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A company has a $150 budget to provide lunch for its 20 employees. The options are to provide either roast beef sandwiches, which cost $5 apiece, or tuna sandwiches, which also cost $5 apiece. The company also wants to use the entire budget. Suppose r represents the number of roast beef sandwiches it provides and t represents the number of tuna sandwiches. Which statement is correct?
The company can provide lunch for all 20 employees and use the entire budget because there is a solution to the system of equations r minus t = 20 and 5 r + 5 t = 150.
The company can provide lunch for all 20 employees and use the entire budget because there is a solution to the system of equations r + t = 20 and 5 r + 5 t = 150.
The company cannot provide lunch for all 20 employees and use the entire budget because there is no solution to the system of equations r minus t = 20 and 5 r + 5 t = 150.
The company cannot provide lunch for all 20 employees and use the entire budget because there is no solution to the system of equations r + t = 20 and 5 r + 5 t = 150.

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Answer:

D.) The company cannot provide lunch for all 20 employees and use the entire budget because there is no solution to the system of equations r+t=20 and 5r+5t=150

Step-by-step explanation:

The given information says that the total amount of lunch bought should equal $150 when both options cost $5:

[tex]5r+5t=150[/tex]

It also says that the food should feed all 20 employees:

[tex]r+t=20[/tex]

This is now a system. Solve by substitution.

Solve the second equation for r. Use inverse operations to isolate the variable by subtracting t from both sides:

[tex]r+t-t=20-t\\\\r=20-t[/tex]

Now insert this value of r into the first equation:

[tex]5(20-t)+5t=150[/tex]

Simplify the equation. Use the distributive property:

[tex]5(20)+5(-t)+5t=150\\\\100-5t+5t=150[/tex]

Cancel the terms:

[tex]100\neq 150[/tex]

100 does not equal 150, so there is no solution to the system.

:Done

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