+10 Question 1: Will and Grace are canoeing on a lake. Will rows at $50$ meters per minute and Grace rows at $20$ meters per minute. Will starts rowing at $2$ p.m. from the west end of the lake, and Grace starts rowing from the east end of the lake at $2{:}05$ p.m. If they always row directly towards each other, and the lake is $3400$ meters across from the west side of the lake to the east side, at what time will the two meet?
Question 2: A plane flies from Penthaven to Jackson and then back to Penthaven. When there is no wind, the round trip takes $4$ hours and $40$ minutes, but when there is a wind blowing from Penthaven to Jackson at $50$ miles per hour, the trip takes $4$ hours and $48$ minutes. How many miles is the distance from Penthaven to Jackson?
Question 3: Find all values of $x$ such that $\sqrt{4x^2} + \sqrt{x^2} = 6$.
Question 4: For a certain value of $k$, the system \(\begin{align*} x + y + 3z &= 10, \\ 4x + 2y + 5z &= 7, \\ kx + z &= 3 \end{align*}\) has no solutions. What is this value of $k$?
+208 For number 1) Will started 5 minutes before Grace, so 50*5=250 meters
So they have 3400-250=3150 meters when they start approaching each other
.
They approach at 70m/minute, so 3150/70 = 45
So then they meet at 2:50 p.m
+208 The answer for number 1 is 2:50 pm my bad for not higlighting
+208 For number 2)
Say plane speed is P mph
Distance is D miles.
Time for round trip is: D/P + D/P = 2D/P
4hr 40min = 14/3 hr
2D/P = 14/3
D/P = 7/3
=> P = 3D/7
Time taken (with wind) = D/(P+50) + D/(P-50)
Time taken: 4hr 48min = 24/5 hrs
=> D/(P+50) + D/(P-50) = 24/5
5D/(P+50) + 5D/(P-50) = 24
(5DP-250D)/(P^2-2500) + (5DP+250D)/(P^2-2500) = 24
10DP/(P^2-2500) = 24
10DP = 24P^2-24*2500
5DP = 12P^2-12*2500
P = 3D/7
=> 5D*3D/7 = 12*(9D^2/49 -2500)
Multiply 49 on both sides of the equation:
5D*21D = 108D^2-2500*12*49
105D^2 = 108D^2-2500*12*49
3D^2 = 2500*12*49
D^2 = 2500*4*49
D = 50*2*7 = 700
So the distance from Penthaven to Jackson is 700 miles
+208 For number 3)
\(\sqrt{4x^2}+\sqrt{x^2}=6\)
\(|2x|+|x|=6\)
Say 2x+x = 6
3x = 6
x = 2
Then there is the other case:
2x+x = -6
3x = -6
x = -2
So the values of x would be -2 and 2
+208 For number 4)
\(x+y+3z=10\)
\(4x+2y+5z=7\)
\(kx+z=3\)
Multiply 2 on both sides of the 1st equation:
\(2x+2y+6z=20\)
Subtract it to the 2nd equation:
\(2x-z=-13\)
\(kx+z=3\)
multiply -1 to \(kx+z=3\) on both sides:
\(-kx-z=-3\)
in order for this system to have no solutions
k has to be -2 which gets us: \(2x-z=-3\)
This equation can't have 2 values since it is linear,
so this proof tells us, k = 2
