Can someone please help me with these 2 questions?

[ally1784]

Hi, can someone help me please?

[p0kem0n21]

Hi, can someone help me please?

Hey again. I'd love to solve the questions for you, but I'd doubt that would be super helpful for you. So I ask, where exactly are you struggling with these questions? Are you struggling to formulate the necessary equations (I certainly did with the first one haha)? Or is there some other issue? This helps us know where exactly you need help.

[Danzorr]

Hi, can someone help me please?

To do these questions, it primarily focuses on the setting of variables and then producing a simultaneous equation to find answer.

For the first question, you want to begin by reading the whole question and seeing all the information you can gain from it. Then you would note some variables. I set x = cheaper ticket price and y = expensive ticket price. Then you would want to produce a relationship between these two variables. From reading the question, it says that the more expensive ticket price would be $30 more than the cheaper one, so using the variables of x and y which was set before, we get:

y = x + 30
x = y - 30

Now, continuing onto the next part of the equation, it provides more vital information, stating that the group can purchase 10 more of the cheaper tickets than the expensive tickets for $1800. However, to create the simultaneous equation, we're missing one more variable, which is the number of tickets purchased. So we have to create a new variable being, m = number of tickets purchased.

Now, it is possible to create the simultaneous equation to answer the question.

For the first equation, we can begin with the cheaper ticket. We would do (m + 10)*x = 1800. We would add 10 to the m as we are dealing with the cheaper ticket, and it states that we are able to buy 10 more of the cheaper tickets than the expensive tickets at $1800.
For the second equation, we have m*y = 1800. However, we want to make it so there are only two variables in the equation, so we substitute y = x + 30 from earlier. The two equations we now have are

(m + 10)*x = 1800
m*(x+30) = 1800

Now, we just have to solve these two equations which can be done by CAS or by hand. Once we have solved, we end up having two sets of solutions, one being x = -90 and m = -30, and the other one being; x = 60 and m = 20. The number of tickets purchased nor the price of the tickets can't be negative, so the answer would have to be x = 60 and m = 20.. So the cost of the cheaper ticket is $60. Now we just need to substitute in $60 to y = x + 30  and we would end up with y = 90. So the cost of the more expensive ticket would be $90.

Then we can just check our answer by doing 20*90 = $1800 and 30*60 = 1800. We got 20 from m = 20 from earlier.

I'd recommend trying the second question on your own first and if you come across any issues, just reply and I can provide you with assistance.