can someone help me with these algebra equations?

1)
(x + 2)/3 = 8/15
x + 2 = 3(8/15)
x + 2 = 8/5
x = 8/5 - 2
5x = 5(8/5 - 2)
5x = 8 - 10
x = -2/5 (-0.4)

2)
18/x = 6
18 = 6(x)
6x = 18
x = 18/6
x = 3

3)
5/7 = 10/(x + 2)
5(x + 2)/7 = 10
5*x + 5*2 = 7(10)
5x + 10 = 70
5x = 70 - 10
5x = 60
x = 60/5
x = 12

1) (x + 2)/ 3 = 8 /15
(x + 2)..... = 8 * 3 / 15 -----( 3 goes in right side so that we multiply)
x + 2 ......= 8 / 5 ( 3 divide by 15 = 5)
x..............= 8/5 - 2 ...... (2 goes in right side and we sub)
x..............= -2/ 5 .........( you calculate by using factorization or LCM method)

2) 18/ x = 6
6x = 18 ( x goes in another direction so we multiply)
x = 18/ 6 ( 6 goes in another direction so we divide)
x = 3 ( 18 divide by 6 = 3)


3) 5/7 = 10/x + 2
5/7 - 2 = 10/x
-9 /7 .....= 10/x ( you factorise )
- 9x/7 .....= 10 (x goes in another direction)
- 9x ........= 10*7
x ...........= -70 / 9

i hope it helps you to understand or remember them now!! :)

Just remember that whenever you're doing this sort of algebra question, the 1st thing you need to do is to try removing the fractions (that's what my maths teacher says :-D)

1) (x+2)/3 = 8/15

To remove the fractions you'll have to find the LCM (lowest common multiple) for all the denominators, in this case it's 15:

5(x+2)/3*5 = 8/15
(5x+10)/15 = 8/15

Now you got the same denominators, you can simply remove them to get some nice whole numbers:

5x+10 = 8
5x = -2
x = -2/5


2) 18/x = 6
(18*x)/x = 6*x
18 = 6x
x = 3


3) 5/7 = 10/x+2
[5(x+2)]/[7(x+2)] = [10(7)]/[7(x+2)]
5x+10 = 70
5x = 60
x = 12

one way to start is to get rid of the fractions

for #1, multiply both sides by 15
#2, multiply both sides by x
#3, multiply both sides by 7(x+2)