2x^2+3xy-4y2(4)+3(2)(-4)-(4)(-4)8-24+16=0
yes 3xy=6
3x + 4y = 04y = -3xy = -0.75xy = mx + b with b being the y interceptSince: y = -0.75x + 0 , the y-intercept is 0.
(x + 6y)(x - 3y)
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
3xy - 3xy = 0
2x^2+3xy-4y2(4)+3(2)(-4)-(4)(-4)8-24+16=0
no
No, it is not.
3xy-2x-4=0,... so taking the dirividive of y using newtons it would be(3y)(3xy') -2=0 which could be called9 xyy'-2=0.. remember you Are trying to isolate the y prime. so bring the 2 over first.9xyy'=2.....then divide by 9xyy'=(2)/(9xy).....so then just plug in your variables and presto
yes 3xy=6
3x + 4y = 04y = -3xy = -0.75xy = mx + b with b being the y interceptSince: y = -0.75x + 0 , the y-intercept is 0.
(x + 6y)(x - 3y)
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
Whooop!
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)
They are: (3, 1) and (-11/5, -8/5)