Concept: Solving Multi-Step Equations

Transcription

Equations – Section 4 Part B: Solving Multi-Step Equations
Concept: Solving Multi-Step Equations
Name:
 You should have completed Equations – Section 4 Part A: Solving Multi-Step
Equations before beginning this handout.
Warm Up
Solve each multi-step equation below. Show all your steps and make sure you check to
see if your solution is correct.
1.
-t)
+11)
4(t - 2) - (t + 3) = t - 1
4t - 8 - t
3t
3t - t
2t
2t - 11
÷2)
Check:
L.S. =
=
=
=
=
2.
-2.9x)
-5.4)
÷2.1)
Check:
L.S. =
=
=
=
- 3
- 11
- 11
- 11
+ 11
2t
2t
2
t
4(t - 2) - (t + 3)
4(5 - 2) - (5 + 3)
4(3) - (8)
12 - 8
4
5 x + 5.4
5 x - 2.9x + 5.4
2.1x + 5.4
2.1x + 5.4 - 5.4
2.1x
2.1x
2.1
x
5x + 5.4
5(-3) + 5.4
-15 + 5.4
-9.6
=
=
=
=
=
=
=
t - 1
t - 1
t - t - 1
-1
-1 + 11
10
10
2
= 5
R.S.
= t - 1
= 5 - 1
= 4
L.S. equals R.S., the solution is t = 5.
= 2.9 x - 0.9
=
=
=
=
=
=
2.9 x - 2.9x - 0.9
- 0.9
- 0.9 - 5.4
- 6.3
- 6.3
2.1
-3
R.S.
=
=
=
=
2.9 x - 0.9
2.9 (-3) - 0.9
-8.7 - 0.9
-9.6
L.S. equals R.S., the solution is x = -3.
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1
COMPUTER COMPONENT
Instructions:
Login to UMath X
Hover over the strand: Equations
Select the section: Solving Multi-Step Equations
NOTE: You will need to use the Menu feature of the program (found on the left side of
your screen) in order to get to the lesson where you left off.


Work through all Sub Lessons of the following Lessons in order:
 Summary
 Literal Equations
As you work through the computer exercises, you will be prompted to
make notes in your notebook/math journal.
NOTES:
Fill in the following:
1. Two ways to solve an equation.
(a) Solve and equation with algebra tiles.
(b) Solve an equation algebraically.
2. To keep a balance balanced, you must perform the same operation
to both sides.
3. You know when you have a solution when:
(a) There is one variable tile on one side of the equation (tile solution).
(b) There is one variable on one side of the equation (algebraic solution).
4.
Combine like terms if they are on the same side of the equation.
5.
Equations with fractions, require you to first multiply each side (three words)
by a common denominator (two words).
This keeps the equation balanced.
2
6. Use the original equation to check your answer by substituting
your solution for the variable . Check each side
of the equation. Your solution is correct if you have the same value on each side.
Literal Equations
7. Are the following perimeter equations the same? Why or why not.
2
1
+
and
-2W)
P
P – 2W
P – 2W
= 2L + 2W
= 2L + 2W - 2W
= 2L
÷2)
P – 2W
2
=
P – 2W
2
= L
1
2L
21
SAME
The perimeter equations are the same but in different forms. The first (1) is useful for
finding P when the length and width are known. The second (2) is useful for finding L
when the Perimeter and width are known.
Solving Linear Equations: (You use similar steps to solve literal equations as you do for
equations with one variable)
Solve 4 x + 2 y = 16 for y
 The equation has two different variables, x and y
 Determine the variable you have to solve for: y.
 You will need to isolate the variable to solve the literal equation.
3
Fill in the blanks.
Literal Equation
Similar Equation
4 x + 2 y = 16
35 + 8y = 11
 You will need to isolate the variable to solve the literal equation.
4x
-4x
+ 2 y = 16 - 4x
35 - 35
+ 8y = 11 - 35
 Simplify
2 y = 16 -
4x
8y =
- 24
 Isolate y
2 y = 16 - 4x
2
2
8y =
8
-24
8
y =
-3
= 16 - 4x
2
2
 Simplify
y = 8 - 2x
4
Literal Equations:
Use the Frayer Diagram to demonstrate your understanding of the meaning of the word
“Literal Equations”. First fill in examples and then the non- examples. Using these,
determine the characteristics of “Literal Equations”. With the information in the chart,
write your definition of “Literal Equations”.
(Answers will vary) Example:
Frayer Diagram
Definition
Characteristics
-
At least two variables
equation
An equation that is expressed by
means of at least 2 different
variables.
Literal
Equations
Examples
Non-Examples
P = 2L + 2W
V = BH
3
2x + 3 = 1
5y
A = l ×w
4(t - 2) - (t + 3) = t - 1
d = 2r
5
OFF COMPUTER EXERCISES
1. Solve the following equations. (Remember to show your work and check your
answers.)
(a)
9 + 3(m - 4) = 5m + 1
Expand:
Simplify:
-3m)
9 + 3m - 12
3m - 3
3m - 3m - 3
-3
-3 - 1
-4
-4
2
-2
-1)
÷2)
=
=
=
=
=
=
=
=
5m
5m
5m
2m
2m
2m
2m
2
m
+
+
+
+
1
1
3m + 1
1
1 - 1
Check:
L.S. =
=
=
=
=
9 + 3(m - 4)
9 + 3 ( -2 - 4 )
9 + 3(-6)
9 - 18
-9
R.S.
=
=
=
=
5m + 1
5 (-2) + 1
-10 + 1
-9
L.S. equals R.S., the solution is m = -2.
(b)
3 m - 4 ( m + 6 ) = 2 ( m + 2 ) - 13
Expand:
-2m)
+24)
÷-3)
3 m - 4m
-1m
-1 m - 2m
-3m
-3 m + 24
- 24
- 24
- 24
- 24
- 24
-3m
-3m
-3
m
=
=
=
=
=
=
=
=
2m + 4 - 13
2m - 9
2m - 2m - 9
-9
-9 + 24
15
15
-3
-5
Check:
L.S. =
=
=
=
=
3m - 4(m + 6 )
R.S. = 2(m + 2) - 13
3m - 4(-5 + 6 )
= 2(-5 + 2) - 13
15 - 4 (1)
= 2 (-3) - 13
-15 - 4
= -6 - 13
-19
= -19
6
(c)
3 - 2 ( x + 4 ) = -3 ( 1 - 2 x ) + 14
Expand:
+2x)
-11)
÷8)
3 - 2x - 8
- 2x - 5
- 2x + 2x - 5
-5
- 5 - 11
- 16
- 16
=
=
=
=
=
=
=
3 +
6x
6x
8x
8x
8x
8x
8
-2
=
x
6 x + 14
+ 11
+ 2x + 11
+ 11
+ 11 - 11
8
Check:
L.S. =
=
=
=
=
3 3 3 3 -1
(d)
2(x + 4)
2(-2 + 4)
2(2)
4
R.S.
= -3(1 - 2 x) + 14
= -3(1 - 2 (-2)) + 14
= -3(1 + 4) + 14
= -3 ( 5 ) + 14
= -15 + 14
= -1
L.S. equals R.S., the solution is x = -2.
6 + 3(m - 4) = 6m - 3
Expand:
-3m)
+3)
÷3)
6 + 3m - 12
3m - 6
3m - 3m - 6
-6
-6 +3
-3
-3
=
=
=
=
=
=
=
6m - 3
6m - 3
6 m - 3m - 3
3m - 3
3m - 3 + 3
3m
3m
3
-1
=
m
3
Check:
L.S. =
=
=
=
=
6 + 3(m - 4)
6 + 3(-1 - 4)
6 + 3(-5)
6 - 15
-9
R.S.
=
=
=
=
6m - 3
6(-1) - 3
-6 - 3
-9
7
(e)
2n - 3
2
=
-n - 1
4
2n - 3
21
=
1
2 (2n – 3)
=
-n - 1
4n – 6
=
-n - 1
4n + n – 6
=
-n + n - 1
5n – 6
=
-1
5n – 6 + 6
=
-1 + 6
5n
=
5
5n
5
=
=
5
5
n
=
1
R.S.
= -n - 1
4
= -(1) - 1
4
= -2
4
= -1
2
Clear Fraction:
×4)
Expand:
+n)
+6)
÷5)
2
4
4
-n - 1
41
Check:
L.S. = 2n - 3
2
= 2(1) - 3
2
= 2 - 3
2
= -1
2
L.S. equals R.S., the solution is n = 1.
8
(f)
×30)
30×
6
3
5
-
x
3
=
x
2
3
5
-
x
3
=
x × 30
2
=
x × 30
2
10
30 × 3
5
1
+10x)
1
1
18 - 10x
=
15x
18 - 10x + 10x
=
15x
18
=
18
25
=
25x
25
18
25
=
x
÷25)
Check:
LS
=
15
- 30 × x
3
3 - 1 × 18
5
3 1 25
=
3 × 5 - 6
5 × 5 25
=
15 - 6
25
25
=
9
25
6
+ 10x
25x
RS
=
18 × 1
25
2
=
9
25
L.S. equals R.S., the solution is x = 18.
9
g)
3 (2x - 1)
4
×12)
=
5 (2 – 4x)
6
3
2
12 × 3(2x - 1) =
4
12 × 5 (2 – 4x)
6
1
9(2x - 1)
=
10 (2 - 4x)
18x - 9
=
20 - 40x
Expand:
+9)
1
18x - 9 + 9 =
18x
=
18x + 40x =
58x
=
+40x)
÷58)
20 - 40x + 9
29 - 40x
29 - 40x + 40x
29
58x
58
=
29
58
x
=
1
2
Check
L.S.
=
3 (2x - 1)
4
=
3 (2(1) - 1)
4
2
R.S.
=
5 (2 – 4x)
6
=
5 (2 – 4(1) )
6
2
1
2
1
1
=
3 (0)
4
=
5 (0 )
6
=
0
=
0
L.S. equals R.S., the solution is x = 1.
10
h)
×12)
6a - 5 - 2 =
3
12 ×
6a - 5 - 2
3
=
5a - 1
4
12 ×
+
1
3
5a - 1
4
4
3
12 × (6a - 5) - 12 × 2 =
3
12 × (5a - 1)
4
4
1
-15a)
÷9)
Check:
L.S. =
=
=
=
=
=
=
=
6a - 5 - 2
3
6(5) - 5 - 2
3
30 - 5 - 2
3
25 - 2
3
25 - 2 × 3
3
1×3
25 - 6
3
19
3
19
3
+
12 × 1
3
1
24a - 20 - 24
24a - 44
24a - 44 + 44
24a
24a - 15a
9a
9a
9
a
+44)
+ 1
3
=
=
=
=
=
=
=
=
15a
15a
15a
15a
15a
45
45
9
5
R.S.
1
- 3 + 4
+ 1
+ 1 + 44
+ 45
- 15a + 45
=
=
=
=
=
=
=
=
5a - 1 +
4
5(5) - 1 +
4
25 - 1 +
4
24 + 1
4
3
6 + 1
1
3
6×3 +
1×3
1
3
1
3
1
3
1
3
18 + 1
3
19
3
L.S. equals R.S., the solution is a = 5.
11
2. Solve each literal question.
a) Solve: y = m x + b for x
y
y–b
y–b
m
y–b
m
-b)
÷m)
b)
=
=
=
=
mx + b
mx + b - b
mx
m
x
Solve: C = 2 p r + w
C
C - w
C- w
C- w
2r
C- w
2r
-w)
÷2r)
c)
=
=
=
=
=
for p
2pr + w
2pr + w - w
2pr
2pr
2r
p
Solve: P = 2 L + 2 W for L
(Hint: use a diagram.)
W
L
-2W)
L
÷2)
W
d)
P
P - 2W
P - 2W
P - 2W
2
P - 2W
2
=
=
=
=
=
2L + 2W
2L + 2W - 2W
2L
2L
2
L
As you know, P = 2 L + 2 W is the formula for perimeter. If a field has a
width of 25m and a perimeter of 206 m, find the length by using your answer in
(c). (Hint: use a diagram.)
25m
L
P = 2L + 2W
= 206 m
12
P - 2W
2
206 - 2(25)
2
206 - 50
2
156
2
78
=
L
=
L
=
L
= L
= L
Therefore the length of the field is 78m.
13

Concept: Solving Multi-Step Equations

Transcription

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