Solve this word problem (linear systems) by elimination:
The mean attendance at the Winnipeg Folk Festival for 2006 and 2008 was 45,265.
The attendance in 2008 was 120 more than the attendance in 2006.
What was the attendance in each year?
Answer (3)
Your equation would be:
(x+(x+120))/2=45265
If you set it equal to x, your answer is
45,205
To solve the linear system, define variables for 2006 and 2008 attendance, set up two equations based on the given information, use elimination to solve for one variable, and substitute back to find the other. The solution reveals the attendance in 2006 was 45,205 and in 2008 was 45,325.
The question involves solving a linear system by elimination to find the attendance in two different years. First, let's define two variables, let x be the attendance in 2006 and y be the attendance in 2008.
According to the given information, we have two equations:
The mean (average) attendance for the two years is 45,265, so (x + y)/2 = 45,265.
The attendance in 2008 is 120 more than the attendance in 2006, so y = x + 120.
To solve by elimination, let's multiply the first equation by 2 to eliminate the denominator: x + y = 90,530. Now we have a new system:
x + y = 90,530
y = x + 120
Substitute the second equation into the first: x + (x + 120) = 90,530. Simplify this to 2x + 120 = 90,530. Subtract 120 from both sides to get 2x = 90,410. Divide by 2 and we find x = 45,205. To find y, plug x back into the second equation: y = 45,205 + 120, and we get y = 45,325.
Therefore, the attendance in 2006 was 45,205 and the attendance in 2008 was 45,325.
The attendance at the Winnipeg Folk Festival was 45,205 in 2006 and 45,325 in 2008. This was determined by setting up two equations representing the average attendance and the relationship between the two years. Using substitution, we found the values for each year.
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