Year 9 Algebra Help.

[ramieshoa]

Hey guys,
I’m currently in year 9 and doing year 10 maths right now. I have no clue how to question 6 and 7 at all. I have attempted numerous times and no result. Please help me if you can, would mean the world. ?

 

[blamocur]

Do you know the formulas for the areas of rectangles and triangles ?

 

[JeffM]

Name things.

Let’s do a problem similar to #6.

Problem.

The area of a rectangle is $y^2 + 4y - 21$, and the width is $y + 7$.

What is the height of the rectangle?

Solution.

We know as a matter of general information that

$$W = \text {rectangle’s width, } H = \text {rectangle’s height, and} A = \text = \text {rectangle’s area} \implies$$
$$A = W * H \implies H = \dfrac{A}{W}.$$
So in this specific case $H = \dfrac{y^2 + 4y - 21}{y + 7} = \dfrac{(y + 7)(y - 3)}{y + 7} = y - 3.$

What is confusing you here is that you are not getting a specific number for an answer. So what?

 

[pka]

Hey guys,
I’m currently in year 9 and doing year 10 maths right now. I have no clue how to question 6 and 7 at all. I have attempted numerous times and no result. Please help me if you can, would mean the world. ?
View attachment 29713

For #6 I would note that area equals length x height so:
$4x^2+12x=(2x)(2x+6)$. Therefore ${\bf h}=2x+6$.

For #7 $\text{Area}~$ $= \left(\frac{1}{2}\right)\left(14\cdot y\right)\left({\bf h}\right)=\\35y^2-21 y=(7\cdot y)(?)$ $$ $$ $$