70+ pages kx 2y 5 3x y 1 1.5mb. 3x y - 1 0. A_1a_2 b_1b_2 ie k3 24 k 32 Thus for all real values of k other than 32 the given system. For this we must have. Read also solution and understand more manual guide in kx 2y 5 3x y 1 This happens iff their slopes are equal but y-intercepts are different.
Prev Question Next Question. The given equations are of the following form.
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A 1 x b 1 y c 1 0 and a 2 x b 2 y c 2 0. Asked Apr 27 in Linear Equations by Gargi01 251k points pair of linear equations in two variables. Kx 2y 3 0 and 3x y 5 0. A 1 x b 1 y c 1 0. To ask Unlimited Maths doubts download Doubtnut from - httpsgoogl9WZjCW For what value of k do the equations kx-2y 3 and 3x y 5 represent two. Thus for these graph lines to intersect at a unique point we must have.
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The given system of equations is. It is of the form a_1x b_1y c_1 0 a_2x b_2y c_2 0 Where a_1 k b_1 2 c_1 -5 And a_2 3 b_2 1 c_2 -1 Where a_1 k b_1 2 c_1 -5 And a_2 3 b_2 1 c_2 -1 1 The given system will have a unique solution if a_1a_2 b_1b_2 k3 21 k 6 So the given system of equations will have a unique. The system of these two equations has no solution means the lines in the plane corresponding to the equations are distinct and parallel.
Here is all you have to to learn about kx 2y 5 3x y 1 Mathbeginvmatrix k 2 3 1 endvmat. Hence the graph lines will intersect at all real values of k except 6. And are the equivalent lines in slope intercept form. Kx 2y 5 3x y 1.
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