no
If: x-2y = 1 then x = 2y+1 If: 3xy -y^2 = 8 then 3*(2y+1)*y -y^2 = 8 So: 6y^2 +3y -y^2 = 8 or 5y^2 +3y -8 = 0 Factorizing the above: (y-1)(5y+8) = 0 meaning y = 1 or y = -8/5 Solutions by substitution are: (3, 1) and (-11/5, -8/5)
The two equations are: 1) 3xy - y² = 8 2) x - 2y = 1 Make x the subject of (2): x - 2y = 1 → x = 2y + 1 substitute for x in (1) and solve for y: 3xy - y² = 8 → 3(2y + 1)y - y² = 8 → 6y + 3y - y² - 8 = 0 → 5y + 3y - 8 = 0 → (5y + 8)(y - 1) = 0 → 5y + 8 = 0 → y = -8/5 or y - 1 = 0 → y = 1 and substitute into 2 to find the corresponding x: y = -8/5 → x = 2(-8/5) + 1 = -11/5 y = 1 → x = 2(1) + 1 = 3 → the solutions are the ordered pairs (-11/5, -8/5) {or (-2.2, -1.6)} and (3, 1)
3xy
f(x) = 2*(x-3)*(x+2)/(x-1) for x ≠1
no
(x + 6y)(x - 3y)
x = -1 y = 2 2x3 - 3xy = 2 (-1)3 - 3(-1)(2) = 2 (-1) - (-6) = -2 + 6 = 4 Suggestion: Please be careful around problems like this until you have some more experience.
2(2)2 + 3(2)(- 4) - 4(- 4)2 8 - 24 - 64 = - 80 ======
It is the same as: 12y^2 -3xy
-3xy is a monomial where x and y are variables and -3 is the coefficient of the monomial.If we know the value of y, we have to substitute it for y, and we say that -3(y-value) is the coefficient of x.
1,3,x,y,3x,3y,xy,3xy
They are: (3, 1) and (-11/5, -8/5)
3xy + 3y2 = 3y (x + y)
If: x-2y = 1 then x = 2y+1 If: 3xy -y^2 = 8 then 3*(2y+1)*y -y^2 = 8 So: 6y^2 +3y -y^2 = 8 or 5y^2 +3y -8 = 0 Factorizing the above: (y-1)(5y+8) = 0 meaning y = 1 or y = -8/5 Solutions by substitution are: (3, 1) and (-11/5, -8/5)
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)
x = 2 so x squared = 4 and 2x squared = 2 times 4 = 8; y = 4 so xy = 2 times 4 = 8 and 3xy = 3 times 8 = 24; y squared = 4 times 4 = 16 and 4y squared = 4 times 16 = 64; so 8 + 24 - 64 = 32 - 64 = -32