Answer:
(- 1, - 2 )
Step-by-step explanation:
Given the 2 equations
x - 3y = 5 → (1)
5x - 2y = - 1 → (2)
Rearrange (1) expressing x in terms of y by adding 3y to both sides
x = 5 + 3y → (3)
Substitute x = 5 + 3y in (2)
5(5 + 3y) - 2y = - 1 ← distribute left side
25 + 15y - 2y = - 1
25 + 13y = - 1 ( subtract 25 from both sides )
13y = - 26 ( divide both sides by 13 )
y = - 2
Substitute y = - 2 in (3) for corresponding value of x
x = 5 + (3 × - 2) = 5 - 6 = - 1
Solution is (- 1, - 2 )
if those are fractions being five and two thirds etc... I believe it would be 17/6. if you convert to improper fractions and then get a common denominator to subtract that should do it. Hope that helps.
Answer:
34 classes
Step-by-step explanation:
First, subract 180 from 1200. You should get 1020. then do 1020/30. So 34 classes can watch the performance of 180 students.
A system of equations is good for a problem like this.
Let x be the number of student tickets sold
Let y be the number of adult tickets sold
x + y = 200
2x + 3y = 490
x = 200 - y
2(200 - y) + 3y = 490
400 - 2y + 3y = 490
400 + y = 490
y = 90
The number of adult tickets sold was 90.
x + 90 = 200 --> x = 110
2x + 3(90) = 490 --> 2x + 270 = 490 --> 2x = 220 --> x = 110
The number student tickets sold was 110.
A. 1
b. 1
c. 1
d. 1/2
2. 1/2
3. Yes they did break it because there are 3 wholes.
