Some Math Help please

[aspiringantelope]

Hey I need some help on these
Linear equations
a) x+4=\frac{3}{2}x
b) \frac{2\left(1-2x\right)}{3}-2x=-\frac{2}{5}+\frac{4\left(2-3x\right)}{3}
c) \frac{4y-5}{2}-\frac{2y-1}{6}=y

Solve for x
a) \frac{a}{x}+\frac{b}{x}=1
b)\frac{x}{a}+\frac{x}{b}=1
c) ax+b=cx

And harder ones - (solve x)
a) \frac{b-cx}{a}-\frac{a-cx}{b}+2=0
b) \frac{a}{x+a}+\frac{b}{x-b}=\frac{a+b}{x+c}

Hopefully these equations show up.
And please if could, show step by step solution?
Thanks

[aspiringantelope]

WHATTTTTTTTT
anyone know how to make them show the equation?? I don't know how to type equations (

[RuiAce]

Your LaTeX is fine, you just gotta enclose them in these brackets:

Code: [Select]

\( LaTeX goes here. \)


[aspiringantelope]

Ok thanks  wow never knew these stuff (testing one)
a) \(  x+4=\frac{3}{2}x \)
b) \(  \frac{2\left(1-2x\right)}{3}-2x=-\frac{2}{5}+\frac{4\left(2-3x\right)}{3} \)
c) \(  \frac{4y-5}{2}-\frac{2y-1}{6}=y \)

Solve for x
a) \( \frac{a}{x}+\frac{b}{x}=1 \)
b) \( \frac{x}{a}+\frac{x}{b}=1 \)
c) \(  ax+b=cx \)

And harder ones - (solve x)
a) \(  \frac{b-cx}{a}-\frac{a-cx}{b}+2=0 \)
b) \(   \frac{a}{x+a}+\frac{b}{x-b}=\frac{a+b}{x+c} \)

[Bri MT]

A good place to start is gathering the x terms on one side.
So for a this would be

x - 3x/2 + 4 = 0

Then we collect like terms
-x/2 + 4 = 0

You might already see that x=8 is the solution

If you don't see that yet,  add x/2 to each side, then multiply each side by two. 

b looks a bit trickier, but it isn't as bad as it seems.

We can multiply the whole equation by 3 to get rid of most of the fractions if you'd rather not work with them.
Once you've expanded out the brackets,  you can then collect like terms (like we did for part a). From there,  it should be fairly simple to find the solution. 

I'm leaving that here since I'm on mobile,  but hopefully this helps you get a start

If you're still having trouble,  reply back with what you've tried and you'll be helped through

[Chelsea f.c.]

Kudos to you for learning latex early... I got a maths degree and never learnt 

[aspiringantelope]

Kudos to you for learning latex early... I got a maths degree and never learnt 

It I actually haven't learnt it, I just went on a website and copied the equation because I couldn't get it in equation form here.

[aspiringantelope]

A good place to start is gathering the x terms on one side.
So for a this would be

x - 3x/2 + 4 = 0

Then we collect like terms
-x/2 + 4 = 0

You might already see that x=8 is the solution

If you don't see that yet,  add x/2 to each side, then multiply each side by two. 

b looks a bit trickier, but it isn't as bad as it seems.

We can multiply the whole equation by 3 to get rid of most of the fractions if you'd rather not work with them.
Once you've expanded out the brackets,  you can then collect like terms (like we did for part a). From there,  it should be fairly simple to find the solution. 

I'm leaving that here since I'm on mobile,  but hopefully this helps you get a start

If you're still having trouble,  reply back with what you've tried and you'll be helped through

And thanks a lot!!!!!
I will attempt them now

[RuiAce]

I'll put up a solution for the last one as well, more or less because that one takes considerably more effort than the previous lot.




Note that difference of two squares was used.
\begin{align*} \therefore cx &= ax - bx - ab\\ -ax+bx+cx &= -ab\\ -(a-b-c)x &= -ab\\ x&= \frac{ab}{a-b-c} \end{align*}
Hopefully I didn't mess that up - with any luck it should be ok

[aspiringantelope]

I'll put up a solution for the last one as well, more or less because that one takes considerably more effort than the previous lot.




Note that difference of two squares was used.
\begin{align*} \therefore cx &= ax - bx - ab\\ -ax+bx+cx &= -ab\\ -(a-b-c)x &= -ab\\ x&= \frac{ab}{a-b-c} \end{align*}
Hopefully I didn't mess that up - with any luck it should be ok

Wow! That's right btw! (checked the answers as I have to study for a LOTE exam tomorrow ) btw, just wondering how long that one question took you to work out? (Not type it on the site)

Thanks a lot wow!

[aspiringantelope]

Can someone please help me with 
\(\frac{4y-5}{2}-\frac{2y-1}{6}=y \)
I keep getting y = 4 but answer says 7/2

[Lear]

Can someone please help me with 
\(\frac{4y-5}{2}-\frac{2y-1}{6}=y \)
I keep getting y = 4 but answer says 7/2

Recheck your usage of brackets when minusing the fractions

[AlphaZero]

I suspect you may have expanded the second set of brackets incorrectly. (Mind the negative in the second term!)

Edit 1: as pointed out by miniturtle, I multiplied by 12 to highlight how the brackets matter. You should treat the minus sign in front of the brackets as a coefficient of \(-1\). (Or you could just flip the sign of every term inside the brackets - but I don't believe you should think about it like this). But now that you've noted your mistake, I've changed the first step to "multiply both sides by 6" as it leads to a cleaner solution.

[aspiringantelope]

I suspect you may have expanded the second set of brackets incorrectly. (Mind the negative in the second term!)

Couldn't you multiply both sides by 6?

[Bri MT]

Couldn't you multiply both sides by 6?

You could (this is what I did but I decided not to post my answers since Dan was done by the time I was ready to)
I suspect that Dan used 12 to highlight the implied bracket around the numerator


Both approaches are valid