Question 248028: Find the greatest possible pair of integers such that one integer is twice the other and their sum is less than 30.
I tried x+2x<30
3x<30
x<10
but I know that is not correct and I can not understand what I am forgetting to do in my calculation, please help, thank you
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! You have two integers, say, x and y.
Let y = 2x, which means it is twice the value of x.
We are told to solve x + y < 30.
So we start with the equivalent equation:
x + 2x = 30
x = 10
y = 2x = 20
.
So the two largest integers that solve the equivalent equation x+y = 30 are x=10 and y=20.
BUT
We are solving an inequality, not an equation. So, looking back, we see that x+y = 30 is NOT an acceptable answer. The requirement is that are x+y < 30.
.
Since we are still dealing only with integers, we can subtract 1 from x, which makes x = 9.
Then we can find y = 2x = 18
.
So, our proposed answer is:
x =9
y = 18
.
Checking our work to make sure x+y < 30...
x + y = 9 + 18 = 27 < 30. Which is fine.
.
Note that there is an intuitive value in the 27 in that we subtracted 1 from x, so we subtracted 2*1 from y, or 3 in total from 30.
.
A non-algebraic, brute force approach could be used with a problem like this. Simply set up table and plug in values to find the maximum value of x + 2x that is less than 30. BUT if you have to show your algebra teacher your work, the "brute force" approach will not earn much, if any, credit.
.
Done.
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