2/9/17 Week 5 Thursday daily sheet: Multi-step equations
Daily aims:
 1. I can solve multi-step equations that require using the distributive property, combining variables and/or
constants on a single side, and collecting variables from both sides of the equation.
Before lesson
1) Solve: 4a + 3 + 3a = 20 + 4
During lesson
2) Solve: 4(2x – 5) + 15 = 3
3) Solve: 2x + 5 = 6x  7
4) Solve: 2(x + 4)  3x = 5x + 2 + 4
D. Stark 1/16/2017
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KEY
2/9/17 Week 5 Thursday daily sheet: Multi-step equations
Before lesson
During lesson
1) Solve: 4a + 3 + 3a = 20 + 4
First, simplify everything you can on each side
separately. Clean up your own neighborhood first
before you start doing things to both sides.
Combine like terms that are living on the same side
of the tracks—variables with variables, constants
with constants.
4a + 3 + 3a = 20 + 4
4a + 3a
+3=
20 + 4
7a + 3 = 24 [Now it’s a regular 2-stepper.]
7a + 3 = 24
7a = 21
7a = 21
a=3
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2) Solve: 4(2x – 5) + 15 = 3
A useful starting point is the distributive property.
That will let you get rid of the parentheses so
maybe you can combine some terms.
4(2x – 5) + 15 = 3
4(2x) – 4(5) + 15 = 3
Here’s an alternative method that doesn’t use the
distributive property. Things won’t always work out
so nicely, but sometimes they do:
4(2x – 5) + 15 = 3
4(2x – 5) + 15 = 3
4(2x – 5) = 12
4(2x – 5) = 12
Now simplify as much as you can on each side by
itself:
8x – 20 + 15 = 3
8x – 5 = 3 [Now it’s a regular 2-step equation.]
2x – 5 = 3
2x – 5 = 3
8x – 5 = 3
2x = 2
2x = 2
[The 4 is multiplied by the whole left
side, so it’s OK to divide by it. You’re
not cancelling inside the parentheses
where the subtraction is.]
8x = 8
8x = 8
x=1
x=1
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3) Solve: 2x + 5 = 6x  7
Alternatively, if you really like the variable always to
be on the left all along, you can use this method.
Sometimes you’ll find variables on both sides of the You’ll just have to deal with more negative
numbers.
equation. You’ll need to collect them all on 1 side.
(I tend to get variables all on 1 side first and then
2x + 5 = 6x  7
constants all on the other side, but there’s no
reason you have to use that order.)
 4x + 5 =  7
If you want to avoid dealing with negative numbers
any more than necessary, collect the variables on
 4x + 5 =  7
the side that has the most of them—in this case, on
the right.
 4x =  12
2x + 5 = 6x  7
5 = 4x  7
5 = 4x  7
 4x =  12
x=3
12 = 4x
12 = 4x
3=x
or
x=3
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4) Solve: 2(x + 4)  3x = 5x + 2 + 4
2(x + 4)  3x = 5x + 2 + 4
2x + 8  3x = 5x + 2 + 4
 x + 8 = 5x + 6
 x + 8 = 5x + 6
8 = 6x + 6
8 = 6x + 6
2 = 6x
6x = 2 [You can flip left & right sides of the = any time.]
x=
𝟏
𝟑
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