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The pair of linear equations x + 2y = 5, 3x + 12y = 10 has –
- a)Unique solution
- b)No solution
- c)More than two solutions
- d)Infinitely many solutions
Correct answer is option 'A'. Can you explain this answer?
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The pair of linear equations x + 2y = 5, 3x + 12y = 10 has –a)Un...
No solution.
To see why, let's first put the equations in standard form:
x - 2y = 5
3x - 12y = 10
Now we can use elimination to try to solve for x and y. Let's multiply the first equation by 3 to get rid of the x coefficient:
3x - 6y = 15
3x - 12y = 10
Now we can subtract the first equation from the second:
0x - 6y = -5
Simplifying, we get:
y = 5/6
But if we try to substitute this value of y into the first equation, we get:
x - 2(5/6) = 5
x - 5/3 = 5
x = 20/3
However, if we substitute these values of x and y into the second equation, we get:
3(20/3) - 12(5/6) = 10
20 - 10 = 10
This is a contradiction, since we started with two equations that were supposed to be true at the same time. Therefore, there is no solution to this system of equations.
To see why, let's first put the equations in standard form:
x - 2y = 5
3x - 12y = 10
Now we can use elimination to try to solve for x and y. Let's multiply the first equation by 3 to get rid of the x coefficient:
3x - 6y = 15
3x - 12y = 10
Now we can subtract the first equation from the second:
0x - 6y = -5
Simplifying, we get:
y = 5/6
But if we try to substitute this value of y into the first equation, we get:
x - 2(5/6) = 5
x - 5/3 = 5
x = 20/3
However, if we substitute these values of x and y into the second equation, we get:
3(20/3) - 12(5/6) = 10
20 - 10 = 10
This is a contradiction, since we started with two equations that were supposed to be true at the same time. Therefore, there is no solution to this system of equations.
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The pair of linear equations x + 2y = 5, 3x + 12y = 10 has –a)Unique solutionb)No solutionc)More than two solutionsd)Infinitely many solutionsCorrect answer is option 'A'. Can you explain this answer?
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