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QUESTION OF THE DAY
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Question of the Day (12Apr17)
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00
:
00
What is the number of solution sets (
x
,
y
) for the equation
x
^{2}

y
 − 10
x

y
 = −9, given that
x
and
y
are both integers?
OPTIONS
1)
0
2)
2
3)
4
4)
8
5)
10
Discuss
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Solution
We need to find integral solutions for
x
^{2}

y
 − 10
x

y
 = −9
∴ 
y
 × (
x
^{2}
− 10
x
) = −9
Now, the ways to factorize −9 into integers are: 1 × (−9), (−1) × 9 and (−3) × 3.
∴ The first term 
y
 and the second term (
x
^{2}
− 10
x
) must take on these values in turn. Thus we have six cases.
Case 1:

y
 = 1 and (
x
^{2}
− 10
x
) = −9
This gives 2 values for
y
as +1 and −1. We then have to solve
x
^{2}
− 10
x
+ 9 = 0
This gives
x
= 9 and
x
= 1 as the two solutions.
We have got 4 solution sets from this case: (9, 1), (9, −1), (1, 1) and (1, −1).
Case 2:

y
 = −9 and (
x
^{2}
− 10
x
) = 1
∵ 
y
 cannot be negative.
∴ We reject this case.
Case 3:

y
 = −1 and (
x
^{2}
− 10
x
) = 9
∵ 
y
 cannot be negative.
∴ We reject this case.
Case 4:

y
 = 9 and (
x
^{2}
− 10
x
) = −1
This gives 2 values for
y
as +9 and −9. We then have to solve
x
^{2}
− 10
x
+ 1 = 0
This does not give an integral solution for
x
. So this case does not give us a valid solution set.
Case 5:

y
 = −3 and (
x
^{2}
− 10
x
) = 3
∵ 
y
 cannot be negative.
∴ We reject this case.
Case 6:

y
 = 3 and (
x
^{2}
− 10
x
) = −3
This gives 2 values for
y
as +3 and −3. We then have to solve
x
^{2}
− 10
x
+ 3 = 0
This does not give an integral solution for
x
. So this case does not give us a valid solution set.
∴ The total number of solution sets are just those we got from case 1. They are 4 in number.
Hence, option 3.
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