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Transcript
Solving Linear Equations
Using
Multiplication/Division
Solving 1-Step Equations
To find the solution to an equation we must
isolate the variable.
We isolate the variable by performing
operations that will eliminate (cancel) the
other numbers from the expression.
The Multiplication Property of Equality says
that we can multiply the same number on
both sides of an equation without changing
the solution to the equation.
Multiplication Property of Equality
Take the equation: 3x = 15
Our goal is to multiply a number on both sides
of the equation that will isolate the x by
canceling the coefficient of x, 3.
Since 3 and x are multiplied, we don’t want to
cancel 3 to a zero, we want to cancel 3 to a
value of one. (Perform an operation that will
make 3x become 1x = x.)
Multiplication Property of Equality
What number, when multiplied by 3, will give
us one?
1
3
3
1
Reciprocals multiply to 1.
The Multiplication Property of Equality says we
can multiply 1/3 to the equation as long as we
multiply it on both sides.
3x  15
1
1
 3 x  15
3
3
Multiplication Property of Equality
See what happens:
Multiply 1/3 on both
sides of the equal sign.
1
1
 3 x  15
Inverse Property
3
3
of Mult. says that
reciprocals
1 15
multiply to 1.
1 x  
3
1
Identity Property of
Mult. Says that 1x = x.
15
x
5
3
Multiply
fractions.
Checking the Solution
Always check the solution by substituting the
value back into the original equation.
3x = 15
3 • 5 = 15
15 = 15
The sentence should be true.
Solving by Multiplication
Multiply both sides of the equation by the
number that will cancel with the coefficient of
the variable. This will isolate the variable by
making its coefficient the number 1.
Reciprocals multiply to 1. The reciprocal of a
whole number is 1 over the number.
Examples
2
Multiply 3/2 on both
1) k  8
sides of the equal sign.
3
3 2
3
 k  8
Check: 2 k  8
2 3
2
3
3 8
2
1 k  
12  8
2 1
3
24
24
8
k
 12
3
2
88 √
Examples
2) 4p  2
Multiply 1/4 on both
sides of the equal sign.
1
1
 4 p   (2)
4
4
1 2
1 p  
4 1
2
1
p

4
2
Check: 4 p  2
1 

4 
 2
 2 
4
 2
2
2  2 √
Examples
w 2
w/5 is the same thing as 1/5 • w,
3)

multiply both sides by 5/1.
5 11
5 w 5 2
  
w 2

Check:
1 5 1 11
10
1 w 
11
10
w
11
5 11
1 10 2
 
5 11 11
10 2

55 11
2
2
√

11 11
Try These!
1)
2)
3)
4)
2
2
x
9
11
 4w  22
r
 6
3
8
6y 
3
Solutions!
2
2
1)
x
9
11
9 2
9 2
 x 
2 9
2 11
18
1 x 
22
9
x
11
2)  4w  22
 1   4w   1  22
 4 
 4  1
22
1 w  
4
11
w
2
Solutions!
r
3)  6
3
3 r 3 6
  
1 3 1 1
18
1 r 
1
r  18
8
4) 6y 
3
1
1 8
 6y  
6
6 3
8
1 y 
18
4
y
9