# Thread: area word problem

1. ## area word problem

Q:if a rectangle were 3 in longer and 1 in narrower it would contain 5 sq.in more than it does now; but if it were 2 in shorter and 2 in wider its area would remain unchanged. What are dimensions?

I understand that the question leads to a simultaneous eq.
it says it would contain 5 sq in
does it mean that it equals 5
therefore $(x+3)(y-1) =5$
i formed this simultaneous eq

$(x+3)(y-1) =5$
$(x-2)(y+2) =5$
but it was wrong

I think i didnt interrupt the question probably
please tel me where i am going wrong

thank ou

2. ## Re: area word problem

Originally Posted by bigmansouf
Q:if a rectangle were 3 in longer and 1 in narrower it would contain 5 sq.in more than it does now; but if it were 2 in shorter and 2 in wider its area would remain unchanged. What are dimensions?

I understand that the question leads to a simultaneous eq.
it says it would contain 5 sq in
does it mean that it equals 5
therefore $(x+3)(y-1) =5$
i formed this simultaneous eq

$(x+3)(y-1) =5$
$(x-2)(y+2) =5$
but it was wrong

I think i didnt interrupt the question probably
please tel me where i am going wrong

thank ou
you want

$(x+3)(y-1) =x y + 5$
$(x-2)(y+2) =x y$

3. ## Re: area word problem

Originally Posted by bigmansouf
Q:if a rectangle were 3 in longer and 1 in narrower it would contain 5 sq.in more than it does now; but if it were 2 in shorter and 2 in wider its area would remain unchanged. What are dimensions?

I understand that the question leads to a simultaneous eq.
it says it would contain 5 sq in
does it mean that it equals 5
The problem doesn't say "it would contain 5 sq in". It says "it would contain 5 sq in more".
With sides x and y, in inches, the original area was xy sq. in. "5 sq in more" would be xy+ 5.

therefore $(x+3)(y-1) =5$
i formed this simultaneous eq

$(x+3)(y-1) =5$
$(x-2)(y+2) =5$
but it was wrong

I think i didnt interrupt the question probably
please tel me where i am going wrong

thank ou