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)
no
3xy
There are only the two terms, so 8xy
They are: (3, 1) and (-11/5, -8/5)
x - 2y = 1 → x = 1 + 2y 3xy - y² = 8 Substitute the first equation into the second equation and subtract 8 from both sides: 3(1 +2y)y - y² = 0 3y +6y² - y² - 8 = 0 5y² + 3y - 8 = 0 When factored: (5y +8)(y -1) = 0 Therefore: y = -⁸/₅ or y = 1 Substituting the above values into the linear equation gives the solutions as: (3, 1) and (-¹¹/₈, -⁸/₅)
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
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)
no
No, it is not.
yes 3xy=6
2x^2+3xy-4y2(4)+3(2)(-4)-(4)(-4)8-24+16=0
3xy - 3xy = 0
(x + 6y)(x - 3y)
Whooop!
Assuming the two equations are: 3xy - 3 = 0 ............................................. 1 2xy = 0 ............................................. 2 Then from eq 1 xy = 1 from eq 2 xy = 0 That is not possible. Or, assuming the equation is 3xy - 3 + 2xy = 0 Then 5xy = 3 or y = 3/5x Substitute for any value of x to get the value of y