Question

Find the value of k , if (-1,1) is a solution of the equation (k+2)x-(3 k-2)y-3=0

Answer 1

Given:-

(-1,1) = (x,y) and

(k+2)x-(3 k-2)y-3=0

SOLUTION:-

here (-1,1) is solution of equation

=> (k + 2)x - (3k - 2)y - 3 = 0

that means (x,y) = (-1,1)

•°• x = -1 and y = 1

substituting these values in the given equation

•°• (k + 2) × -1 - (3k - 2) × 1 - 3 = 0

-k - 2 - (3k - 2) - 3 = 0

-k - 2 - 3k + 2 - 3 = 0

-4k - 3 = 0

4k = - 3

k = -3/4

The value of k is -3/4

More information:-

The above equation ia linear equation in two variable.

An equation is said to be linear equation in two variables if it is written in the form of ax+by+c=0, where a,b & c are real numbers and the coefficients of x and y, i.e a and b respectively, are not equal to zero.