3
2x^2+3xy-4y2(4)+3(2)(-4)-(4)(-4)8-24+16=0
2sinx+1 equals 0
For y - 2y - 3y equals 0, y equals 0.
yes 3xy=6
Suppose x3-4x = 0. To solve, factor: x3-4x = x(x2-4) = x(x+2)(x-2) = 0 Now, a product equals 0 if and only one or more of the factors equals 0, so set each factor to 0 and solve. The roots are 0,-2 and +2.
3xy - 3xy = 0
2x^2+3xy-4y2(4)+3(2)(-4)-(4)(-4)8-24+16=0
3xy-4x + 15y-20 Sol: As there are no number are same. So -4x-3xy +15y-20 -------------------- -4x-3xy+15y-20
2sinx+1 equals 0
For y - 2y - 3y equals 0, y equals 0.
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)
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
Suppose x3-4x = 0. To solve, factor: x3-4x = x(x2-4) = x(x+2)(x-2) = 0 Now, a product equals 0 if and only one or more of the factors equals 0, so set each factor to 0 and solve. The roots are 0,-2 and +2.
many solutions