177k views
5 votes
In a theater, there are 4 long rows of chairs,each with an equal number of chairs There are 5 ahort rows of chairs,each with an equal number of chairs On a crowded day, the ushers added 2 chairs to each long row and 3 chairs to each short row if x represents the number of chairs in each long row and y represents the number of chairs in each short row,the total number of chairs in the theater is given by the expression 2)Sly 3) An equivalent expression is

User Thymo
by
6.4k points

2 Answers

7 votes

Answer:

An equivalent expression is 4x + 5y + 23

Explanation:

There are 4 long rows of chairs and 3 short rows of chairs.

Let x be the number of chairs in long rows and y in short rows.

So total number of chairs in theater

C = 4x + 5y

Now on a crowded day 2 chairs were added in long rows so number of chairs in long rows becomes

= 4 (x + 2)

Similarly ushers add 3 chairs add in short rows so total number of chairs in short row = 5 ( y + 3 )

Now total chairs in theater

C' = 4 ( x + 2 ) + 5 ( y + 3 )

= 4x + 8 + 5y + 15

= 4x + 5y + 23

C' = C + 23 [Since C = ( 4x + 5y ) ]

Number of chairs added were 23 as compared to normal days.

User Yevgeniy
by
6.4k points
7 votes
There are 4 long rows and 5 short rows in the theater. x represents the number of chairs in each long row and y represents the number of chairs in each short row.

So, total number of chairs in 4 long rows= 4x
Total number of chairs in 5 short rows = 5y
Total number of chairs in the theater on a normal day = 4x + 5y

When 2 chairs are added to each long row, the number of chairs will change to (x+2).
So, total number of chairs in 4 long rows will be = 4(x+2)

When 3 chairs are added to each short row, the number of chairs will change to (y+3)
So, total number of chairs in 5 short rows will be = 5(y+3)

Thus, total number of chairs in the theater in rush day = 4(x+2) + 5(y+3)
= 4x + 8 + 5y + 15
= 4x + 5y + 23

Thus we can say the number of chairs increase by 23 as compared to a normal day.
User Paulo Rosa
by
6.7k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.