Transcript File - Mrs. Malinda Young, M.Ed

Solving Equations
With Variables on Both Sides
&
Special Cases
Algebra 1
Glencoe McGraw-Hill
JoAnn Evans
Some equations have variables on both sides of
the equal sign. To solve such equations, collect
like variables on the same side.
In the equation shown below, the term 9x has a greater
coefficient than the term 4x. You can avoid a negative
coefficient by collecting variables on the right.
4x  12  9x  8
4x
4x
12  5x  8
8
8
20  5x
5
5
4x
4x  12  9x  8
9x
9x
5x  12  8
12 12
But it
5x  20
works
5
5
this way
x4
too.
Solve the equation by collecting like variables
on one side. Check the solution.
7x  19  2x  55
2x
2x
9x  19  55
19 19
9x  36
9
9
x4
7  4   19  2  4   55
28  19  8  55
Solve the equation by collecting like variables
on one side. Check the solution.
4 1  x   3x  2  x  1 
4  4x  3x  2x  2
4  x  2x  2
2x 2x
4  x  2
4
4
x  6



4 1   6   3  6   2  6   1
4  7    18   2  5 
28   18   10

Solve the equation by collecting like variables
on one side. Check the solution.
1
12x  16   10  3  x  2 

4
1


12x

16

10

3
x

2


4
3x  4  10  3x  6


3x  4  16  3x
3x
3x
6x  4  16
4 4
6x  12
x 2
Check the solution
with mental math.
Try this one on your own:
7  2  x   5x
14  7x  5x
7x 7x

 7 
7
7 2     5 
 6 
 6

 12  7  
7
7      5 
 6
 6  6 
5
7
7    5 
 6
 6
35 35

6
6
14  12x
12 12
7
x
6
Don’t be reluctant to check
a fractional solution.
It’s just arithmetic!
A special type of equation
3  4  x   2x  12  x
12  3x  3x  12
3x  12  3x  12
3x
3x
12  12
commutative property of addition
When you try to solve an
identity, you end up with a
statement that is always true.
When the two sides of an equation are identical it’s called
an identity. What happens if you continue solving?
An identity will have
INFINITELY MANY SOLUTIONS.
When the two sides of an equation are
identical, any number substituted in as a
solution will make a true statement.
All real numbers are solutions of an identity.
Write:
All Real Numbers
Another special type of equation
8x   4   2    8x
8x   4  2  8x
8x  6  8x
8x
8x
Untrue
statement!
6  0
When all variable terms cancel out and what’s
left is an untrue statement of equality, there
is no solution to the equation.
Write:
No Solution
Solve the equation, if possible. Determine
whether it has one solution, no solution, or is an
identity and thus all real numbers are solutions.
5(x  3)  3x  15  2x
5x  15  5x  15
 5x
 5x
15  15
It is an identity. All real numbers are
solutions of this equation.
5(x  3)  3x  10  2x
5x  15  5x  10
 5x
 5x
15  10
An untrue statement
is left, so there is
no solution to this
equation. No value of
the variable will make
the equation true.
5(x  3)  3x  5  x
5x  15  4x  5
 4x
 4x
x  15  5
 15  15
x  10
This equation has
one solution.
x = -10
1
6(2m  7)  (18  12m)
3
12m  42  6  4m
 4m
 4m
8m  42  6
 42 42
8m  48
8
8
m6
One solution!
8(5c  2)  10(32  4c)
40c  16  320  40c
 40c
 40c
1
4( y  20)  (20 y  400)
5
4 y  80  4 y  80
16  320
No solution!
All real numbers!
Extra practice problems:
1.
2.
3.
4.
2  2  3x   3  3  x   4
 5x  9    3x  13  2 11  x 
14   6  3n   4n  n
3y  2  y  19   9y  3  9  y 
1.
2.
3.
4.
x  1
all real numbers
no solution
y  1