Download Math 84 Activity – Solving Equations

1
Math 84
SOLVING LINEAR EQUATIONS
Activity #
(With sections 3.2, 3.4, 4.8, 5.6 )
Name:________________________________ Team Member #1:_______________________
Team Member #2:______________________ Team Member #3:______________________
An equation is equality of algebraic expressions. Ex. 5x + 2
5x + 2
=
= 3x – 10 .
3x – 10
Take away 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 do 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

C
C
, for any real number C, C  0 (why?)……Division property of equality
A linear equation in the variable x is one that can be expressed as ax  b  c , where a, b, c are real
numbers with a  0 . The following are examples of linear equations (identify a, b, c)
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
A solution of an equation is a number that makes the equation a true statement upon substitution for
the variable.
2
EX. To determine whether – 2 is a solution of – 3x +
if we get a TRUE statement.
– 3x
+ 10 = 16
– 3(– 2) + 10 =
6 + 10 =
16 =
10 = 16, we substitute (– 2) for x and see
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
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
x  0   68
x   68
3)
1
x   12
5
, subtraction
x  84  84  16  84
, addition
x  100
4) 5 x   12
5x
12

5
5
multiplication
, division
12
x  
5
In the following note that dividing by a fraction is the same as multiplying by the reciprocal
3
1
9
y 
5)
6) 3w 
2
4
2
2 3
2 1
3w
9  1
 y   …….explain
     ……….explain
3 2
3 4
3
2  3
1
3
y
w   …………...explain
…….explain
6
2
1
5  x  5  ( 12)
,
5
x   60
7) x  ( 8)  11
8) 15  x  24
15  ( 15)  x  24  15
x  49
x  8  8  11  8
x  19
Observe that we have been performing the opposite operation(s) to isolate the variable
3
Now try the following.
1) Determine whether (– 110) is a solution of
x
1

220
2
2) Solve the following. Explain each step and check whether your solution is correct
x = – 22
a) 11 +
c) 11x
e)
=
– 22
11
x   22
5
b) x – 11 = – 22
d)
x
  22
11
f) 
5
x  25
11
*In the following you need to decide which property to use first – will be explained next. Try now
anyway and check your answer – discuss with team members
g) 1  x   11
h) 1  2 x   7
4
Using Multiple properties of equality in solving a linear equation (Again our aim is to isolate the
variable).
A step by step guide to solve any linear equation.
1st ) Simplify each side separately
(If there are fractions, multiply both sides (every term of the equation) by the LCD of all the fractions
to clear fractions. If there are decimals, multiply by the appropriate power of ten to clear decimals.)
2nd ) If the variable appears on both sides, remove it from one side using addition or subtraction
property. A better choice is removing the one with the lesser numerical coefficient (why?)
3rd ) Now ISOLATE the variable by performing the opposite operations on the variable in a
reverse order.
EXAMPLES: Solve the following equations.
1) 2 x  17  20
We observe that x is first multiplied by 2 and then 17 is added (how do we know this?)
Therefore, we will first subtract 17 from both sides, and then divide both sides by 2.
2 x  17  17  20  17
2x  3
2x
3

2
2
3
x 
2
2)
5
11
4
x

7
2
3
5
11


4
42  x 
  42  
2
7
 3
42  4
5 
 11 
42  x   42   
3
7 
2
6  5  x  21 11  14  4
30 x  231  56
30 x  231  231  56  231
30 x  287
30 x
287

30
30
287
x 
30
………….1st we clear the fractions; LCD = 42
................(Why?)
………….(Why?)
………… Think !, Do we first add or divide (why?)
5
Exercises: Solve the following equations. Justify your steps and check your 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
4

5) 2  x    x 
 x
3
3

7)
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