中一我maths問題,,,,方程的問題

Let X be the money Susan has, 4X be the money Helen has.

X + 4X = $40
5X = $40
X = $8

Therefore, Susan has $8, Helen has 4x8=$32


Let Y be the money Michael has, 3Y be the money Joyce has.

3Y - 30 - (6x3) = $12
3Y - 30 - 18 = $12
3Y = 60
Y= 20

Michael had $20

1.
Set Helen has $x and Susan has $y.

Then x+y=40 (a) and x=4y (b).

State x in terms of y and put it into equation (a), 4y+y=5y=40.
So y=40/5=8 and x=4y=4x8=32.

Therefore, Helen has $32 while Susan has $8.

2.
Set Joyce had $x and Michael had $y.

Then x=3y (a) and
x-30-6*3=12.(b)

In equation (b), x-30-6*3=x-30-18=x-48=12 or x=12+48=60

Substitute x into equation (a), 60=3y or y=60/3=20

So Michael had $20.



By the way, some correction for your typing in your question as follow:

"S"usan, ho"w"

Let Helen has $4x and Susan has $x.

4x+x=40
x=8

so, Helen has $32 and Susan has $8.

Let Joyce had $3x and Michael had $x.
3x-30-6*3=12
x=20.
so, Micheal had $20.

Let $a be the money Susan has,,
Then the money Helen has is $4a
a + 4a = 40
　　5a = 40
　　a = 8
Susan has $ 8
Helen has $ 32


Joyce had a sum of money of ( 30 + 6x3 + 12 ) = $60
Let the money Michael had = b
3b = 30 + 3x3 + 12
3b = 60
b = 20
The amout of money Michael had was $20