Download M84 Act 9 Solving Linear Equations

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Math 84
Activity # 9
Your Name: ___________________________
“Solving Linear Equations”
An equation is equality of two algebraic expressions.
Example.
5x + 2
5x + 2
= 3x – 10 .
=
3x – 10
Subtract 3x from each side and record what you have on each side. Then subtract 2 from each side and
record the result. Now divide both sides by 2. What is the result?
We may think of an equation as a balance beam which we want to keep balanced all the time. It
remains balanced if we take the same action on both sides. Therefore, given an equation A = B, we
can perform the following operations on both sides.
1) A + C = B + C , for any real number C …………… Addition Property of Equality
2) A – C
= B – C , for any real number C …………… Subtraction Property of Equality
3) A x C = B x C , for any real number C ………….. Multiplication Property of Equality
( but we don’t want to multiply by 0 – why? )
4)
A
B
, for any real number C ≠ 0 (why?) …… Division Property of Equality
=
C
C
Task 1:. A linear equation, in the variable x, is one that can be expressed as ax + b =
c , where a,
b, c are real numbers. Identify a, b, c in the linear equations below.
5x + 7 = 10
1 – x = 3
– 2x + 3 = 5
1 −
1
x =
0
4
–3
+ 11x = 2
4
x = 5
3
2
5
x +
=
1
3
4
0.5 x − 1.2 =
3.8
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A solution of an equation is a number that makes the equation a true statement upon substitution for
the variable.
Example. To determine whether – 2 is a solution of
see if we get a TRUE statement.
– 3x
+ 10
– 3(– 2) + 10
6 + 10
16
– 3x + 10 = 16, we substitute (– 2) for x and
= 16
= 16
= 16
= 16 ……….TRUE
Therefore – 2 is a solution (it is actually the solution) of the equation – 3x + 10 = 16
Try: Determine whether 3 is a solution of the equation 3x + (– 9) = 18
3.2 Using The Properties Of Equality To Solve Linear Equations.
We will use the appropriate property of equality to isolate the variable.
Examples: Solve the following equations: (Indicate the property used in each step).
1) x + 84 = 16
2) x – 84 = 16
x + 84 − 84 = 16 − 84
, subtraction
− 68
x +0 =
x = − 68
3)
1
x =
5
− 12
x − 84 + 84 = 16 + 84
, addition
x = 100
4) 5 x =
− 12
5x
−12
=
5
5
multiplication
, division
12
x = −
5
Task 2: In problems 5 through 8, which operation is used to solve for the variable?
3
1
9
5)
6) −3w =
y =
2
4
2
2 3
2 1
−3w
9  1
⋅ y =⋅
=
⋅ − 
3 2
3 4
−3
2  3
3
1
___________
____________
w = −
y=
2
6
1
5 ⋅ x = 5 ⋅ ( −12)
,
5
x = − 60
7) x + ( −8) =11
x − 8 + 8 = 11 + 8
x = 19 _____________
8) −15 + x =
24
15 + ( −15) + x= 24 + 15
x = 49
___________________
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Observe that we have been performing the opposite operation to isolate the variable.
Task 2: Now try the following.
x
1
1) Determine whether (– 110) is a solution of
=
220
2
2) Solve the following. Explain each step and check whether your solution is correct.
a) 11 + x
= – 22
b) x – 11 = – 22
c) 11x
– 22
d)
e)
11
x =
5
=
− 22
x
=
11
f) −
− 22
5
x =
25
11
*In the following, you need to decide which property to use first. Check your answer and also discuss
with team members.
g) 1 − x =− 11
h) 1 − 2 x =
−7
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Exercises: Solve the following equations. Justify your steps and check the solution.
1) −3x + 2 =− 22
2)
4
2
x −
=
0
3
3
1
7
11
− x =
−
2
3
6
4)
1
1
x − x =
−x
3
2
3)
1

5) 2  x −  + x =
3

7)
4
+ x
3
7
2
11
1
x− x +
=
12
5
5
5
5
9) 3( x – 5 ) + 2
=
4(x – 1 ) + 1
6) 0.3x + 1.5 ( x − 1) =
2.1
8) −2 x +
1
+ 4 x =− 2 + 11x
2
10) 1 – x = x – 1