40,000 + 0.07(120,000) =
40,000 + 8,400 =
$ 48,400 per year <===
Answer would be B. Sorry if my handwriting is messy or if I didn’t explain enough
The solution of the system can be x - 3y = 4 only if both the equations can be simplified to x - 3y = 4.
This will mean that both the equations will result in the same line which is x - 3y = 4 and thus have infinitely many solutions.
Second equation is:
Qx - 6y = 8
Taking 2 common we get:
(Q/2)x - 3y = 4
Comparing this equation to x- 3y = 4, we can say that
Q/2 = 1
So,
Q = 2
Therefore, the second equation will be:
2x - 6y = 8
A)
Let x represent the cost of 1 student, and y the cost of 1 teacher.
B)
In the first group, there's 25 students and 2 teachers. Their total cost is $97.50
So 25x + 2y = 97.50
In the second group, there's 32 students and 3 teachers. Their total cost is $127
So 32x + 3y = 127
We get the following system of equations:
25x + 2y = 97.50 (1)
32x + 3y = 127 (2)
C)
25x + 2y = 97.50 (1)
32x + 3y = 127 (2)
In equation (1)
25x + 2y = 97.50
25x + 2y - 2y = 97.50 - 2y
25x = 97.50 - 2y
25x / 25 = 97.50/25 - 2y/25
x = 3.9 - (2/25)y
In equation (2), let's replace x by its algebraic value
32x + 3y = 127
32(-2/25y + 3.9) + 3y = 127
11/25y + 124.8 = 127
11/25y + 124.8 - 124.8 = 127 - 124.8
11/25y = 2.2
(11/25y) / (11/25) = 2.2 / (11/25)
y = 5
x = -2/25y + 3.9
x = -2/25 * 5 + 3.9
x = 3.5
So the cost of each student is $3.5, and the cost of each teacher is $5.
Hope this helps! :)
