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3 years ago

Solve this system of linear equations. Separate

Mathematics

1 answer:

alex41 [277]3 years ago

5 0

Answer:

y = 7 and x =3

Step-by-step explanation:

use elimination method

so we gonna eliminate y first

9x = 1 - 4y

-7x = -7 + 4y

7| 9x = 1 - 4y

9| -7x = -7 + 4y

63x = 7 - 28y

-63x = -63 + 36y

add eqtn 1 to eqtn2

you will get

0 = -56 + 8y

56 = -56 + 56 + 8y

56 = 8y

56÷8 = 8y÷ 8

7 = y

also eliminate x as we have done wth y

4| 9x = 1 - 4y

4| -7x = -7 + 4y

36x = 4 - 16y

-28x = -28 + 16y

add the two equations

u will get

8x = -24 + 0

8x = -24

8x÷8 = 24÷ 8

x = 3

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Answer:

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The total number of people at a football game was 5600. Field-side tickets were 40 dollars and end-zone tickets were 20 dollars.

rewona [7]

Answer:

1100 field-side tickets and 4500 end-zone tickets.

Step-by-step explanation:

Let x represent number of field side tickets and y represent number of end-zone tickets.

We have been given that the total number of people at a football game was 5600. We can represent this information in an equation as:

$x+y=5600...(1)$

$y=5600-x...(1)$

We are also told that Field-side tickets were 40 dollars and end-zone tickets were 20 dollars.

Cost of x field side tickets would be $40x$ and cost of y end-zone tickets would be $20y$ .

The total amount of money received for the tickets was $134000. We can represent this information in an equation as:

$40x+20y=134000...(2)$

Upon substituting equation (1) in equation (2), we will get:

$40x+20(5600-x)=134000$

$40x+112000-20x=134000$

$20x+112000=134000$

$20x+112000-112000=134000-112000$

$20x=22000$

$\frac{20x}{20}=\frac{22000}{20}$

$x=1100$

Therefore, 1100 field side tickets were sold.

Upon substituting $x=1100$ in equation (1), we will get:

$y=5600-1100$

$y=4500$

Therefore, 4500 end-zone tickets were sold.

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