Maths help.

[cover drive man]

I always get stuck on these type of equations, could anyone find me an easy way to work them out.

The equations;

5x-2=x+10

5x-2=3x+1

3(x+4)=x-5

5(x-2)=2(x+4)







Shameful I know, but I have my reasons.

[Steulen]

dude, these are way too simple.

move all X's to one side of the equation and all loose terms to the other, and bingo.

for instance, the first one:

5x - 2 = x + 10
5x -x = 10 + 2
4x = 12
x = 3

[cover drive man]

Also this simultaneous equation:

5x+3y=8
3x-y=9




I know this is pathetic but still.

[GIMH]

5x=x+8

5x=3x+3

3x+7=x

5x=2x+18



I have done some quick cancelling out for you that may well be wrong as I am only on a quick break. Let me know if that helps, it should make things a bit simpler.

basically the first thing you need to do is balance the equation. If you have something saying x+ 7= 3x+19, take 7 off both, ie x=3x+12. Thinking about it now it seems a ridiculously bad example, but hopefully you see what I mean

All this being said, there are plenty of maths geeks here better than me

[cover drive man]

Originally Posted by Steulen

dude, these are way too simple.

move all X's to one side of the equation and all loose terms to the other, and bingo.

for instance, the first one:

5x - 2 = x + 10
5x -x = 10 + 2
4x = 12
x = 3

Thanks, and I know these are easy but when you've been off school for months you forget things.

[GIMH]

There is nothing pathetic about being stuck, or needing help, cdm

[GIMH]

And it seems my method is wrong, jeez shows you how lnog it's been since I learned these things

[silentstriker]

Instead of asking us to do the homework, you should post the question, and then show your attempt at a solution.

There are steps to solving systems of equations. List them, and try to follow them. Show your work, and then ask when you are stuck. You're not going to learn how to do it by having us do your homework.

[cover drive man]

Originally Posted by Steulen

dude, these are way too simple.

move all X's to one side of the equation and all loose terms to the other, and bingo.

for instance, the first one:

5x - 2 = x + 10
5x -x = 10 + 2
4x = 12
x = 3

I tried that on the second one I did.

5x-3x=-2+1
2x=-1
x=-0.5


Thats wrong though, what did I do wrong?

[Steulen]

these are a little harder, because with two unknowns you have to balance both equations in such a way that one of the unknowns gets cancelled out.

So:

5x + 3y = 8
3x - y = 9

let's first move all non-x terms to the right of the equations:

5x = 8 - 3y
3x = 9 + y

then make sure both x terms are equal, by multiplying the first equation by 3,, and the second equation by 5:

15x = 24 - 9y
15x = 45 + 5y

this means that 24-9y = 45+5y, as they both equal 15x.

you can then do the same thing as with the earlier equation where you only has to solve X:

-9y - 5y = 45 -24
-14y = 21
y = -1.5

Then substitute this value for y in any of the two original equations to find x;

5x + (3*-1.5) = 8
5x - 4.5 = 8
5x = 12.5
x = 2.5

[Steulen]

Originally Posted by cover drive man

I tried that on the second one I did.

5x-3x=-2+1
2x=-1
x=-0.5


Thats wrong though, what did I do wrong?

if you move -2 to the other side, it becomes +2.

[cover drive man]

Originally Posted by Steulen

if you move -2 to the other side, it becomes +2.

So,

5x-3x=2+1
2x=3
x=1.5

[cover drive man]

Originally Posted by cover drive man

3(x+4)=x-5

5(x-2)=2(x+4)

What about these ones, do you expand the brackets?

[silentstriker]

Remember PEMDAS. Try it out, and post your attempt. We'll help you out, but you should try it first. We'll correct what you get wrong, it's the best way to learn.

[cover drive man]

I got this

3(x+4)=x-5
3x+12=x-5
3x-x=-5-12
2x=-17
x=-8.5