Question 242805: One number is eleven more than twice another. If their sum is decreased by nine, the result is seventeen. Find the numbers Found 2 solutions by unlockmath, oberobic:Answer by unlockmath(1688) (Show Source):
You can put this solution on YOUR website! Hello,
Let's establish one number represented by x. The other number will be 2x+11 Then we can add these and subtract 9 to equal 17. Here's how it will look.
[(2x+11) + x] -9 = 17 Now we can combine like terms.
3x+2=17 Subtract 2 from both sides.
3x=15 Divide 3 into both sides.
x = 5
RJ Toftness
www.math-unlock.com
You can put this solution on YOUR website! In the setup to the problem, always try to minimize the unknowns.
x = one number
It is 11 more than twice another number.
y = another number
x = 2y + 11
...
Now the second part says if the sum of x + y is decreased by 9 (that is: x + y - 9), the result is 17.
x + y - 9 = 17
Add 9 to both sides
x + y = 17+9 = 26
...
Now we have to solve x and y
...
Subtracting y from both sides of the equation above.
x + y - y = 26 - y
x = 26 - y
...
Substituting in the other equation.
x = 2y + 11
26 - y = 2y + 11
Adding y to both sides
26 = 3y + 11
Subtracting 11 from both sides
15 = 3y
Dividing both sides by 3
5 = y
...
y = 5
...
Recall
x + y = 26
Substituting for y
x + 5 = 26
Subtracting 5 from both sides
x = 21
Check your work (even if you're tired)...
Does x = 2y + 11?
21 = 2(5) + 11?
Yes it does.
Does x + y - 9 = 17?
21 + 5 - 9 = 17?
Yes it does.
So we're done.
x = 21
y = 5
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