Check whether (2 root 2, 3 root 2) is a solution of 3y-2x = 1 or not
The correct answer and explanation is:
Let’s check whether the point (22,32)(2\sqrt{2}, 3\sqrt{2}) satisfies the equation 3y−2x=13y – 2x = 1.
Given:
- x=22x = 2\sqrt{2}
- y=32y = 3\sqrt{2}
- Equation: 3y−2x=13y – 2x = 1
Step-by-Step Calculation:
Substitute x=22x = 2\sqrt{2} and y=32y = 3\sqrt{2} into the equation: 3(32)−2(22)=13(3\sqrt{2}) – 2(2\sqrt{2}) = 1
Simplifying each term:
- 3(32)=923(3\sqrt{2}) = 9\sqrt{2}
- 2(22)=422(2\sqrt{2}) = 4\sqrt{2}
Now substitute these values back into the equation: 92−42=19\sqrt{2} – 4\sqrt{2} = 1
Combine like terms: (92−42)=52(9\sqrt{2} – 4\sqrt{2}) = 5\sqrt{2}
Thus, the left-hand side simplifies to 525\sqrt{2}, and we need to check if this equals 1.
Since 52≠15\sqrt{2} \neq 1 (as 525\sqrt{2} is a numerical value greater than 1), the equation does not hold true.
Conclusion:
The point (22,32)(2\sqrt{2}, 3\sqrt{2}) is not a solution to the equation 3y−2x=13y – 2x = 1.