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?
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?
I d'nt understand this part
so 5J + 6G = 150
and 5J + 5G = 140
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?
Thanks so much
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.
To solve this problem, we can set up a system of equations.
Let's represent the number of jellybeans as 'j' and the number of gums as 'g'. We are given the following information:
1. Julia can buy 3 jellybeans for 5 cents, so the cost of each jellybean is 5/3 cents.
2. Julia can buy a piece of gum for 2 cents.
3. Julia has spent a total of 50 cents.
4. Julia has a total of 28 items.
Based on this information, we can set up the following equations:
Equation 1: (5/3)j + 2g = 50
This equation represents the cost of the jellybeans and gums that Julia bought, which must equal 50 cents.
Equation 2: j + g = 28
This equation represents the total number of jellybeans and gums that Julia bought, which must equal 28 items.
To solve this system of equations, we have a few options. One method is to isolate one variable in one equation and substitute it into the other equation.
Solving for j in Equation 2, we get:
j = 28 - g
Substituting this expression for j in Equation 1, we get:
(5/3)(28 - g) + 2g = 50
Next, we can solve this equation for g to find the value of g. Once we have g, we can substitute it back into Equation 2 to find the value of j.
By solving these equations, we should be able to determine how many of each kind of candy Julia has. I will calculate the solution for you.