x = 5 x-2y = 6 => x = 6+2y 6+2y = 5 2y = 5-6 2y = -1 y = -1/2 and x = 5 Point of intersection: (5, -1/2)
4
x = y+3 x+2y = 9 => x = 9-2y y+3 = 9-2y 2y+y = 9-3 3y = 6 y = 2 and x = 5
x - 2y=0 so x = 2y 9y - 4x=1 Substituting for x, 9y-4*(2y) = 1 or 9y-8y = 1 so that y = 1 then x = 2y = 2*1 = 2
3x+2y+x
x + 2y = x + 2y Y and X aren't compatibles.
5
x-2y=0 x=2y The solution set is the set of all (x,y) such that x=2y
x2 - 4y2 = 16∴ (x - 2y)(x + 2y) = 162y - x = 2∴ x = 2y - 2∴ ([2y - 2] - 2y)([2y - 2] + 2y) = 16∴ (y - 1 - y)(y - 1 + y) = 16∴ -1(2y - 1) = 16∴ 1 - 2y = 16∴ -2y = 15∴ y = -7.52y - x = -2∴ -15 - x = -2∴ x = -13So the point of intersection is (-13, -7.5)
x = 5 x-2y = 6 => x = 6+2y 6+2y = 5 2y = 5-6 2y = -1 y = -1/2 and x = 5 Point of intersection: (5, -1/2)
This means that 2y - 3 = 2y + 4 which is not possible.
x = 11 - 2y and 14 - 2y. This means that 11 = 14. Is there a typo in the question?
(x^2 + 2y)(x^3 - 2xy + y^3) = x^2(x^3 - 2xy + y^3) + 2y(x^3 - 2xy + y^3) Now, let's distribute each term: = x^2 * x^3 - x^2 * 2xy + x^2 * y^3 + 2y * x^3 - 2y * 2xy + 2y * y^3 Now, simplify each term: = x^5 - 2x^3y + x^2y^3 + 2x^3y - 4xy^2 + 2y^4 Now, combine like terms: = x^5 + x^2y^3 - 4xy^2 + 2y^4 So, the expanded form of (x^2 + 2y)(x^3 - 2xy + y^3) is x^5 + x^2y^3 - 4xy^2 + 2y^4.
x+2y = 7 2y = -x+7 y = -0.5x+3.5
solve for x -x-2y=6 -x=2y+6 x=-2y-6 plug in y=0 x=-6 when y is 0 therefore the x intercept is (-6,0)
-2y=x+8 is the answer
2y(x+2)=