Int 2 Maths Problem

cool!

If you need help with anything else, I'd be more than glad to help!

thank you so much!

Reply 21

OP

Omg I'm so confused with simultaneous equations...
Can someone show how to continue this please! Thanks.

4x+2y=13 X3
5x+3y=17 X2

12x+6y=39
10x+6y=34

(The signs are the same, so you subtract the equations from each other?)

2x=6
x=3

12+2y=13
2y=13-12
2y=1
y=1/2

So, x=3 and y=0.5

However, the answers say that x=2.5 and y=1.5?

Please help someone my prelim is on Thursday!

Reply 22

You have just made a silly mistake when subtracting the equations.

39-34\neq6

Reply 23

OP

namedeprived

You have just made a silly mistake when subtracting the equations.

39-34\neq6

Lmao pretend that never happened. Omg I've made so many pointless mistakes today. 6^2 = 6?, 14x+2x=15x and now this.
Thanks ha!
Could you please do me a quick favour and change the subject of this equation to x, c, and then a for me. It's just so I can see what happens when different parts move to the opposite side of =.

x/c+a=b

Ty.

Reply 24

Is that

\dfrac{x}{c+a}=b

or

\dfrac{x}{c}+a=b

?

Reply 25

Quintro

...

Is that

\dfrac{x}{c+a}=b

or

\dfrac{x}{c}+a=b

?

Reply 26

OP

namedeprived

Is that

\dfrac{x}{c+a}=b

or

\dfrac{x}{c}+a=b

?

It's the second one sorry.

Reply 27

Quintro

It's the second one sorry.

Ok, rearranging for x. We first want to get rid of the +a by subtracting it from both sides, giving

\frac{x}{c}=b-a

We know have a fraction with the numerator equal to x. We want to keep this numerator and get rid of the denominator from the LHS of the equation, so we multiply both sides by c, giving

x=c(b-a)

, which you could multiply out if you wanted.

From this rearranging for c is a fairly simple single step.

When rearranging for (a) I would suggest rearranging from the original equation.

If you need any more help just ask.

Reply 28

OP

namedeprived

Ok, rearranging for x. We first want to get rid of the +a by subtracting it from both sides, giving

\frac{x}{c}=b-a

We know have a fraction with the numerator equal to x. We want to keep this numerator and get rid of the denominator from the LHS of the equation, so we multiply both sides by c, giving

x=c(b-a)

, which you could multiply out if you wanted.

From this rearranging for c is a fairly simple single step.

When rearranging for (a) I would suggest rearranging from the original equation.

If you need any more help just ask.

Omg thank you so much! You have no idea how much you helping me out here and how much I appreciate it.

I'm a bit confused with surds as when my class done them in school I was absent and I'm trying to learn on my own but these books have terrible examples.

Could you please walk me through these 2 questions please.. I'm really sorry I don't know how to write the square root symbol:

2 root 5 + root 20 - root 45
Express as a surd in it's simplest form.

2 root 3 X root 6
Express as a surd in it's simplest form.

Thanks

Reply 29

Quintro

Omg thank you so much! You have no idea how much you helping me out here and how much I appreciate it.

I'm a bit confused with surds as when my class done them in school I was absent and I'm trying to learn on my own but these books have terrible examples.

Could you please walk me through these 2 questions please.. I'm really sorry I don't know how to write the square root symbol:

2 root 5 + root 20 - root 45
Express as a surd in it's simplest form.

2 root 3 X root 6
Express as a surd in it's simplest form.

Thanks

If you want to know how to use the square root symbol and make your equations look nicer click here.

Now the most important rule you'll need to know about surds is that

\sqrt{a}\times\sqrt{b}=\sqrt{ab}

Now for your first equation can you see that 5,20 and 45 are all multiples of 5. This suggests that we will be able to get them all in a form involving

\sqrt5

For the first one we have

2\sqrt{5} + \sqrt{5\times4} - \sqrt{5\times9}

This equals

Spoiler

Try the second one yourself. Surds are one of the hardest things at Sg/Int 2 to get your head around imo, so don't worry if you're finding it difficult.

Reply 30

OP

Can you show me how..

\sqrt {5\times9}=3\sqrt{5}

Please..
Wow that latex thing is nice :biggrin:

Reply 31

Quintro

Can you show me how..

\sqrt {5\times9}=3\sqrt{5}

Please..
Wow that latex thing is nice :biggrin:

\sqrt{5\times9}=\sqrt{9} \times \sqrt{5}=3 \times \sqrt{5}

You're right, looks so much nicer thantrying to represent it with ordinary text. :smile:

Reply 32

OP

I get it now! Oh my god thank you I thought I was never going to get it.
Is the answer to the second one:

6\sqrt 2

?

Reply 33

Quintro

I get it now! Oh my god thank you I thought I was never going to get it.
Is the answer to the second one:

6\sqrt 2

?

Thanks for the rep btw.

Reply 34

OP

namedeprived

Thanks for the rep btw.

Nb. Oh yeah I hit a question the other day where you had to find the height of a cuboid when given the length, the breadth and the volume.

LBH=V

So when I'm changing the subject of this, am I allowed to divide both sides by both L and B at the same time?

Does this mean H=V/LB ?

Reply 35

Quintro

Nb. Oh yeah I hit a question the other day where you had to find the height of a cuboid when given the length, the breadth and the volume.

LBH=V

So when I'm changing the subject of this, am I allowed to divide both sides by both L and B at the same time?

Does this mean H=V/LB ?

Yes, you can divide through by anything if you have an equation(provided it isn't 0!), as long as you make sure you do the same to both sides.

Reply 36

OP

namedeprived

Yes, you can divide through by anything if you have an equation(provided it isn't 0!), as long as you make sure you do the same to both sides.

Nice that's really useful. I think the following type of question is one of the only ones I don't know how to do out of all my past papers now. Can you show me how to work out the following please...

\frac{3}{(x+1)} - \frac{1}{(x-2)}

As a fraction in it's simplest form.

Where x can't be -1 or 2.

Reply 37

Quintro

Nice that's really useful. I think the following type of question is one of the only ones I don't know how to do out of all my past papers now. Can you show me how to work out the following please...

\frac{3}{(x+1)} - \frac{1}{x-2}

As a fraction in it's simplest form.

Where x can't be -1 or 2.

What do you have to have in order to take away one fraction from another?

Reply 38

OP

The same denominator, but I don't know how I would achieve that. I know it's probably something REALLY basic lol, but I can't think for some reason.

Reply 39