# math

this one is kind of hard

julia can buy 3 jellybeans for 5 cent. she can buy a piece of gum for 2 cent. if julia spends 50 cent in all and gets 28 items, how much of each kind of candy does she have?

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1. Say J = no. of jellybeans
and G = no. of pieces of gum

then J + G = 28 (items)
and 5J/3 + 2G = 50 (cents)
so 5J + 6G = 150
and 5J + 5G = 140

You've now got two equations which differ by exactly 1 x G. Can you finish it off?

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2. I d'nt understand this part

so 5J + 6G = 150
and 5J + 5G = 140

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3. They come from multiplying the previous two equations by 5 and 3 respectively. What you're trying to do is get two equations that you can use to eliminate one of the two variables, so after I multiplied the equation that reads 5J/3 + 2G = 50 by 3 to clear the fraction, I then needed to multiply the equation that reads J + G = 28 by 5 to get the same multiplier for J. By good luck, that leaves me with 6G in one equation, and 5G in the other. Is that okay?

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4. Thanks so much

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5. What I *could* have done instead would be to just multiply the first equation by 2 to make the first one have the same number of Gs as the second:

J + G = 28 (items)
5J/3 + 2G = 50 (cents)

Multiplying the first one by 2 would give me:
2J + 2G = 56
5J/3 + 2G = 50
so J/3 = 6

If you solve this for J and then feed that answer back to solve for G, you should get the same answer as before.

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