Year 9 Algebra Help.
- Thread starter ramieshoa
- Start date
Name things.
Let’s do a problem similar to #6.
Problem.
The area of a rectangle is [imath]y^2 + 4y - 21[/imath], and the width is [imath]y + 7[/imath].
What is the height of the rectangle?
Solution.
We know as a matter of general information that
[math]W = \text {rectangle’s width, } H = \text {rectangle’s height, and} A = \text = \text {rectangle’s area} \implies[/math]
[math]A = W * H \implies H = \dfrac{A}{W}.[/math]
So in this specific case [imath]H = \dfrac{y^2 + 4y - 21}{y + 7} = \dfrac{(y + 7)(y - 3)}{y + 7} = y - 3.[/imath]
What is confusing you here is that you are not getting a specific number for an answer. So what?
Let’s do a problem similar to #6.
Problem.
The area of a rectangle is [imath]y^2 + 4y - 21[/imath], and the width is [imath]y + 7[/imath].
What is the height of the rectangle?
Solution.
We know as a matter of general information that
[math]W = \text {rectangle’s width, } H = \text {rectangle’s height, and} A = \text = \text {rectangle’s area} \implies[/math]
[math]A = W * H \implies H = \dfrac{A}{W}.[/math]
So in this specific case [imath]H = \dfrac{y^2 + 4y - 21}{y + 7} = \dfrac{(y + 7)(y - 3)}{y + 7} = y - 3.[/imath]
What is confusing you here is that you are not getting a specific number for an answer. So what?
pka
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- Jan 29, 2005
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For #6 I would note that area equals length x height so:
[imath]4x^2+12x=(2x)(2x+6)[/imath]. Therefore [imath]{\bf h}=2x+6[/imath].
For #7 [imath]\text{Area}~[/imath] [imath]= \left(\frac{1}{2}\right)\left(14\cdot y\right)\left({\bf h}\right)=\\35y^2-21 y=(7\cdot y)(?)[/imath] [imath][/imath] [imath][/imath] [imath][/imath]