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secondary 3 | E Maths
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Ria
Ria

secondary 3 chevron_right E Maths chevron_right Singapore

As you can see I have shown you two different answers The one on top as my teacher’s answer and the one below is the one I tried to solve and I don’t understand what the answers are different is my teachers answer (the TOP one)wrong?I don’t understand why the second line the whole equation changes 15-7y-2y^2 suddenly becomes 2y^2+7y-15 I totally don’t understand the change in the second line for the plus and minus signs and my last answer you can see is y=5 while my teachers one is y=-5 someone please help and explain to me very clearly step-by-step thank you very very much!!

Date Posted: 5 years ago
Views: 253
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Marissa
Marissa's answer
1 answers (A Helpful Person)
1st
Hi Ria, there are some errors in your workings. Otherwise, your answer would match your teacher's.
Ria
Ria
5 years ago
Hi thank you so much for your help but I still don’t get it for the cross multiplication 3×y=3y and -2y×5=-10y and 3y plus -10y is equal to -7y in the cross multiplication! No I don’t get why we have to have a negative sign on both numbers 3 and 5 in the cross multiplication? Will you please explain to me thank you so much
Marissa
Marissa
5 years ago
Hi, while doing this cross multiplication the last row must be the same as your equation. As you can see, from your working, your end equation is 2y^2+15-7y but the equation you have written at the beginning is -2y^2+15y-7y. So there is a need for a negative sign on either 2y or y.

For your question on why we have to put -3 and -5, after putting the negative sign on y in the first column, you realise that the only way you can get -7y in the last row and last column is by putting -3 and -5.

It can only be both positive or both negative. In this case, it cannot be both positive as you will not get the -7y in the end.

Hope this answers your question.