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Simplify the following expressions.
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Part a) Four π§ minus π§ plus nine π§.
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Another way of saying βsimplifyβ when weβre talking about algebraic expressions is to say βcollect like terms.β
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What that means is just to gather all of the terms that are alike.
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In this first expression, every single term is a π§, so all we need to do is add or subtract as necessary.
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Remember, when addition and a subtraction occur in the same sum, we simply go from left to right.
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Four π§ minus π§ is three π§.
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Careful though, a common mistake here is to think that the answer is four.
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In fact, what itβs saying is that we have four π§s and then we take one of these π§s away.
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So we must be left with three π§.
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Then we add three π§ to the nine π§, which is 12π§.
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Four π§ minus π§ plus nine π§ is 12π§.
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Part b) π multiplied by four multiplied by seven multiplied by ππ.
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When multiplying terms, we can rewrite the expression somewhat.
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Multiplication is what we call commutative, which means that we can perform it in any order.
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We can rewrite this expression as four multiplied by seven multiplied by π multiplied by ππ.
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Once we have the numbers together, we can do the easy part of the multiplication.
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Four multiplied by seven is 28.
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We try to avoid using multiplication signs when we have algebraic expressions.
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So π multiplied by π multiplied by π can simply be written as πππ.
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π multiplied by four multiplied by seven multiplied by ππ simplifies to 28πππ.
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Now a little note about mathematical convention, the number should always be at the front of the expression where possible.
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The letters though can go in any order, though we do tend to keep them alphabetical.
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Part c) Two π¦ cubed minus π¦ cubed minus three π¦ cubed.
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Just because each term has a power doesnβt really change anything.
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A common mistake here is to think that we need to cube the numbers in front of the letters.
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In fact, since indices are calculated before multiplication, the question is simply saying that we have two lots of π¦ cubed.
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Then we subtract one π¦ cubed.
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Then we subtract another three of these π¦ cubed.
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Once again, weβll just go from left to right.
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If we have two π¦ cubes and then we subtract one π¦ cubed, weβre left with one π¦ cubed.
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We donβt actually write the number one.
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We simply write that as π¦ cubed.
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π¦ cubed minus three π¦ cubed is a little trickier, and we can use a number line to help us work it out.
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Letβs imagine weβre subtracting three from one.
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To do this, we start at the number one.
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We move down the number line three spaces.
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That takes us to negative two.
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Since one minus three is equal to negative two, one π¦ cubed minus three π¦ cubed is equal to negative two π¦ cubed.
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And this means that our expression two π¦ cubed minus π¦ cubed minus three π¦ cubed is equal to negative two π¦ cubed.