What is the value of k for which the system of equations x 2y 3 and 3x Ky 1 has a unique solution?

The given system of equations:x + 2y = 3⇒ x + 2y - 3 = 0                           ….(i)And, 5x + ky + 7 = 0                   …(ii)These equations are of the following form:`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`where, `a_1 = 1, b_1= 2, c_1= -3 and a_2 = 5, b_2 = k, c_2 = 7`(i) For a unique solution, we must have:∴ `(a_1)/(a_2) ≠ (b_1)/(b_2) i.e., 1/5 ≠ 2/k ⇒ k ≠ 10`Thus for all real values of k other than 10, the given system of equations will have a unique solution.(ii) In order that the given system of equations has no solution, we must have:`(a_1)/(a_2) = (b_1)/(b_2 )≠ (c_1)/(c_2)``⇒ 1/5 ≠ 2/k ≠ (−3)/7``⇒ 1/5 ≠ 2/k and 2/k ≠ (−3)/7``⇒k = 10, k ≠ 14/(−3)`Hence, the required value of k is 10.

There is no value of k for which the given system of equations has an infinite number of solutions.

On comparing the ratios  

 and without drawing them, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincide.

5x - 4y + 8 = 0, 7x + 6y - 9 = 0

Comparing the given equations with standard forms of equations a1x + b1y + c1, = 0 and a2x + b2y + c2

 = 0 we have,

Thus, the lines representing the pair of linear equations are intersecting.


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