`[3x+2y=5],[2x-3y=7]`

{:(3x + 2y = 4),(2x - 3y = 7):}

The equation of a line which is parallel to the line common to the pair of lines given by 6x^2-x y-12 y^2=0 and 15 x^2+14 x y-8y^2=0 and at a distance of 7 units from it is 3x-4y=-35 5x-2y=7 3x+4y=35 2x-3y=7

The equation of a line which is parallel to the line common to the pair of lines given by 6x^2-x y-12 y^2=0 and 15 x^2+14 x y-8y^2=0 and at a distance of 7 units from it is 3x-4y=-35 5x-2y=7 3x+4y=35 2x-3y=7

Simplify: [5-3x+2y-(2x-y)]-(3x-7y+9)

Given the four lines with the equations x+2y-3=0 , 3x+4y-7=0 , 2x+3y-4=0 , 4x+5y-6=0 , then

If [[y+2x,5],[-x,3]]=[[7, 5],[-2 ,3]], find the value of y

The equation 2x+y=5, x+3y =5, x-2y=0 have :

Solve : {:(x+y=7xy),(2x-3y=-xy):}

The lines 2x-3y=5 and 3x-4y=7 are diameters of a circle of are 154 sq. units. Taking pi=22/7 , show that the equation of the circle is x^(2)+y^(2)-2x+2y=47