 5
33 = 5 p + 10
Subtract 10
23 = 5 p
Divide by 5
23
5
p = 4.6
p=
Example 3
Solve for x
Multiply by 12
Simplify
Subtract 4x and 9
x + 3 x !1
=
4
3
" x + 3%
" x ! 1%
12 $
= 12 $
'
# 4 &
# 3 '&
3(x + 3) = 4(x ! 1)
3x + 9 = 4x ! 4
!x = !13
x = 13
Chapter 5: Multiplication and Proportions
37
Problems
Solve each proportion.
1.
x
3
=
20 5
2.
x!4 2
=
12
3
3.
100 5
=
40 2x
4.
6x x ! 3
=
5
10
5.
8x 13
=
7
2
6.
3! x x ! 7
=
4
10
7.
6x 35
=
11
4
8.
6+x 4!x
=
2
8
9.
9 ! x 24
=
6
2
10.
3x 24
=
10
9
11.
4x x ! 2
=
5
7
12.
6x 42
=
7
3
13.
4x 81
=
7 15
14.
x
5
=
100 20
15.
x
7
=
100 35
Answers
1.
12
2.
12
3.
1
4.
! "0.27
5.
!5.69
6.
! 4.14
7.
! 16.04
8.
!4
9.
!63
10.
! 8.89
11.
! "0.43
12.
16.3
13.
9.45
14.
25
15.
20
38
Algebra Connections Parent Guide