You can extend the techniques you have learned for solving
systems of two equations to solve systems of three equations.
Students are baking shortbread, croissants, and pound cakes.
• One group of students used a total of 3,525 grams of flour to make
5 batches of shortbread, 4 pound cakes, and 2 batches of croissants.
• Another group of students used 1,020 grams of flour to make one
batch each of shortbread and croissants and one pound cake.
• There is the same amount of flour in 4 pound cakes as there is
in 2 batches of croissants.
1. Let s = the amount of flour in a batch of shortbread, let c = the
amount of flour in a batch of croissants, and let p = the amount
of flour in a pound cake. Write an equation for each of the
3 statements above.
2. Multiply one equation, then add or subtract it with another
equation you wrote in Exercise 1 to eliminate the variable p.
3. Use the third equation and your result from Exercise 2 to write a
system of two equations with the variables p and c. Solve your system
and find the values for p and c.
4. What is the value of s? Explain what you did.
PLEASE HELP GIVE FULL DETAIL I TRIED BUT I WAS WRONG HE CHECKED IT! HELp PLEASE!!
1. Let's break down the information given and write an equation for each statement:
First statement: One group of students used a total of 3,525 grams of flour to make 5 batches of shortbread, 4 pound cakes, and 2 batches of croissants.
We can represent this statement as:
5s + 4p + 2c = 3,525
Second statement: Another group of students used 1,020 grams of flour to make one batch each of shortbread and croissants and one pound cake.
We can represent this statement as:
s + p + c = 1,020
Third statement: There is the same amount of flour in 4 pound cakes as there is in 2 batches of croissants.
We can represent this statement as:
4p = 2c
2. To eliminate the variable p, we can multiply the third equation by 2 so that the coefficients of p in both the second and third equations become the same. Let's do that:
2 * (4p) = 2 * (2c)
8p = 4c
Now, we can subtract this equation from the second equation to eliminate p:
s + p + c - 8p + 4c = 1,020
s - 7p + 5c = 1,020
3. We now have a system of two equations:
Equation 1: 5s + 4p + 2c = 3,525
Equation 2: s - 7p + 5c = 1,020
Let's solve this system to find the values for p and c.
Multiply Equation 2 by 4 to eliminate s:
4s - 28p + 20c = 4,080
Now, add this equation to Equation 1:
5s + 4p + 2c + 4s - 28p + 20c = 3,525 + 4,080
9s - 24p + 22c = 7,605
We now have a system of two equations:
Equation 1: 9s - 24p + 22c = 7,605
Equation 2: s - 7p + 5c = 1,020
Using these two equations, we can solve for p and c.
4. Since we have solved for p and c in the previous step, we can substitute these values back into any of the original equations to find the value of s. Let's use the second equation:
s + p + c = 1,020
Substituting the values we found for p and c, we have:
s + 255 + 204 = 1,020
s + 459 = 1,020
Now, let's solve for s:
s = 1,020 - 459
s = 561
So, the value of s is 561.
1. Let's write the equations based on the given information:
First statement: One group of students used a total of 3,525 grams of flour to make 5 batches of shortbread, 4 pound cakes, and 2 batches of croissants.
The amount of flour in 5 batches of shortbread is 5s.
The amount of flour in 4 pound cakes is 4p.
The amount of flour in 2 batches of croissants is 2c.
So, the equation for the first statement is: 5s + 4p + 2c = 3525.
Second statement: Another group of students used 1,020 grams of flour to make one batch each of shortbread and croissants and one pound cake.
The amount of flour in one batch of shortbread is s.
The amount of flour in one batch of croissants is c.
The amount of flour in one pound cake is p.
So, the equation for the second statement is: s + c + p = 1020.
Third statement: There is the same amount of flour in 4 pound cakes as there is in 2 batches of croissants.
The amount of flour in 4 pound cakes is 4p.
The amount of flour in 2 batches of croissants is 2c.
So, the equation for the third statement is: 4p = 2c.
2. We can multiply the third equation by 2 to get 8p = 4c.
Now, we can subtract the second equation (s + c + p = 1020) from the first equation (5s + 4p + 2c = 3525) to eliminate the variable p.
(5s + 4p + 2c) - (s + c + p) = 3525 - 1020
4s + 3p + c = 2505
3. We can substitute the value of c from the third statement (4p = 2c) into the equation we obtained by eliminating p.
4s + 3p + (4p/2) = 2505
4s + 3p + 2p = 2505
4s + 5p = 2505
Now we have a system of two equations:
s + c + p = 1020 ---(Equation A)
4s + 5p = 2505 ---(Equation B)
We can solve this system using any method like substitution, elimination, or matrix method.
Let's solve the system using the substitution method. Solve Equation A for s in terms of c and p:
s = 1020 - c - p
Substitute this value of s in Equation B:
4(1020 - c - p) + 5p = 2505
4080 - 4c - 4p + 5p = 2505
4080 - 4c + p = 2505
p = 1425 - 4080 + 4c
p = -2655 + 4c ---(Equation C)
Now, we have two equations: Equation B and Equation C.
4s + 5p = 2505 ---(Equation B)
p = -2655 + 4c ---(Equation C)
4. To find the value of s, we need to substitute the values of p and c into either Equation A or Equation C.
Let's use Equation C to substitute p:
p = -2655 + 4c
Substitute this value of p in Equation A:
s + c + (-2655 + 4c) = 1020
s - c + 3c = 1020 + 2655
s + 2c = 3675
s = 3675 - 2c
Now, we have the value of s in terms of c. It means that s depends on c. Therefore, without the value of c, we cannot find the exact value of s.