You're a manager in a company that produces rocket ships. Machine A and Machine B both produce cockpits and propulsion systems. Machine A and Machine B produce cockpits at the same rate, and they produce propulsion systems at the same rate. Machine A ran for 26 hours and produced 4 cockpits and 6 propulsion systems. Machine B ran for 56 hours and produced 8 cockpits and 12 propulsion systems. We use a system of linear equations in two variables. Can we solve for a unique amount of time that it takes each machine to produce a cockpit and to produce a propulsion system?

For Machine A, we have 4 cockpits; this would take 4c time. We also have 6 propulsion systems; this would take 6p time. Together it takes 26 hours; this would give us

4c+6p=26.

For Machine B, we have 8 cockpits; this would take 8c time. We also have 12 propulsion systems; this would take 12p time. Together it takes 56 hours; this would give us

8c+12p=56

This gives us the system

[tex]\left \{ {{4c+6p=26} \atop {8c+12p=56}} \right.[/tex]

To solve this, we want the coefficients of one of the variables to be the same. To make the coefficients of c the same, we can multiply the top equation by 2:

[tex]\left \{ {{2(4c+6p=26)} \atop {8c+12p=56}} \right. \\\\\left \{ {{8c+12p=52} \atop {8c+12p=56}} \right.[/tex]

To cancel c, we will subtract the second equation from the first one. However, this also cancels p:

[tex]\left \{ {{8c+12p=52} \atop {-(8c+12p=56)}} \right. \\\\0+0=-4\\0=-4[/tex]

We get a statement that is not true; thus the system has no solution, and we cannot solve for a unique amount of time.

wifilethbridge wifilethbridge · Answer 2 · 2019-02-13T12:34:08+01:00

Answer:

Step-by-step explanation:

Given :

Machine A ran for 26 hours and produced 4 cockpits and 6 propulsion systems.

Machine B ran for 56 hours and produced 8 cockpits and 12 propulsion systems.

To Find : Can we solve for a unique amount of time that it takes each machine to produce a cockpit and to produce a propulsion system?

Solution :

Let x be the time taken to produce 1 cockpits

So, time taken to produce 4 cockpits = 4x

So, time taken to produce 8 cockpits = 8x

Let y be the time taken to produce 1 Propulsion

So. time taken to produce 6 propulsion = 6y

So. time taken to produce 12 propulsion = 12y

Now we are given that Machine A ran for 26 hours and produced 4 cockpits and 6 propulsion systems.

⇒[tex]4x+6y =26[/tex] --a

We are also given that Machine B ran for 56 hours and produced 8 cockpits and 12 propulsion systems.

⇒[tex]8x+12y =56[/tex] --b

Solving equation a and b by substitution method

Substitute the value of x from equation in b

⇒[tex]8(\frac{26-6y}{4}) +12y =56[/tex]

⇒[tex]52-12y+12y =56[/tex]

⇒[tex]52 =56[/tex]

We get a statement that is not true; thus the system has no solution, and we cannot solve for a unique amount of time.