Mrs. julien’s and mrs. castillejo’s classes are selling cookie dough for a school fundraiser. customers can buy packages of macadamia nut chip cookie dough and packages of triple chocolate cookie dough. mrs. julien’s class sold 25 packages of macadamia nut chip cookie dough and 30 packages of triple chocolate cookie dough for a total of $221.25. mrs. castillejo’s class sold 5 packages of macadamia nut chip cookie dough and 45 packages of triple chocolate cookie dough for a total of $191.25. (a) write the system of equations that model the problem. be sure to explain which equation represents which situation. (b) find the cost of each type of cookie. show your work. (c) explain which method you used to solve the system and why you chose that method.

Part (a): System of equations:

(1) 25m + 30c = 221.25 ↔ Mrs. Julien’s class sold 25 packages of macadamia nut chip cookie dough and 30 packages of triple chocolate cookie dough for a total of $221.25

(2) 5m + 45 c = 191.25 ↔ Castillejo’s class sold 5 packages of macadamia nut chip cookie dough and 45 packages of triple chocolate cookie dough for a total of $191.25

Part (b): Cost of each type of cookie:

- Macadamia nut chip cookie doug (m): $4.33 (the package)

- Triple chocolate cookie dough (c): $ 3.77 (the package)

Part (c): The method

The method used is elimination.

Reason: it was very convenient to eliminate one variable, by using the addtion property of the equalities (full explanation below).

Step-by-step explanation:

This kind of problems is known as word problems and are aimed to develop algebraic skills.

The first thing to do is to translate the word statements into mathematical (algebraic) statements, using numbers and letters (variables).

These are the steps in this case:

1) Variable names:

a)  Number of packages of macadamia nut chip cookie dough: m

b) Number of packages of triple chocolate cookie dough: c

2) First word statement:

- Mrs. Julien’s class sold 25 packages of macadamia nut chip cookie dough and 30 packages of triple chocolate cookie dough for a total of $221.25:

- Equivalent algebraic statement: 25m + 30c = 221.25

That is the first equation

3) Second word statement:

- Castillejo’s class sold 5 packages of macadamia nut chip cookie dough and 45 packages of triple chocolate cookie dough for a total of $191.25

- Equivalent algebraic statement: 5m + 45 c = 191.25.

With that we have answered the part (a). The system of equations is:

(1) 25m + 30c = 221.25

(2) 5m + 45 c = 191.25

4) Part (b) Find the cost of each type of cookie.

This means that you have to solve the system, i.e. find the values of the variables m and c.

a) Multiply the equation (2) by 5 ⇒ 25m + 225c = 956.25

b) Rewrite the system:

(1) 25m + 30c = 221.25

(3) 25m + 225c = 956.25

c) Subtract equation (1) from equation (3):

195c = 735

⇒ c = 735 / 195 = 3.77

d) Substitute the value of c in the equation (1) and solve for m:

25m + 30c = 221.25

25m + 30(3.77) = 221.25

25m + 113.1 = 221.25

25m = 221.25 - 113.1

25m = 108.15

m = 108.15 / 25

m = 4.33

Hence, the costs of each type of cookie are:

- Macadamia nut chip cookie doug (m): $4.33 (the package)

- Triple chocolate cookie dough (c): $ 3.77 (the package)

5) Part (c): Explain which method you used to solve the system and why you chose that method.

The method used is named elimination.

This method uses the addition property to eliminate one variable and solve an equation with one variable.

To use this method, you usually multiply one of the equations by a suited coefficient in order to it can be eliminated when the two equations are added (addition property of the equalities).

I chose this method because, normally, the solution involves easy steps to calculate the values of the variables.

1. f(x)=x²+10x+16

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=1(as, a> 0 the parabola is open upward), b=10. by putting the values.

-b/2a = -10/2(1) = -5

f(-b/2a)= f(-5)= (-5)²+10(-5)+16= -9

so, vertex = (-5, -9)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+16, we get point (0,16).

now find x-intercept put y=0 in the above equation. 0= x²+10x+16

x²+10x+16=0 ⇒x²+8x+2x+16=0 ⇒x(x+8)+2(x+8)=0 ⇒(x+8)(x+2)=0 ⇒x=-8 , x=-2

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

2. f(x)=−(x−3)(x+1)

by multiplying the factors, the general form is f(x)= -x²+2x+3.

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(as, a< 0 the parabola is open downward), b=2. by putting the values.

-b/2a = -2/2(-1) = 1

f(-b/2a)= f(1)=-(1)²+2(1)+3= 4

so, vertex = (1, 4)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+3, we get point (0, 3).

now find x-intercept put y=0 in the above equation. 0= -x²+2x+3.

-x²+2x+3=0 the factor form is already given in the question so, ⇒-(x-3)(x+1)=0 ⇒x=3 , x=-1

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

3. f(x)= −x²+4

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(as, a< 0 the parabola is open downward), b=0. by putting the values.

-b/2a = -0/2(-1) = 0

f(-b/2a)= f(0)= −(0)²+4 =4

so, vertex = (0, 4)

now, find y- intercept put x=0 in the above equation. f(0)= −(0)²+4, we get point (0, 4).

now find x-intercept put y=0 in the above equation. 0= −x²+4

−x²+4=0 ⇒-(x²-4)=0 ⇒ -(x-2)(x+2)=0 ⇒x=2 , x=-2

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

4. f(x)=2x²+16x+30

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=2(as, a> 0 the parabola is open upward), b=16. by putting the values.

-b/2a = -16/2(2) = -4

f(-b/2a)= f(-4)= 2(-4)²+16(-4)+30 = -2

so, vertex = (-4, -2)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+30, we get point (0, 30).

now find x-intercept put y=0 in the above equation. 0=2x²+16x+30

2x²+16x+30=0 ⇒2(x²+8x+15)=0 ⇒x²+8x+15=0 ⇒x²+5x+3x+15=0 ⇒x(x+5)+3(x+5)=0 ⇒(x+5)(x+3)=0 ⇒x=-5 , x= -3

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

5. y=(x+2)²+4

the general form of parabola is y=a(x-h)²+k , where vertex = (h,k)

if a> 0 parabola is opened upward.

if a< 0 parabola is opened downward.

compare the given equation with general form of parabola.

-h=2 ⇒h=-2

k=4

so, vertex= (-2, 4)

as, a=1 which is greater than 0 so parabola is opened upward and the graph has minimum.

hope this : )

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step-by-step explanation: