Question

Each of 36 students at a school play bought either a cup of orange juice or a sandwich. A cup of orange juice costs $1 and a sandwich costs $3. The total amount collected was $76. How many students bought orange juice, and how many bought a sandwich?

Let x represent the number of students who bought a cup of orange juice and y represent the number of students who bought a sandwich. Then the problem can be represented by this system of equations:

x + 3y = 76
x + y = 36

Answer the questions to solve the problem.

1. Explain what you should do with the two equations to eliminate one of the variables. (2 points)

2. Find the value of y. (3 points)

3. Find the value of x. (2 points)

4. Interpret the solution and check the values in the system. (3 points)

Answer 1

The values of x and y in the system of equations are found to be 16 and 20, respectively. This means that 16 students bought orange juice and 20 students bought a sandwich, with the solution satisfying both equations.

Step-by-step explanation:

To solve the system of equations for the number of students who bought orange juice and sandwiches, we must first eliminate one of the variables. We can do this by subtracting the second equation from the first. This will eliminate the variable x because x minus x is 0. The modification results in a new equation leading to the determination of y.

Subtracting the second equation from the first gives us:

x + 3y = 76
x + y = 36

2y = 40

By dividing both sides by 2, we find:

y = 20

Now that we have the value of y, we can find x by substituting y back into the second original equation:

x + y = 36
x + 20 = 36

x = 16

The solution indicates that 16 students bought orange juice and 20 bought a sandwich. To check the values, we substitute x and y into the original equations:

• 16 + 3(20) = 76
• 16 + 20 = 36

Both equations are true, thus our solution is correct.

Answer 2

1. Refer to the explanation below.

2. y = 20

3. x = 16

4. 16 students bought a cup of orange juice and 20 students bought a sandwich. Both values are correct (refer to work below).

Step-by-step explanation:

To solve the given problem involving a system of linear equations, we can use the method of elimination to find the number of students who bought orange juice (represented as 'x') and the number of students who bought a sandwich (represented as 'y').

Given system of equations:

`\left\{\begin{array}{ccc}x+3y=76& \dots (1)\\x+y=36& \dots (2)\end{array}\right`

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1. What should you do to eliminate one of the variables.
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To "eliminate" one of the variables, we can manipulate the equations to have coefficients that are opposites for one of the variables and then add or subtract the equations from each other.

Equation (1) and (2) have a varible 'x' with the same coefficient of "1." We can simply subtract equation (2) from equation (1). This will eliminate the varible 'x,' making an easy equation to solve for 'y'. We will see how this works below.

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2. Finding the value of 'y'.
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We have,

`\left\{\begin{array}{ccc}x+3y=76& \dots (1)\\x+y=36& \dots (2)\end{array}\right`

Subtracting the equations gives us:

`\Longrightarrow [x+3y=76]-[x+y=36]\\\\\\\\\Longrightarrow (x-x)+(3y-y)=(76-36)\\\\\\\\\Longrightarrow 2y=40`

Dividing by 2 gives us the value of 'y':

`\Longrightarrow y=(40)/(2) \\\\\\\\\therefore \boxed{y=20}`

Thus, the value of 'y' is 20.

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3. Finding the value of 'x'.
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Now that we have the value for 'y', we can substitute it back into one of the original equations to find 'x'. Using equation (2):

`\Longrightarrow x+y=36 \ \dots (2); y=20\\\\\\\\\Longrightarrow x+20=36\\\\\\\\\Longrightarrow x=36-20\\\\\\\\\therefore \boxed{x=16}`

Thus, the value of 'x' is 16.

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4. Interpreting the solution and checking the solution.
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The solution can be interpreted as follows: 16 students bought a cup of orange juice, and 20 students bought a sandwich.

To check the solution, we substitute the values of 'x' and 'y' into both original equations:

Equation (1):

`\Longrightarrow x+3y=76 \ \dots (1)\\\\\\\\\Longrightarrow 16+3(20)=76\\\\\\\\\therefore 76 = 76 \ \checkmark`

Equation (2):

`\Longrightarrow x+y=36 \ \dots (2)\\\\\\\\\Longrightarrow 16+20=36\\\\\\\\\therefore 36=36 \ \checkmark`

The check confirms that our values of x = 16 and y = 20 are correct. Thus, our solution meets the criteria given in the problem.