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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)

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Q: What are the solutions to the simultaneous equations of x -2y equals 1 and 3xy -y2 equals 8?
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Related questions

What are the solutions to the simultaneous equations of x -2y equals 1 and 3xy -y squared equals 8?

They are: (3, 1) and (-11/5, -8/5)


How do you solve the simultaneous equations of x -2y equals 1 and 3xy -y squared equals 8?

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)


What are the solutions to the simultaneous equations of x -2y equals 1 and 3xy -y squared equals 8 showing all appropiate work with answers?

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


What are the solutions to the simultaneous equations of 3xy -y squared equals 8 and x -2y equals 1?

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)


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