Find two consecutive even integers such that twice the larger is 14 less
than 5 times the smaller.

help me

You can set up a system of equations with x being the smaller number and y being the larger number. So the first equation would be 2y=5x-14 and the second equation would be x=y-2. You can solve by substitution and substitute y-2 into the x in the first equation to get 2y=5(y-2)-14. You can solve this out and it ends up simplifying to y=8. So you know the larger number is 8. So since the larger number is 8 and they are consecutive even integers, the smaller number must be 6. So the two numbers would be 6 and 8.