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### Course: Algebra (all content)>Unit 5

Lesson 5: Manipulating expressions with unknown variables

# Manipulating expressions using structure (example 2)

Given that a+b=2a, Sal finds an expression equivalent to b-a.

## Want to join the conversation?

• Why do we manipulate expressions in math and what is it used for in reality?
• We manipulate expressions for several reasons. Here are three of them.

One reason is to make the expression easier to work with. Sometimes an expression might be written in a more complicated way than is necessary. We can manipulate the expression to change it into a simpler form.

A second reason is to highlight an important aspect of the expression. For example, I might have an equation for a line written as "3x - 4y = 5". If I write the equation this way, I can quickly find the x-intercept and the y-intercept. But it's not so easy to see what the slope is. I can manipulate that equation to rewrite t as "y = 3/4(x) - 5/4". If I do, now I can easily see that the slope is 3/4.

A third reason is to get the expression to fit a formula. For example, the formula for difference of squares is "a^2 minus b^2 equals (a+b) times (a-b)". That's a great tool for getting rid of some pesky exponents. But the tool only works if you have two quantities that are both squares of something.

Could I use the difference of squares on an expression like "x^2 + 12x + 32"? At first, it doesn't look like a difference of squares. But maybe I can manipulate the expression to make it fit the formula for difference of squares. Let's try.

x^2 + 12x + 32
x^2 + 12x + 36 - 4 [replacing "32" with "36 - 4"]
(x^2 + 12x + 36) - 4 [associative property]
(x+6)^2 - 4 [factoring the perfect square trinomial]
(x+6)^2 - 2^2 [replacing "4" with "2^2"]

Now it's in the right format for difference of squares. "(x+6)" is the "a", and "2" is the "b".

(x+6 + 2)(x+6 - 2) [difference of squares]
(x+8)(x+4) [subtraction]

By manipulating the expression, I was able to turn "x^2 + 12x + 32" into "(x+8)(x+4)". That's good, because now I can easily find the two places where a graph of the expression will cross the x-axis.
• Couldn't this also be solved in this way:
a+b=2a
a+b-a=2a-a (subtract a from both sides)

Next, if b=a we can either subtract b from both sides or a from both sides
Example of subtracting a from both sides : b-a=a-a thus b-a =0, or
subtracting b from both sides of the equation :
b-b=a-b thus a-b=0
a-b is the answer becuase it is equal to b-a
Good thinking - Nicely done.

Now, I am not too sure what you mean by "no negatives involved", or why you think that is even a concern, but be that as it may, remember there are many different ways to arrive a a solution. Some solutions may appeal more to you than others; there is always something to be gained by looking at a problem from as many perspectives as possible.
• i'm still quite confused by this 'manipulating expressions using structure' module.Can somebody explain to me how to solve this question again?
• Why did Sal multiply both sides by -1 to get the answer?
• Remember that an equation is a balance. It means literally 'equivalent' (an association).

So when you say: a = 2 you can also say that 2 = a , because they are equivalent.

Now imagine that: If you put a rock in one plate of a balance, you need to put a rock on the other plate as well to keep the balance plates stabilized at the same height.

That said, if I have an equation like: "a = 2", and multiply 'a' by '-1', I need to make sure that '2' gets multiplied by '-1' as well in order to maintain the balance, so "-1*a = -1*2" is the same thing as "a = 2".

What Saul did was just the same (just to make the equation looks like some of the available choices, first he found that 'b' was equal to 'a', given that he found that "b-a = 0", but there were no choice like "b-a" or "0", so he multiplied both sides by -1 to get "-b+a=0" or "a-b=0".
• 5a2+4b3 , when a = 3 and b = 2
• 1) Replace each variable with its given value.
2) Follow order of operations rules (PEMDAS) to complete the math. Do exponents, then multiply, then add.

Give it a try. Hope this helps.
(1 vote)
• Am I confused or Sal? when its written b=a, its clear that even a-b=0 consequently b-a will be 0 too. Do we have to multiply it -1 in Grid In questions ?
• I never would've guessed to multiply by negative 1. Does this make me incompetent?
(1 vote)
• No, not at all. . . . AND . . . now you know a new method to try if you come across a similar situation.
• this uses the same format as linear equations but in variables?
b-a=0 why didn't he just swap them because a=b and b=a?
(1 vote)
• You can't swap them because subtraction is not commutative. Of course since they are equal, swapping them won't change the answer, but it's still breaking the math rules, which makes your proof incomplete/incorrect.
• I have looked at all the tips and all the questions. So now I'm going to ask this question, how did you come up with multiplying by negative one and is that legal? If it is legal, why is it legal?
(1 vote)
• When you have an equation, you can manipulate the parts by operations including multiplying by -1 as long as you do it on both sides, but with a 0, multiplying anything just gets zero
Lets do it another way mathematically
If we start from b - a = 0 , I can do opposites of subtracting b and adding a
b - a = 0
-b+a -b+a
0 = -b + a since addition is commutative, I can switch paces for a and b
0 = a - b
Since both are equal to zero, they must be equal to each other,
b - a = a - b
So multiplying by a negative 1 is equivalent to moving all the terms to the other side by subtraction or addition, and multiplying by -1 on both sides is much easier than to move everything