Question

The pair of equations x+y-40=0 and x-2y+14=0 have​

Answer 1

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Answer 2

Answer:

There will be a unique solution for this equations.

Step-by-step explanation:

Given that x + y -40 = 0 and x - 2y +14 = 0

The general equations are a1x+b1y+c1 = 0, and a2x+b2y+c2 =0

x + y -40 = 0  ----------(1)

a1 = 1 , b1 = 1, c1 = -40

x - 2y + 14 = 0 -----------------(2)

a2 = 1, b2 = -2, c2 = 14.

From the given equations we have,

$\frac{a1}{a2} = \frac{1}{1}=1$

$\frac{b1}{b2} = \frac{1}{-2} =-\frac{1}{2}$

$\frac{c1}{c2} = \frac{-40}{-14} = \frac{20}{7}$

Therefore, $\frac{a1}{a2}$ ≠ $\frac{b1}{b2}$

Therefore, there will be a unique solution for this equations.

Your question is incomplete, but most probably your full question is:

Question: "The pair of equations x+y-40 = 0 and x-2y+14 = 0 have at

(A) unique solution (B) Exactly two solutions (c) Infinitely many solutions (d) No solution."

That is option(A) is correct.