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3x-y=2 first you subtract the 3x to the other side.
-y=-3x+2 Now you divide each side by -1
y=3x-2
2x + 5y = 2-3x - y = -3from the second equation:-y = -3 + 3xy = 3 - 3xsubstitute this value in the first equation:2x + 5(3-3x) = 22x + 15 - 15x = 2-13x + 15 = 2-13x = 2 - 15-13x = -13so, x = 1Now, solve for y :y = 3 - 3xy = 3 - 3 (1)y = zero
If: x-2y = 1 and 3xy -y^2 = 8 Then: x = 1+2y and 3(1+2y)y -y^2 = 8 Hence: 3y+6y^2 -y^2 = 8 => 5y^2 +3y -8 = 0 Solving the above quadratic equation: y =1 or y = -8/5 Solutions by substitution are: when y=1 then x=3 and when y=-8/5 then x=-11/5
2(2)2 + 3(2)(- 4) - 4(- 4)2 8 - 24 - 64 = - 80 ======
If: x -2y = 1 then x = 1+2y If: 3xy -y^2 = 8 then 3(1+2y)y -y^2 = 8 So: 3y+6y^2 -y^2 = 8 => 3y+5y^2 -8 = 0 Solving the above quadratic equation: y = 1 or y = -8/5 By substitution the points of intersection are at: (3, 1) and (-11/5, -8/5)
[ 8x - 2 = 14 ] is the equation.[ x = 2 ] is the solution to the equation.