Question

solving systems of linear equations by elimination
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.

part a
explain what you should do with the two equations to eliminate one of the variables.

Explanation:

To eliminate one variable (either \(x\) or \(y\)) in the system \(

$$\begin{cases}x + 3y=76\\x + y = 36\end{cases}$$

\), we can use the elimination method. Since the coefficient of \(x\) is \(1\) in both equations, we can subtract the second equation from the first equation (or vice - versa). Subtracting the second equation (\(x + y=36\)) from the first equation (\(x + 3y = 76\)) will eliminate the \(x\) variable. The operation would be \((x + 3y)-(x + y)=76 - 36\). This works because when we subtract the two equations, the \(x\) terms (\(x-x\)) will cancel out, leaving an equation only in terms of \(y\). Alternatively, we could also multiply one of the equations by \(- 1\) and then add the two equations together. For example, multiply the second equation \(x + y=36\) by \(-1\) to get \(-x - y=-36\), and then add it to the first equation \(x + 3y=76\). When we add \((x + 3y)+(-x - y)=76+(-36)\), the \(x\) terms (\(x+(-x)\)) will cancel out.

Answer:

To eliminate a variable (e.g., \(x\)), we can subtract the second equation \(x + y = 36\) from the first equation \(x+3y = 76\) (or multiply the second equation by \(- 1\) and add it to the first equation). This will cancel the \(x\) - terms, leaving an equation in terms of \(y\) (e.g., \((x + 3y)-(x + y)=76 - 36\) or \((x + 3y)+(-x - y)=76-36\)).