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grandymaker [24]
3 years ago
5

A discount airline sells a certain number of tickets, x, for a flight for $90 each. It sells the number of remaining tickets, y,

for $250 each. For a particular flight, the airline sold 120 tickets and collected a total of $27,600 from the sale of those tickets. Which system of equations represents this relationship between x and y?
Mathematics
1 answer:
yKpoI14uk [10]3 years ago
7 0

Answer:

x+y= 120

90x+250y=27,600

Step-by-step explanation:

x= number of certain tickets

y= number of remaining tickets

The statement indicates that for a particular flight the airline sold 120 tickets which means that the sum of the different tickets is 120. The first equation is:

x+y= 120

Then, the statement says that certain number of tickets costs $90 and the remaining tickets cost $250 and that for a particular flight, the airline collected $27,600. From this, you can say that the number of certain tickets, x, multiply for its price that is $90 plus the number of the remaining tickets, y, multiply for its price that is $250 would be equal to $27,600. The equation would be:

90x+250y=27,600

According to this, the answer is that the system of equations that represents this relationship between x and y is:

x+y= 120

90x+250y=27,600

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Read 2 more answers
Need help with my homework ​
Volgvan

Answer:

\displaystyle y=\frac{16-9x^3}{2x^3 - 3}

\displaystyle y=-\frac{56}{13}

Step-by-step explanation:

<u>Equation Solving</u>

We are given the equation:

\displaystyle x=\sqrt[3]{\frac{3y+16}{2y+9}}

i)

To make y as a subject, we need to isolate y, that is, leaving it alone in the left side of the equation, and an expression with no y's to the right side.

We have to make it in steps like follows.

Cube both sides:

\displaystyle x^3=\left(\sqrt[3]{\frac{3y+16}{2y+9}}\right)^3

Simplify the radical with the cube:

\displaystyle x^3=\frac{3y+16}{2y+9}

Multiply by 2y+9

\displaystyle x^3(2y+9)=\frac{3y+16}{2y+9}(2y+9)

Simplify:

\displaystyle x^3(2y+9)=3y+16

Operate the parentheses:

\displaystyle x^3(2y)+x^3(9)=3y+16

\displaystyle 2x^3y+9x^3=3y+16

Subtract 3y and 9x^3:

\displaystyle 2x^3y - 3y=16-9x^3

Factor y out of the left side:

\displaystyle y(2x^3 - 3)=16-9x^3

Divide by 2x^3 - 3:

\mathbf{\displaystyle y=\frac{16-9x^3}{2x^3 - 3}}

ii) To find y when x=2, substitute:

\displaystyle y=\frac{16-9\cdot 2^3}{2\cdot 2^3 - 3}

\displaystyle y=\frac{16-9\cdot 8}{2\cdot 8 - 3}

\displaystyle y=\frac{16-72}{16- 3}

\displaystyle y=\frac{-56}{13}

\mathbf{\displaystyle y=-\frac{56}{13}}

8 0
3 years ago
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