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The pair of linear equations x + y + 10 = 0 and x + y – 7 = 0 has:
  • a)
    One solution
  • b)
    Infinitely many solutions
  • c)
    No solutions
  • d)
    Two solutions
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The pair of linear equations x + y + 10 = 0 and x + y – 7 = 0 ha...
We have a1, a2 the coefficients of x2,b1 and b2 coefficients of x and c1 and c2 the constant terms.So,a1a2=b1b2c1c2which is a case of parallel lines which which never meet. So there are no solutions obtainable for these equations.
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Most Upvoted Answer
The pair of linear equations x + y + 10 = 0 and x + y – 7 = 0 ha...
Understanding the Given Equations
The equations provided are:
1. \(x + y + 10 = 0\)
2. \(x + y - 7 = 0\)
To analyze the relationship between these two equations, we can rewrite them in a more familiar form.
Rearranging the Equations
- Equation 1:
\(x + y = -10\)
- Equation 2:
\(x + y = 7\)
Identifying the Nature of Solutions
Both equations represent straight lines in a 2D plane:
- Line 1: All points \((x, y)\) that satisfy \(x + y = -10\).
- Line 2: All points \((x, y)\) that satisfy \(x + y = 7\).
Analyzing the Lines
- Both lines are parallel because they have the same slope (coefficient of \(x\) and \(y\) is 1 in both cases).
- Since they have different y-intercepts (\(-10\) and \(7\)), they will never intersect.
Conclusion
Since the two lines are parallel and do not intersect, they do not share any common points. Therefore, the system of equations has:
- No solutions.
Hence, the correct answer is option 'C'.
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Community Answer
The pair of linear equations x + y + 10 = 0 and x + y – 7 = 0 ha...
Yes option C is correct as From both equation A1/A2=B1/B2but C1 ≠C2 (where A1 and A2 coffecient of X^2 B1 and B2 are the coffecient of x and C1 And C2 are constant term.) This relation is true for No solution.
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The pair of linear equations x + y + 10 = 0 and x + y – 7 = 0 has:a)One solutionb)Infinitely many solutionsc)No solutionsd)Two solutionsCorrect answer is option 'C'. Can you explain this answer?
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