To solve the equation ( xy^0 + 5x + 2y - 3 = 0 ), note that ( y^0 = 1 ), simplifying the equation to ( x + 5x + 2y - 3 = 0 ) or ( 6x + 2y - 3 = 0 ). Rearranging gives ( 2y = 3 - 6x ), leading to ( y = \frac{3 - 6x}{2} ). This represents a linear relationship between ( x ) and ( y ).
5x-2y=-3 -2y=-5x-3 y=(5/2)x+3/2 slope is 5/2
3(5x-2y)=18 5/2x-y=-1
x = -1.2, y = -3
Put in point slope/function form. 5X + 2Y + 3 = 0 5X + 2Y = - 3 2Y = - 5X - 3 Y = - 5/2X - 3/2 slope = - 5/2 ---------------- Y intercept = - 3/2 -------------------------
To find the expression for ((5x - y)(3x^2 - 3xy + 2y^2)), you need to distribute (5x) and (-y) across each term in the second polynomial. This results in: [ 5x(3x^2) - 5x(3xy) + 5x(2y^2) - y(3x^2) + y(3xy) - y(2y^2) ] Simplifying this gives: [ 15x^3 - 15x^2y + 10xy^2 - 3x^2y + 3xy^2 - 2y^3 ] Combining like terms results in: [ 15x^3 - 18x^2y + 13xy^2 - 2y^3 ]
5x - 4y ≥ -203x - 2y ≤ -8y ≥ -3
5x-2y=-3 -2y=-5x-3 y=(5/2)x+3/2 slope is 5/2
3(5x-2y)=18 5/2x-y=-1
x = -1.2, y = -3
Put in point slope/function form. 5X + 2Y + 3 = 0 5X + 2Y = - 3 2Y = - 5X - 3 Y = - 5/2X - 3/2 slope = - 5/2 ---------------- Y intercept = - 3/2 -------------------------
To find the expression for ((5x - y)(3x^2 - 3xy + 2y^2)), you need to distribute (5x) and (-y) across each term in the second polynomial. This results in: [ 5x(3x^2) - 5x(3xy) + 5x(2y^2) - y(3x^2) + y(3xy) - y(2y^2) ] Simplifying this gives: [ 15x^3 - 15x^2y + 10xy^2 - 3x^2y + 3xy^2 - 2y^3 ] Combining like terms results in: [ 15x^3 - 18x^2y + 13xy^2 - 2y^3 ]
For 5x+y=1, you would subtract 5x from each side, so you would get y=1-5x For 3x+2y=2, you would subtract 3x from each side, and then divide by 2. 2y=2-3x y=1-(3/2)x
No
(3, 2)
3x+2y=20 5x-2y=1 (3x+2y) + (5x-2y)= 20+1 3x+5x= 21....... 8x=21.........x=2 5/8 Subsitute x=3(2 5/8)+ 2y=20........... 7 7/8 +2y=20.........2y= 20- 7 7/8....... y=6 1/16
9+2y=3 2y=-6 y=-3
5x - 3y -7x + 2y +3x = (5-7+3)x + (-3+2)y = x - y