### Present Remotely

Send the link below via email or IM

• Invited audience members will follow you as you navigate and present
• People invited to a presentation do not need a Prezi account
• This link expires 10 minutes after you close the presentation

Do you really want to delete this prezi?

Neither you, nor the coeditors you shared it with will be able to recover it again.

# Solving Equations with Variables on Both Sides

Solving with Variables on Both Sides (Including No Solution and Infinite Solutions) for Algebra 1
by

## Rob Frederick

on 4 September 2012

Report abuse

#### Transcript of Solving Equations with Variables on Both Sides

Solving Equations
with

on
Application The long-distance rates of two phone companies are shown in the table. Phone Co. Charges Company A 36 cents plus 3 per min
Company B 6 cents per min Application (part two) "When is 36 cents plus 3 cents per minute
the same as 6 cents per minute?" To solve this problem we need to break it down. Let's sum up what we have into one sentence: Now what can we do with this? Answer: Let's make it an algebraic expression.
Then, we can solve for the missing piece Application (part three) When
is 36
cents plus 3 cents
per minute the same
as 6 cents
per minute ? Now that we have an algebraic equation,
we can solve for m... Solving with Variables on Both Sides Answer: Let's make it an algebraic expression.
Then, we can solve for the missing piece Get like terms
onto one side
and combine. Then solve
just like any
other equation. m = 12 A call 12 minutes
long will be the same
price on either company. ? ? ? Identity (Infinitely Many Solutions) Consider this: Condradiction (No Solutions) Also observe what happens here:
x + 4 - 6x = 6 - 5x - 2
Notice anything strange? Since this equation is ALWAYS true
for ANY value of x, it is called an
and has infinitely many solutions. identity For an identity,
all real numbers
are solutions.
-8x + 6 + 9x = -17 + x
Since this equation is NEVER true
for ANY value of x, it is called a
there is no value of x
that will make the
equation true. OTHER EXAMPLES 7k = 4k + 15 k = 5 5x - 2 = 3x + 4 x = 3 2(y + 6) = 3y y = 12 3 - 5b + 2b = -2 - 2(1 - b) b = 1.4 Don't forget:

STAY ORGANIZED!