`2x+y=7` and `3x-2y=0`

Solve:(1)/(2(2x+3y))+(12)/(7(3x-2y))=(1)/(2)(7)/(2x+3y)+(4)/(3x-2y)=2 where 2x+3y!=0 and 3x-2y!=0

Equation of circle through (5,0) , two of whose diameters are x + 2y = 7 and 3x - y = 0 , is

Find the greatest value of x^(3)y^(4) if 2x+3y=7 and x>=0,y>=0 .

Find the equation of the line passing through the point of intersection of the lines x+5y+7=0 and 3x+2y-5=0 (a) parallel to the line 7x+2y-5=0

Find the equation of the line passing through the point of intersection of the lines x+5y+7=0 and 3x+2y-5=0 (b) perpendicular to the line 7x + 2y - 5 = 0

The area of the parallelogram formed by the lines 2x-3y+a=0, 3x-2y-a=0, 2x-3y+3a=0 and 3x-2y-2a=0 in square units , is

If the equations of the diagonals of the parallelogram formed by the lines,2x-y+7=0;2x-y-5=0;3x+2y-5=0 and 3x+2y+4=0 are ax+by-5=0 and px+qy+1=0. Find the value of (a+b+p+q)

If the lines 2x+3y+1=0 and 3x-y-4=0 lie along diameters of a circle of circumference 10 pi ,then the equation of the circle is (A) x^(2)+y^(2)-2x+2y-23=0 (B) x^(2)+y^(2)+2x-2y-23=0 (C) x^(2)+y^(2)+2x+2y-23=0 (D) x^(2)+y^(2)-2x-2y-23=0

The equation x+2y=3,y-2x=1 and 7x-6y+a=0 are consisten for