WEBVTT
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Part a) Factorise 49 plus 14π§.
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To factorise an algebraic expression like the one we have here means to write it as the product of terms or expressions which multiply together to give the original expression.
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We need to look for common factors between the terms and then take these common factors out.
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First, letβs look at the number part of each term.
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We have 49 and 14.
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We need to find the highest common factor of these two numbers.
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49 and 14 both appear in the seven times table.
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49 is seven multiplied by seven, and 14 is seven multiplied by two.
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And this is the highest common factor of these two numbers.
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So we can take seven out the front of our bracket.
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Usually, we would then look at the letter part of each term.
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But in fact, only one of these two terms has any letters at all.
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So this tells us that weβve already taken out the highest common factor of these two terms.
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Next, we need to work out what goes inside our bracket.
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And this needs to be the numbers or terms that we multiply seven by to give the original expression.
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First, we need to make 49, and seven multiplied by seven is 49 as weβve already said.
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So we just need a seven for the first term in the bracket.
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Thereβs an addition sign between the two terms.
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And then we need to make 14π§.
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Well, seven multiplied by two gives 14, so seven multiplied by two π§ will give 14π§, so this is the second term in the bracket.
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Looking at the terms inside our bracket, so thatβs seven and two π§, these terms have no common factors other than one.
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So this tells us that we fully factorised our expression.
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And our answer is seven multiplied by seven plus two π§.
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We can of course perform a quick check by expanding the bracket out again, and it gives 49 plus 14π§.
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Part b) Simplify nine π¦ minus three π₯ minus eight π¦ minus seven π₯ plus π¦.
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To simplify an expression of this form means to group like terms together.
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Like terms are those which have the same letters with the same powers.
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For example, three π¦ and seven π¦ are like terms, as they have the same letter π¦ and although it isnβt written they have the same power of π¦, which is one.
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However, three π¦ and seven π¦ squared are not like terms, as although they have the same letters, they have different powers.
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We need to identify the like terms in this question.
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We can see that we have three terms that all have π¦, and this is π¦ to the power of one.
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So we have nine π¦ minus eight π¦ plus π¦.
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So nine π¦ minus eight π¦ just gives one π¦, which remember we always just write as π¦, and then π¦ plus π¦ gives two π¦.
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So all of the π¦ terms can be simplified to just two π¦.
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Next, we look at the π₯ terms.
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And we have negative three π₯ minus seven π₯.
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Well, if we start at negative three and then we subtract a further seven, this makes our value more negative.
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So negative three π₯ minus seven π₯ gives negative 10π₯.
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We canβt simplify this expression any further as our π¦s and π₯s are not like terms.
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So our final answer is two π¦ minus 10π₯.
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Another way to approach this question would be to reorder the terms in the original expression so that all of the like terms are together.
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So here weβve reordered so that we have all the π¦s first and then all of the π₯s.
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If youβre going to do this, you must make sure that you bring the sign in front of each term with it.
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We could then look at the π¦s and the π₯s separately.
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And we would see again that we have two π¦ and negative 10π₯ giving us the same simplified answer of two π¦ minus 10π₯.
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Part c) Solve 49 plus two π€ equals 100π€.
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In this part of the question, weβre being asked to solve an equation.
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So we need to find the value of π€ that makes this equation work.
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We notice first of all that π€ is currently on both sides of the equation.
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So we want to rearrange the equation slightly so that π€ is only on one side.
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We have a larger number of π€s on the right of the equation.
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So weβll collect the π€ terms on this side as this means weβll be working with a positive rather than a negative number of π€s.
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In order to eliminate the positive two π€ on the left of the equation, we need to subtract two π€.
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But whatever we do to one side of the equation we must also do to the other in order to keep the equation balanced.
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So weβre subtracting two π€ from both sides.
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On the left of the equation, we now just have 49.
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And on the right, 100π€ minus two π€ gives 98π€.
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To find the value of one π€, we need to divide the right of the equation by 98, but we also need to do the same on the left.
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So this gives 49 over 98 is equal to π€.
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We can actually simplify this fraction quite a lot as both 49 and 98 can be divided by 49.
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49 divided by 49 is one, and 98 divided by 49 is two.
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So this fraction just simplifies to one over two or one-half.
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Be careful here.
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A common mistake thatβs often made is to think that the solution to an equation needs to be a whole number.
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So instead of dividing 49 by 98, some students will divide the bigger number by the smaller number, so thatβs 98 by 49, and get an answer of two.
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However, this is incorrect.
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So our solution to part c) is that π€ is equal to a half.
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Itβs always a good idea to check our answers where possible.
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So letβs substitute this value that we found for π€ back into each side of the equation.
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On the left, we have 49 plus two π€, which is now equal to 49 plus two multiplied by a half.
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Two multiplied by a half is just one, so we have 49 plus one, which is 50.
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On the right of the equation, we have 100π€, which is now equal to 100 multiplied by a half.
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Remember, multiplying by a half is the same as dividing by two.
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So 100 multiplied by a half is the same as 100 divided by two, which is 50.
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Therefore, we found that both sides of this equation give the value 50 when we substitute our value of π€.
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And this tells us that our solution of π€ equals a half is correct.