सारणिक विधि से हल कीजिये:
`x+2y=7`
`4x-3y +5=0`

5x+2y=4 7x+3y=5

3x-4y=-7,5x-2y=0

Find the orthocenter of the triangle formed by the lines x + 2y = 0, 4x + 3y - 5 =0, 3x + y = 0 .

Three straight lines 2x+11y - 5 = 0 , 24 x + 7y - 20 = 0 and 4x - 3y - 2 = 0

5x+7y+2=0 4x+6y+3=0

Solve for x and y, using substitution method : 2x + y = 7, 4x - 3y + 1 =0

The equation of straight line belonging to both the families of lines (x-y+1)+lambda_1(2x-y-2)=0 and (5x+3y-2)+lambda_2(3x-y-4)=0 where lambda_1, lambda_2 are arbitrary numbers is (A) 5x -2y -7=0 (B) 2x+ 5y - 7= 0 (C) 5x + 2y -7 =0 (D) 2x- 5y- 7= 0

The equation of straight line belonging to both the families of lines (x-y+1)+lambda_1(2x-y-2)=0 and (5x+3y-2)+lambda_2(3x-y-4)=0 where lambda_1, lambda_2 are arbitrary numbers is (A) 5x -2y -7=0 (B) 2x+ 5y - 7= 0 (C) 5x + 2y -7 =0 (D) 2x- 5y- 7= 0

The equation of straight line belonging to both the families of lines (x-y+1)+lambda_1(2x-y-2)=0 and (5x+3y-2)+lambda_2(3x-y-4)=0 where lambda_1, lambda_2 are arbitrary numbers is (A) 5x -2y -7=0 (B) 2x+ 5y - 7= 0 (C) 5x + 2y -7 =0 (D) 2x- 5y- 7= 0