Linear Equation in one Variable
„ FORM: Ax + B = C
(A, B, and C are real numbers; A cannot equal 0)
„ A linear equation
q
is called a first degree
g
equation since the greatest power of a variable
is one.
„ GOAL: To find value(s) for x that make the
equation true. (Solve for x.)
Addition Property of
Equality
Objectives: To identify linear
equations and to use the addition
property of equality.
VOCABULARY Review
Equation -- math sentence with an equal
sign; equal sign divides the equation into two
“sides”
„ Expression – math sentence with no equal
sign
„ Solution of the equation – value(s) for the
variable that makes the equation true
„ Inverse operations – operations that undo
one another; For example: Addition &
subtraction are inverse operations.
„
Properties
„ Addition Property of Equality
What you add to one side of an equation must also
be added to the other side of the equation.
Example: A = B, so A + C = B + C
„ Subtraction Property of Equality
What you subtract on one side of an equation, you
must also subtract the exact same value on the
other side of the equation.
Example: A = B, so A - C = B - C.
Are the given numbers solutions of the given
equations? (Substitute number for variable
and simplify.)
A) 5k = 15; 3
B) r + 6 = 4; -1
5(3) = 15
-1 + 6 = 4
15 = 15 (true)
5 = 4 (false)
Yes, 3 is a solution.
So, -1 is not a solution.
„ C) -8m = -12; 3/2
-8(3/2) = -12
-24/2 = -12
-12 = -12 (true) So 3/2 is a solution.
„
GOLDEN RULE of
SOLVING EQUATIONS
„Do
unto one side off an
equation as you do
unto the other side.
1
EXAMPLES
EXAMPLES
1b) 9p + 1 = 8p – 9
1a) x – 4 = 21
We need to add 4 to both sides.
x – 4 + 4 = 21 + 4
x = 25
CHECK
25 – 4 = 21
21 = 21 (It checks!)
We need to subtract 8p from both sides.
9p + 1 – 8p = 8p – 9 – 8p
p + 1 = -9
Now we need to subtract one from both sides.
p + 1 – 1 = -9 – 1
p = -10
CHECK
9(-10) + 1 = 8(-10) – 9
-90 + 1 = -80 – 9
-89 = -89 (It checks!)
Steps
EXAMPLES
1c) -4t + 5t – 8 + 4 = - 4
We need to combine like terms on the left side.
-4t + 5t = t and -8 + 4 = -4.
This gives us t – 4 = -4.
Now we need to add four to both sides
sides.
t – 4 + 4 = -4 + 4
t=0
CHECK (Substitute 0 in original problem and see if it works.)
-4(0) + 5(0) – 8 + 4 = -4
0 + 0 - 8 + 4 = -4
-4 = -4 (It checks!)
EXAMPLES
1.
Clear the equation of fractions. (Multiply
every term by the LCD to remove fractions.)
2.
Use the Distributive Property to remove
parentheses on each side.
C bi like
Combine
lik terms
t
t gett variable
to
i bl on one
side. (Undo addition or subtraction.)
Solve. (Undo multiplication or division.)
Check your solution by substituting what
you get into the original equation.
3.
4.
5.
EXAMPLES
2a) 2(3 + 2x) = 3x – 4
First we need to distribute the 2.
6 + 4x = 3x – 4
(Now we need to combine like terms, so we will subtract 3x from
both sides.)
6 + 4x - 3x = 3x – 4 - 3x
6 + x = -4 (Now we need to subtract six from both sides.)
6 + x – 6 = -4 – 6
Check: 2(3 + 2(-10)) = 3(-10) – 4
x = -10
2b) 4[2t – (3 – t) + 5] = -(2 + 7t)
4[2t – 3 + t + 5] = -2 – 7t
4[3t + 2] = -2 – 7t
12t + 8 = -2
2 – 7t
19t + 8 = -2
19t = -10
t = -10/19
2(3 + (-20)) = -30 – 4
2(-17) = -34
-34 = -34 (It checks!)
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Fractions or Decimals?
Solving with Fractions or Decimals
„ When an equation contains fractions,
p y EVERY PART by
y the LCD. This
multiply
will eliminate all fractions in the problem.
„ With decimals, multiply EVERY PART by a
power of ten so that all coefficients are
integers. (Don’t forget the “shortcut” for
multiplying by a power of 10.)
EXAMPLES
2x + 5 2x +1 −x + 7
=
+
5
2
2
SOLUTION
LCD = 10
3c )
−x + 7
2x + 5
2x +1
= (10)
+ (10)
5
2
2
2(2 x + 5) = 5(2 x + 1) + 5(− x + 7)
4 x + 10 = 10 x + 5 − 5 x + 35
(10)
EXAMPLES
−x x
+ =1
5 4
SOLUTION
LCD = 20
3a )
3
5
x = x−5
4
6
SOLUTION
LCD = 12
3b)
⎛ −x ⎞
⎛ x⎞
20 ⎜
⎟ + 20 ⎜ ⎟ = 20(1)
⎝ 5 ⎠
⎝4⎠
−4 x + 5 x = 2 0
x = 20
⎛3 ⎞
⎛5 ⎞
12 ⎜ x ⎟ = 12 ⎜ x ⎟ − 12(5)
⎝4 ⎠
⎝6 ⎠
9 x = 10 x − 60
− x = −60
x = 60
Summary
Simplify.
„ Add or subtract.
„ Multiply or divide
divide.
„ Check.
„
4 x + 10 = 5 x + 40
− x = 30
x = −30
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