# Intermediate Algebra question.

The sum of two integers is twenty-four. The difference between four times the smaller integer and nine is three less than twice the larger integer. Find the integers.

Let the two integers be x and y. Let x be smaller than y. so x + y = 24. Deriving the second equation is more tricky. 4x-9 = 2y-3 OR 9-4x = 2y-3 You'l have to try both. Or have you already done that and need help processing the two equations?
ianmoth
23 November 2011
let the integers in question be called x and y for the time being. Let x be the smaller of the two. Now we can write in terms of x and y the three pieces of information given.
mrtmaths
27 November 2011
The sum of the two integers is 24, we therefore say x+y=24 The difference between four times the smaller integer and nine equates to 4x-9 Three les than twice the larger integer simply means 2y-3
mrtmaths
27 November 2011
We can now look at the first equation and make x the subject by subtracting y from both sides.......x = 24-y..........We can now substitute this value of x into 4x-9.......so this becomes 4(24-y)-9.......when we expand brackets we arrive at 87-4y
mrtmaths
27 November 2011
BUT! we have an equation in the question....4x - 9 = 2y - 3........but we legitimately changed 4x-9 to 87-4y......so we can now write our equation with only one variable instead of a more complicated two variable equation..........87-4y = 2y - 3 .....adding three to both sides give 90 - 4y = 2y.......adding 4y to both sides gives 90 = 6y......dividing both by 6 gives y = 15....
mrtmaths
27 November 2011
Now go back to the first equation x + y = 24 and sub in y = 15 .....x + 15 = 24....subtract 15 from both sides ......x = 9. I think that works! x=9 and y = 15
mrtmaths
27 November 2011