[x+y=3],[3x-82y=4]

Solve the following system of equations by matrix method : 5x+7y+2=0 4x+6y+3=0 5x+2y=3 3x+2y=5 3x+4y-5=0 x-y+3=0 3x+y=19 3x-y=23 3x+7y=4 x+2y=-1 3x+y=7 5x+3y=12

The straight lines x+y=0, 3x+y=4 and x+3y=4 form a triangle which is

If [[x-1,2y],[x+y,4]]=[[3x-7,y^2-3],[6,4]] then [[x,y]]=

If [[x+3,4],[y-4,x+y]]=[[5,4],[3,9]] find x and y .

If [[x+3,4],[y-4,x+y]]=[[5,4],[3,9]], then find x and y

Find the values of 'x' and 'y' when : [(x+2y,3y),(4x,2)]=[(0,-3),(8,2)]

The system of equations 3x+2y+z=6,3x+4y+3z=14 and 6x+10y+82=a ,has infinite number of solutions,if a is equal to

The system of equations 3x+2y+z=6,3x+4y+3z=14 and 6x+10y+82=a ,has infinite number of solutions,if a is equal to

1/(3x+y)+1/(3x-y)=3/4 and 1/(3x+y)-1/(3x-y)= -1/4