SOLUTION: If 5 men and 3 boys can reap 23 acres in 4 days and 3 men and 2 boys can reap 7 acres in 2 days , how many boys must assist 7 men in order that they may reap 45 acres in 6 days?

Let m be (an unknown) rate of work for one man, measured in , 
and let b be (an unknown) rate of work for one boy, measured in same units.

Then we have a system of two linear equations in two unknowns m and b

4*(5m + 3b) = 23,   (1)
2*(3m + 2b) =  7.   (2)

Or

20m + 12b = 23,     (1')
 6m +  4b =  7.     (2')

To solve it, multiply (2') by 3 and then distract it from (1'). You will get

20m - 18m = 23 - 3*7, or

2m = 2, hence, m = 1. 

Substitute it into (2'), and you will get 4b = 7 - 6 = 1, hence, b = .

Thus you obtained that 1 man can reap 1 acr in 1 day, while 1 boy can reap  of acr in 1 day.

Then 7 men in 6 days can reap 7*6*1 = 42 acres.

How many boys must assist to reap remaining 45-42 = 3 acres acres in 6 days?

Their number is  =  = 2.

Answer. Two boys.