SOLUTION: There are 35 tickets to be sold for the dance. The number of tickets sold to seniors must be four times the number of tickets sold to juniors.
please give me 2 linear equations
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Question 252604: There are 35 tickets to be sold for the dance. The number of tickets sold to seniors must be four times the number of tickets sold to juniors.
please give me 2 linear equations and a solution.
A diamond today costs 10$ more than twice what it cost last year. The sum of the cost (last year and this year) is 2500$. What is the cost of last years diamond?
i need 2 linear equations and a solution for this question also.
THANK U SO MUCH!
Found 2 solutions by drk, jim_thompson5910:
Answer by drk(1908) (Show Source): You can put this solution on YOUR website!
Let's go 1 question at a time. Here is the first one: There are 35 tickets to be sold for the dance. The number of tickets sold to seniors must be four times the number of tickets sold to juniors.
please give me 2 linear equations and a solution.
Let S = senior, and J = junior.
---
(i):
(ii)
---
by substitution, we get
(iii) 4J + J = 35 - -> J = 7, S = 28.
---
Here is the second question: A diamond today costs 10$ more than twice what it cost last year. The sum of the cost (last year and this year) is 2500$. What is the cost of last years diamond?
i need 2 linear equations and a solution for this question also.
Let T = today and Y = last year
---
(i) T = 2Y + 10
(ii) T + Y = 2500
---
Use substitution to solve.
(2Y + 10) + Y = 2500
3Y = 2490
Y = $830
T = $1670.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
# 1
Let s = # of senior tickets and j = # of junior tickets
Since there are 35 tickets in total, and we're assuming that only junior and senior tickets are sold, this means that . In other words, adding up the two totals will give you the grand total of 35.
Also, because "The number of tickets sold to seniors must be four times the number of tickets sold to juniors", we know that . In English, the number of senior tickets 's' is 4 times the number of junior tickets 'j'.
Start with the first equation.
Plug in
Combine like terms on the left side.
Divide both sides by to isolate .
Reduce.
So there are 7 junior tickets.
Go back to the second equation
Plug in
Multiply
So there are 28 senior tickets.
As a check, take note that 28 is indeed 4 times 7 AND 28+7=35. So we've met our conditions.
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# 2
Let t = cost of diamond today and y = cost of diamond last year
Since "A diamond today costs 10$ more than twice what it cost last year", we can say that (ie double the cost from last year 'y' and add 10 to get the new cost today 't').
Also, because "The sum of the cost (last year and this year) is 2500$", we know that (just add up the two individual costs to get the grand total of $2500)
Start with the second equation.
Plug in
Combine like terms on the left side.
Subtract from both sides.
Combine like terms on the right side.
Divide both sides by to isolate .
Reduce.
So the cost of the diamond last year was $830.
If you want to keep going then...
Go back to the first equation
Plug in
Multiply
Add
So the cost of the diamond today is $1670.
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