Agreater number is seven less than five times a smaller number. if there sum is 41, find the numbers

Let's assume x= greater number and y= smallest number.

Given, a greater number is seven less than five times a smaller number. Which can be transformed into an algebraic equation as follows:

x= 5y-7 ..... (1)

Now the sum of two numbers is 41. So,

x+y= 41 ...(2)

Now we can plug in x=5y-7 in equation (2). So,

5y-7+y=41

6y-7=41

6y-7+7=41+7 Add 7 to each sides of the equation.

6y= 48

Divide 6 to each sides.

So, y=8

Next step is to plug in y=8 in equation (1). So,

x=5(8)-7

=40-7

=33.

So, the numbers are 8 and 33.

they are simplified

step-by-step explanation: