Hey guys I am new around here, and I noticed that the forum categories seem to cater to much higher level maths than mine (yr10), so I think it should be fairly easy to help me ;3.

1)the sum of two numbers is 12 and the sum of their squares is 80. Find the numbers.
2)If a side of a rhombus is 3cm shorter than one of the diagonals and 1cm shorter than the other diagonal, find the length of a side of a rhombus.
3)A rectangle has an area of 224cm² . If the length was decreased by 1cm and the breadth increased by 1cm, the area would be increased by 1cm² . Find the length of the original rectangle.

I have many more similar questions to these, but if I could just get help answering these specific three, I could follow the formula/way and solve the other ones myself.

Got the first one through guess check and improve, but I'd like to know how to actually do it. Would be nicer to know in case it popped up in a test in the future ;3

1) Call the two numbers x and y. Then you have \displaystyle \begin{align*} x + y = 12 \end{align*} and \displaystyle \begin{align*} x^2 + y^2 = 80 \end{align*}. Can you solve these two equations simultaneously for x and y?

2) Call one of the diagonals x and the other y. The diagonals are perpendicular bisectors of the rhombus, so this creates four right angle triangles. The two shorter lengths in one of the triangles are \displaystyle \begin{align*} \frac{x}{2} \end{align*} and \displaystyle \begin{align*} \frac{y}{2} \end{align*}. The side length of the rhombus is the hypotenuse, and it is both \displaystyle \begin{align*} x - 3 \end{align*} and \displaystyle \begin{align*} y - 1 \end{align*}, so by Pythagoras we have \displaystyle \begin{align*} \left( \frac{x}{2} \right) ^2 + \left( \frac{y}{2} \right) ^2 = \left( x - 3 \right) ^2 \end{align*} and \displaystyle \begin{align*} \left( \frac{x}{2} \right) ^2 + \left( \frac{y}{2} \right) ^2 = \left( y - 1 \right) ^2 \end{align*}. Can you solve these equations simultaneously for x and y? If so you can get the side length.

3) The original rectangle has length "l" and width "w", its area is \displaystyle \begin{align*} 224\,\textrm{cm}^2 \end{align*}, so \displaystyle \begin{align*} lw = 224 \end{align*}. The new rectangle has length \displaystyle \begin{align*} l - 1 \end{align*} and width \displaystyle \begin{align*} w + 1 \end{align*}, its area is \displaystyle \begin{align*} 225\,\textrm{cm}^2 \end{align*}, so \displaystyle \begin{align*} \left( l - 1 \right) \left( w + 1 \right) = 225 \end{align*}. Can you solve these two equations simultaneously for l and w?

To be honest, I only learnt stuff like that ^^^ last lesson.
Basically I can do stuff in ax^2+bx+c=0
and the other formula in the picture/link, but this seems to be a bit different, or maybe I just don't get it haha.

So x+y=12
Trying to make it in the form of ax^2+bx+c=0, I do

x+y-12=0

???
The x is not square, so I can't get the factor for that as easily. I try to get the factor for -12, but my teacher said that it has to add up to the middle one, which is y.

Or am I supposed to take a totally different approach to this question?

Thanks for the help!!

EDIT: To think that was the easy question too.. ;/ Here to learn though eh? ;p

I'd write \displaystyle \begin{align*} y = 12 - x \end{align*} and substitute it into \displaystyle \begin{align*} x^2 + y^2 = 80 \end{align*}, giving \displaystyle \begin{align*} x^2 + \left( 12 - x \right) ^2 = 80 \end{align*}. Expand, simplify, and solve the equation.

Thank you very much! I think I get it now, I'll reply back if I need any more help.

I also have another question.

A machine produces 30 articles in t hours. If the machine were to produce 5 more articles each hour, it would take an hour less to produce 30 articles. Find t.

So I have gotten this far. Not very good at deciphering this stuff, or this kind of maths at all really ;/.

30=t
30+5=t/1????

Not totally sure if that is right at all.

Originally Posted by Nhimzo
I also have another question.

A machine produces 30 articles in t hours. If the machine were to produce 5 more articles each hour, it would take an hour less to produce 30 articles. Find t.

So I have gotten this far. Not very good at deciphering this stuff, or this kind of maths at all really ;/.

30=t
30+5=t/1????

Not totally sure if that is right at all.

probably best to start a new thread but no worries.

let's look at the rate that this machine makes things.

if it makes 35 things in t hours it's rate $R =\dfrac{35}{t}$

Now the problem says that if it produces things at rate R it can make
30 things in 1 hr less time than before, i.e. (t-1).

So

$30 = R(t-1)$

$R=\dfrac{30}{t-1}$ and we know already that $R=\dfrac{35}{t}$ so

$\dfrac{30}{t-1}=\dfrac{35}{t}$

$30t = 35(t-1)$

$30t = 35t-35$

$35 =35t - 30t$

$35 = 5t$

$t=7$

Thank you for that. I tried to use the format that you used for that question for my next question, but it seemed a bit wrong, the numbers just seemed whack. It's a maths equation though so it could just be a weird number but I wanted to know if this was right.

Normally a factory produces 400 radios in x days. If the factory were to produce 20 more radios each day, then it would take 10 days less to produce 400 radios. Calculate x.

R=420/x
400=R(x-10)
Cross multiply
420x-4200=400x
420x-400x=4200
20x=4200
x=210

So that would mean that the factory only makes 400 radios in 210 days??

that's correct.

If it makes 420/210 = 2 radios a day it makes 400 in 200 days which is 10 less then the original 210 days.

Pfft. Can't believe I missed that >.<
Thank you once again!!