Download 3.3 Solving Multi

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
Algebra
3.3 Solving Multi-Step Equations
Solve for x
► 1)
1
x  5  10
2
Solve for x
► 1)
1
x  5  10
2
5 5
Solve for x
► 1)
1
x  5  10
2
5 5
1
x  15
2
Solve for x
► 1)
1
x  5  10
2
5 5
1
x  15
2
2  1   15  2
 x   
12   1 1
Solve for x
► 1)
1
x  5  10
2
5 5
1
x  15
2
2  1   15  2
 x   
12   1 1
x  30
Combining Like Terms First
2) 2x – 9x + 17 = -4
-7x + 17 = -4
- 17 - 17
-7x = -21
x=3
Using the Distributive Property
3) 5x + 3(x + 4) = 28
5x + 3x + 12 = 28
8x + 12 = 28
- 12 - 12
8x = 16
x=2
Distributing a Negative…Careful!
4) 2x – 5(x – 9) = 27
2x – 5x + 45 = 27
-3x + 45 = 27
- 45 - 45
-3x = -18
x=6
Multiplying by a Reciprocal First
5)
Multiplying by a Reciprocal First
3
5) 12  x  2
10
Save time…do not distribute!
Multiplying by a Reciprocal First
3
5) 12  x  2
10
Save time…do not distribute!
10
3
 10
12   x  2
3
10
 3
Multiply by the reciprocal.
Multiplying by a Reciprocal First
3
5) 12  x  2
10
Save time…do not distribute!
10
3
 10
12   x  2
3
10
 3
40  x  2
Multiply by the reciprocal.
Multiplying by a Reciprocal First
3
5) 12  x  2
10
Save time…do not distribute!
10
3
 10
12   x  2
3
10
 3
40  x  2
2
2
Multiply by the reciprocal.
Subtract 2 from each side.
Multiplying by a Reciprocal First
3
5) 12  x  2
10
Save time…do not distribute!
10
3
 10
12   x  2
3
10
 3
40  x  2
2
2
38  x
Multiply by the reciprocal.
Subtract 2 from each side.
Three numbers add to 101. The first number is 12 more than
the second number. The third number is nine more than
twice the second number. Find the numbers.
Three numbers add to 101.
a + b + c = 101
The first number is 12 more than the second number.
a = b + 12
The third number is nine more than twice the second number.
c = 2b + 9
Substitute and solve.
(b + 12) + b + (2b + 9) = 101
4b + 21 = 101
4b = 80
b = 20
a = b +12 = 20 + 12 = 32
c = 2b + 9 = 2(20) + 9 = 40 + 9 = 49
a = 32, b = 20, c = 49 They add to 101!!
Homework
P. 148 (#11-43 Odds) 51, 54
Related documents