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In the given questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer.
I. 3x2 – 18x + 15 = 0
II. 4y2 – 84y + 320 = 0
x > y
x ≤ y
x ≥ y
x < y
x = y or relation between x and y can not be determined
Let’s break it down step by step:
- First, solve the first equation: 3x² – 18x + 15 = 0
Divide by 3: x² – 6x + 5 = 0
Factor: (x – 1)(x – 5) = 0, so x = 1 or x = 5
- Now the second equation: 4y² – 84y + 320 = 0
Divide by 4: y² – 21y + 80 = 0
Factor: (y – 16)(y – 5) = 0, so y = 16 or y = 5
Here’s what these numbers tell us:
- Possible x values: 1, 5
- Possible y values: 5, 16
Now, compare all possible pairings:
- If x = 1, y can be 5 or 16 ? x < y
- If x = 5, y can be 5 or 16 ? x = y or x < y
That means x is always less than or equal to y.
So, here’s the bottom line:
Option 2: x = y is correct.
.
For completeness:
- Option 1 (x > y): Never occurs.
- Option 2 (x = y): Always true.
- Option 3 (x = y): Not always true.
- Option 4 (x < y): Not always true (because x = y is possible).
- Option 5 (relation undetermined): We’ve shown the relationship is defined.
Hope that clears it up!
By: Parvesh Mehta ProfileResourcesReport error
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