Download Algebra I

Algebra I
Warmup – Class #11
Name__________________________ Date ______
Use the distributive property to simplify:
1.
1
(x + 6)
2
3.
−
3
(x + 12)
4
2.
4.
2
(x − 10)
5
2
(x − 7)
3
Solve for x.
5.
15 – x = 12
Check:
6.
7x − 40 = − 5
Check:
7.
−3 − 6x = 39
Check:
Algebra I
3.4 notes
variables on both sides
Look at this equation:
Name________________________ Block 11
5x + 2 = 4x + 10
Do you see anything new?
If you said there are variables on BOTH sides of the equals sign, then you are a genius!!!
BUT, when variables are on different sides of the equals sign, they CANNOT BE COMBINED!!!
Instead, imagine that the equation is a balance. You can remove things from each side, as long
as you remove equal amounts. So, remove x’s!! Be lazy!! Remove the smaller amount!!!
(unless you love negative numbers).
5x + 2 = 4x + 10
−4x
−4x
x + 2 = 10
Now, you know what to do!! Subtract 2, and x = 8.
Check: 5(8) + 2 = 4(8) + 10
40 + 2 = 32 + 10 √
New, Updated Directions for Solving Equations!!!
1. Eliminate parentheses (distribute)
2. Combine like terms
3. Use the addition or subtraction property of equality to get the variables on one side
a. (It’s usually easier to move the “smaller” variable term.)
4. Undo addition or subtraction
5. Undo multiplication or division
1.
6x + 3 = 2x  13
2.
4(x  2) + 3x = 2x + 7
3.
6x + 3 = 4x + 19
4.
3(x + 2) + 2x = x + 36
5.
7x + 14 = 5x
(Since 5x is all by itself, you might want
to get rid of the 7x even though it will
result it a negative coefficient).
6.
10 − 2x = 3x + 5
(how can you undo −2x?)
7.
8x + 25 = 13x
8.
3x − 7x + 6 = 2x
Algebra I
3.4 notes
Infinite Solutions and No Solutions
Solve this equation: 3(x + 5) + 3 =
3x + 15 + 3 =
3x + 18 =
−3x
18 =
Name_________________Block 11
18 + 3x
18 + 3x
18 + 3x Both sides appear to be the same!
−3x
18
Both sides ARE the same!!
18 does equal 18: this statement is ALWAYS true. When this happens, we say that there are
infinite solutions . . . . all values of x make the equation true. We call the equation an IDENTITY.
Test it!! Let x = 1.
Now let x = 3
3(1 + 5) + 3 = 18 + 3(1)
3(6) + 3 = 18 + 3
18 + 3 = 18 + 3 Yes!
3(3 + 5) + 3 = 18 + 3(3)
3(8) + 3 = 18 + 9
27 = 27
Yes!
Infinite solutions!! Identity!!
Now, solve this equation:
8x + 3 = 2(4x + 5)
8x + 3 = 8x + 10
−8x
−8x
3 = 10
Uh-oh!! Not true!!
When this occurs, we say the equation has NO SOLUTIONS!! Ø
When you solve an equation in one variable, you can get a single answer, infinite solutions, or
no solution!!!
Classwork/homework is on page 157-158 # 5-15 odd, 19-25 odd, 31, 32, 45. Write the problem
and show all steps.