Let x be the number of child tickets he bought
Let y be the number of adult tickers he bought
① x+y=7 (child tickets+adult ticket=7 tickets in total)
② 2x+4y=24 (price of child tickets+price of adult tickets=$24 in total)
We may simply the second equation since all of the coefficients are divisible by 2.
① x+y=7
② x+2y=12
We can now use elimination by multiplying the second equation by -1.
② -(x+2y=12)
② -x-2y=-12
① x+y=7
② -x-2y=-12
Now putting the equations together,
-y=-5
y=5
x=2
Therefore he bought 2 child tickets and 5 adult tickets
Answer:
uhhhh what grade level is dis?
Step-by-step explanation:
Answer:
Infinite pairs of numbers
1 and -1
8 and -8
Step-by-step explanation:
Let x³ and y³ be any two real numbers. If the sum of their cube roots is zero, then the following must be true:
![\sqrt[3]{x^3}+ \sqrt[3]{y^3}=0\\ \sqrt[3]{x^3}=- \sqrt[3]{y^3}\\x=-y](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E3%7D%2B%20%5Csqrt%5B3%5D%7By%5E3%7D%3D0%5C%5C%20%5Csqrt%5B3%5D%7Bx%5E3%7D%3D-%20%5Csqrt%5B3%5D%7By%5E3%7D%5C%5Cx%3D-y)
Therefore, any pair of numbers with same absolute value but different signs fit the description, which means that there are infinite pairs of possible numbers.
Examples: 1 and -1; 8 and -8; 27 and -27.
Answer:
um, u gotta ask a question