Question 1154026
Let current speed be x mph
Let keith's speed in still water be y mph

Distance travelled in 1 hour = 3 miles
Travelling with current, effective speed is (y + x) mph
So the distance travelled = speed * time
= (y + x) * 1 = (y + x) = 3
Travelling against current, his effective speed is (y - x) mph
At (y - x) mph, he took 1.5 hours for the same distance i.e. 3 miles

Since speed * time = distance, we get the equation

(y - x) * 1.5 = y + x = 3
Which can be broken up as 2 equations in x and y as below

{{{x + y = 3}}} --- (1)
{{{1.5y - 1.5x = 3}}} --- (2)

Multiplying (1) by 1.5
{{{1.5x + 1.5y = 4.5}}} --- (3)

Adding (2) and (3)
{{{3y = 7.5}}}

i.e.
{{{y = 2.5}}}
{{{x = 3 - y = 0.5}}}

Speed in still water = 2.5 mph
Speed of current = 0.5 mph

Check.
With current, effective speed = 2.5 + 0.5 = 3 mph, so time taken to travel 3 miles = 3 / 3 = 1 hour (check)
Against current, effective speed = 2.5 - 0.5 = 2 mph, so time taken to travel 3 miles = 3 / 2 = 1.5 hours (check)

Hope this helps!