Algebra Equation. Help. Please. (1 Viewer)

[clarebear22]

Argh.
Maths is annoying.
Anyone know how to do this equation?
Please help it's driving me insane.
Thanks.

2y - 5 - 2 = 3x + 2
3 _______ 5

 

[lyounamu]

Argh.
Maths is annoying.
Anyone know how to do this equation?
Please help it's driving me insane.
Thanks.

2y - 5 - 2 = 3x + 2
3 _______ 5

You cannot solve the question because you have two variables with nothing else given. Did you mean (2x-5)/3 - 2 = (3x+2)/5 ?

 

[clarebear22]

Never mind!
I figured it out.
So confusing.
Thanks anyway.

 

[PC]

Equation is:

(2x – 5)/3 – 2 = (3x + 2)/5

What I would do first is get all the fractions on to one side.
Add 2 to both sides, then subtract the fraction on the right from both sides.

(2x – 5)/3 – (3x + 2)/5 = 2

Now look for a common denominator on the left. It's 15, so change both the fractions.
In the first we multiply top and bottom by 5 and in the second we multiply top and bottom by 3.

5(2x – 5)/15 – 3(3x + 2)/15 = 2

Because the denominators are the same we can combine the fractions.

[5(2x – 5) – 3(3x + 2)]/15 = 2

Now multiply both sides by 15.

[5(2x – 5) – 3(3x + 2)] = 30

We should be right from there.

[5(2x – 5) – 3(3x + 2)] = 30
10x – 25 – 9x – 6 = 30
x – 31 = 30
x = 61

 

[bored of sc]

You could multiply everything by 15 at the start to destroy the fractions first up.

 

[PC]

Yeah, that's the standard textbook or teacher method. If you can do it, that's the best way, but many students seems to stuff up that first step somewhere.