SOLUTION: > a guy has 120grams worth of candy. If the ten gram sweets are half the number of five gram sweets how many sweets does he have in total? >solve the simultaneous equation I) (

Question 1062128: > a guy has 120grams worth of candy. If the ten gram sweets are half the number of five gram sweets how many sweets does he have in total?
>solve the simultaneous equation
I) (1/x)+(1/y)=7/10
II) (3/x)-(5/y)=1/2
Found 3 solutions by Boreal, josgarithmetic, MathTherapy:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
10 gm sweets are x
5 gm sweets are 2x
10x+5(2x)=120
20x=120
x=6
2x=12
He has 18 sweets, 6 of them 10 gm and 12 of them 5 gm. They add up to 120 gm.
---------------------------
(1/x)+(1/y)=7/10
(3/x)-(5/y)=1/2
Multiply everything by 10xy to clear fractions for the first and 2 xy for the second
10y+10x=7xy
6y-10x=xy
add them
16y=8xy
divide by y
16=8x
x=2
substitute into first
(1/2)+(1/y)=7/10
(1/y)=2/10=1/5y
y=5
Check in the second
3/2-5/5=1/2, and 1.5-1=0.5
(2,5)
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
NUMBER of sweets or candy pieces

x, number of pieces 10 grams each
y, number of pieces 5 grams each

---------Can you understand this much? You need to before continuing.

Substitute for x...

and.
.
.
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Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
> a guy has 120grams worth of candy. If the ten gram sweets are half the number of five gram sweets how many sweets does he have in total?
>solve the simultaneous equation
I) (1/x)+(1/y)=7/10
II) (3/x)-(5/y)=1/2

Let number of 10-gram sweets be T
Then number of 5-gram sweets = 2T
We then get: 10T + 5(2T) = 120
10T + 10T = 120
20T = 120
T, or number of 10-gram sweets =
From this, you should be able to determine how many 5-gram sweets there are, and the total number. You shouldn't need my help for that.
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