Question

A TV Dinner Company Sells Three Types Of Dinners: A Chicken Dinner For $10, A Beef Dinner For $11, Or A Fish Dinner For $12. At One Particular Grocery Store, They Sold 200 Dinners For A Grand Total Of $2138. Required:a. If they sold three times as many chicken dinners as they did fish, then how many of each kind of dinner did they sell?b. Find the equation of the quadratic function that passes through the points (−1, 9), (2, 6), and (3, 17). Write your answer in the form y = ax2+bx+c.

Answer 1

Answer:

a) The company sold 93 Chicken dinners, 76 Beef dinners and 31 Fish dinners.

b) $y=3x^{2}-4x+2$

Step-by-step explanation:

a)

In order to solve part a of the problem, we must start by setting our variables up:

C=# of Chicken dinners

B=# of Beef dinners

F=# of Fish dinners

Next we can use these variables to build our equations. We start by taking the fact that they sold a total of 200 dinners. The sum of all the variables should add up to that so we get:

C+B+F=200

Then the problem tells us the price of each dinner and the total amount of money they made from selling the dinners that day:

"A TV Dinner Company Sells Three Types Of Dinners: A Chicken Dinner For $10, A Beef Dinner For $11, Or A Fish Dinner For $12. For A Grand Total Of $2138"

So we can use this information to build our second equation:

10C+11B+12F=2138

We have two equations now but three variables, so we need an additional equation so we can finish solving this. The problem tells us that:

"they sold three times as many chicken dinners as they did fish"

which translates to:

C=3F

so now we have enough information to finish solving the problem. We can start by substituting the last equation into the previous two equations so we get:

C+B+F=200

3F+B+F=200

4F+B=200

and

10C+11B+12F=2138

10(3F)+11B+12F=2138

30F+11B+12F=2138

42F+11B=2138

So now we can take the two bolded equations and solve them simultaneously. We can solve them by using any method we wish. Let's solve it by substitution.

We start by solving the first equation for B so we get:

B=200-4F

and now we substitute it into the second equation:

42F+11(200-4F)=2138

and now we solve for F

42F+11(200-4F)=2138

42F+2200-44F=2138

-2F=2138-2200

$F=\frac{-62}{-2}$

F=31

So now that we know the value of F, we can find the values of the rest of the variables so we can take the previous equations to figure this out:

B=200-4F

B=200-4(31)

B=200-124

B=76

and finally:

C=3F

C=3(31)

C=93

So

The company sold 93 Chicken dinners, 76 Beef dinners and 31 Fish dinners.

b) For the second part of the problem we need to build a system of equations to find the equation we are looking for. We were given three points:

(-1,9), (2,6) and (3,17)

so we need to substitute each of the points on the given quadratic equation so we get:

$9=a(-1)^{2}+b(-1)+c$

eq.1:  $9=a-b+c$

$6=a(2)^{2}+b(2)+c$

eq. 2:  $6=4a+2b+c$

$17=a(3)^{2}+b(3)+c$

eq. 3:  $17=9a+3b+c$

So now that we have our three equations we can solve them simultaneously to get the values of a, b and c by using the method you feel more comfortable with. Let's solve it by elimination:

So let's take the first equations and let's subtract them:

$9=a-b+c$

$-6=-4a-2b-c$

-------------------------------

3=-3a-3b

And now we repeat the process with the first and third equations:

$9=a-b+c$

$-17=-9a-3b-c$

-------------------------------

-8=-8a-4b

So now we have two new equations we can solve simultaneously, but they can be simplified by dividing the first one into -3 and the second one into -4 so they become:

a+b=-1

2a+b=2

We can now solve them by substitution. Let's solve the first one for a and then let's substitute it into the second equation:

a=-1-b

let's substitute

2(-1-b)+b=2

and solve for b

-2-2b+b=2

-b=4

b=-4

Now we can find a:

a=-1-b

a=-1+4

a=3

and now we can find c:

c=9-a+b

c=9-3-4

c=2

So we can use these answers to build our equation:

$y=ax^{2}+bx+c$

$y=3x^{2}-4x+2$