is that (y*2)? if yes the answer will be:
2-6y^2
y-4y-2-3y-2y+4y+25y+2y-4y+23y+2y+4y-25y-2
Solving for y: 5y = 9 + 2y 5y - 2y = 9 3y = 9 y = 9/3 = 3
-2y+y = -y
1/3 - 2/(2y + 1) = 3/y Provided that y ≠-1/2 and y ≠0, multiply through by 3y(2y+1) to give y(2y + 1) - 6y = 9(2y+1) that is, 2y2 + y - 6y = 18y + 9 or 2y2 - 23y - 9 = 0 The quadratic formula then gives y = [23 ± sqrt{232 - 4*2*(-9)}]/(2*2) = [23 ± sqrt(601)]/4 so that y = -0.3788 or y = 11.8788
To solve the system of equations given by ( x + 7y = 39 ) and ( 3x - 2y = 2 ) using substitution, we can first solve the first equation for ( x ): ( x = 39 - 7y ). Substituting this into the second equation gives ( 3(39 - 7y) - 2y = 2 ). Simplifying this results in ( 117 - 21y - 2y = 2 ), or ( 117 - 23y = 2 ), leading to ( 23y = 115 ) and ( y = 5 ). Plugging ( y = 5 ) back into ( x = 39 - 7(5) ) gives ( x = 4 ). Thus, the solution in ordered pair form is ( (4, 5) ).
Solving by the elimination method: x = 7 and y = 2
Solving by the elimination method gives: x = 3 and y = 2
1
If: 3x-2y = 1 and 3x^2-2y^2+5 = 0 Then by rearranging: 3x = 2y+1 and y^2-2y-8 = 0 Solving the above quadratic equation: y = 4 or y = -2 Bu substitution points of intersection are: (3, 4) and (-1, -2)
2y
If: x-2y = 8 and xy = 24 Then: x = 8+2y and so (8+2y)y = 24 => 8y+2y2-24 = 0 Solving the quadratic equation: y = -6 or y = 2 By substitution points of intersection: (-4, -6) and (12, 2)
-2y = 10 2y = -10 y = -5