Answer: 3 1/2
Hope this helps:)
Answer:
B.
Step-by-step explanation:
DON'T LISTEN TO THE ANSWERS ON HERE, THEY'RE WRONG!!!!
1. If the total baggage fees were equal, you would have to equate the two situations. Suppose x is the weight above the limit on the flight home, while y is the weight above the limit on the flight back to college. Then, the equation would be:
20 + 8x = 26 + 5y
Now, we have two unknowns, but only one independent equation. This is unsolvable unless we make an assumption. Suppose the excess weight for both trips are the same such that x=y, then,
<em>20 + 8x = 26 + 5x</em>
2. Then, we solve for x.
8x - 5x = 26 - 20
3x = 6
x = 6/3 = <em>2 lb</em>
Thus, Elizabeth's bag was 2 lb over the weight limit.
3. The total baggage fee would then be:
20+8(2) = 26 + 5(2) = <em>$36</em>
Answer: (x,y) = (7,-3)
Step-by-step explanation:
-4x - 10y = 2.........(1)
-6x - 10y = -12.........(2)
Subtract (2) from (1)
-4x — (-6x) -10y — (-10y) = 2 — (-12)
-4x + 6x + 0 = 2 + 12
2x = 14
x = 14/2
x = 7.
Substitute 7 for x in (1)
-4(7) - 10y = 2
-28 — 10y = 2
-10y = 2 + 28
-10y = 30
y = 30/-10
y = -3.
(x,y) = (7,-3)
Hope this helps?