The total number of? restaurant-purchased meals that the average person will eat in a? restaurant, in a? car, or at home in a year is 163. The total number of these meals eaten in a car or at home exceeds the number eaten in a restaurant by 13. Twenty more? restaurant-purchased meals will be eaten in a restaurant than at home. Find the number of? restaurant-purchased meals eaten in a? restaurant, the number eaten in a? car, and the number eaten at home.

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Answer:

The number of restaurant-purchased meals eaten in a restaurant is 75, the number eaten in a car is 33 and the number eaten at home is 55.

Step-by-step explanation:

Suppose, the number of meals eaten in a restaurant is [tex]x[/tex] , in a car is [tex]y[/tex] and at home is [tex]z[/tex]

The total number of meals eaten in a restaurant, in a car or at home is given as 163. So, the first equation will be......

[tex]x+y+z=163 .............................(1)[/tex]

The total number meals eaten in a car or at home exceeds the number eaten in a restaurant by 13. So, the second equation will be......

[tex]y+z=x+13 ................................(2)[/tex]

Twenty more restaurant-purchased meals will be eaten in a restaurant than at home. So, the third equation will be......

[tex]x= z+20..................................(3)[/tex]

First, substituting equation (2) into equation (1), we will get........

[tex]x+(x+13)=163\\ \\ \Rightarrow 2x+13=163\\ \\ \Rightarrow 2x=163-13=150\\ \\ \Rightarrow x=\frac{150}{2}=75[/tex]

Now plugging this [tex]x=75[/tex] into equation (3), we will get......

[tex]75=z+20\\ \\ \Rightarrow z=75-20= 55[/tex]

Plugging [tex]x=75[/tex] and [tex]z=55[/tex] into equation (1) .........

[tex]75+y+55=163\\ \\ \Rightarrow y+130=163\\ \\ \Rightarrow y=163-130=33[/tex]

So, the number of restaurant-purchased meals eaten in a restaurant is 75, the number eaten in a car is 33 and the number eaten at home is 55.

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