Given:
t1 = 3.6 h
t2 = 4.5 h
x = speed of boat
y = speed of water
Required:
a) Expression of distance traveled with moving water with 3.6h
Expression of distance traveled with moving water with 4.5h
b) Solve for y
c) Percent of boat's speed is the water current
Solution:
Working formula: distance = velocity*time
a) For travelling downstream, we get the equation
d = (x +y)*3.6
For travelling upstream, we get the equation
d = (x-y)*4.5
b) Setting the distance as equal for travelling upstream or downstream, we arrive at the equation of
(x+y)*3.6 = (x-y)*4.5
3.6x + 3.6y = 4.5x - 4.5y
8.1y =0.9x
y = x/9
c) percentage = 1/9*100% = 11.1%
<em>ANSWERS: a) d = (x+y)*36; d = (x-y)*4.5
</em> <em>b) y = x/9
</em> <em>c) 11.1%</em>
Both fractions have the same denominator so it's going to be easy. First set up your equation:
Add the whole numbers:
Now add the fractions:
add both together:
is an improper fraction so change it:
Since 6 is one more than 5, add 1 to the whole number and subtract the numerator and denominator(6 - 5 = 1) and make the remaining the new numerator. That leaves you with
Your answer is
If we add the equations it looks like
-5y + 8x + 5y + 2x = -18+58
so 10x=40
so x=40/10=4
now let's replace x by 4 in the second equation
5y +2*4=58
5y=58-2*4=58-8=50
so y=50/5=10
so (x, y) = (2, 10)