Question

find the value of k for which the pair of linear equation has infinitely many solutions kx-y=2 and 6x-2y=3​

Answer 1

Answer:

i think u should check this question once .

Answer 2

Answer:

Concept:

A linear equation in 2 factors is an equation in the form ax+by+c, where a,b, and c are real values and a,b are not equal to zero. In a set of linear equations in two variables, we deal with two such equations. The solution of these equations is a point on the line in the equation. A linear equation with two variables x and y is defined as Ax + By + C = 0, where A, B, and C are real values. A pair of linear simultaneous equations variables is a set of two system of equations in the same two variables.

Given:

Find the k value for which the set of linear equations has an infinite number of solutions kx-y=2 and 6x-2y=3​

Find:

find the solution for the given question

Step-by-step explanation:

given equations:

$kx-y=2\\6x-2y=3$

can be written as

$kx-y-2=0\\6x-2y-3=0$

we know that

$a_{1}x+b_{1}y+c_{1}=0 \\a_{2}x+b_{2}y+c_{2}=0$

$\frac{a_{1} }{a_{2} } =\frac{b_{1} }{b_{2} } =\frac{c_{1} }{c_{2} }$

$a1=k\\b1=-1\\c1=-2$

$a2=b\\b2=-1\\c2=-3$

$\frac{k}{6}=\frac{-1}{-2} =\frac{-2}{-3} \\\frac{k}{6}=\frac{1}{2} =\frac{2}{3}$

$\frac{k}{6} =\frac{1}{2} ,and\frac{1}{2}=\frac{2}{3} \\k=\frac{6}{2} \\k=3$

for infinite solution

$\frac{a_{1} }{a_{2} } =\frac{b_{1} }{b_{2} } =\frac{c_{1} }{c_{2} } \\\frac{k}{b} =\frac{-1}{-2}=\frac{-2}{-3} \\\\\frac{1}{2} =\frac{2}{-3} \\\frac{1}{2} =\frac{2}{3}$

When k is not a positive number, the system has an infinite solution.​

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