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.

1. 3.4/4

2. 0.8/7

3. 0.5 x 3

4. 0.9 x 6

5. 10.8

6. 2.35

7. 0.567

8. 9.35

9. 0.02

10. 0.06

11. 0.16

12. 0.42

13. 0.63

14. 5.6

15. 8.1

16. 9.9

17. 0.28

18. 0.48

19. 0.27

20. 0.49

21. 0.45

22. 0.72

23. 0.9

24. 0.88

25. 0.75

26. 0.16

27. 0.328

28. 0.495

29. 0.72

30. 1.068

the equivalent expression are always the hardest ones but you look at the gcf also

step-by-step explanation: