y=-2x+8
If: x-2y = 1 then x = 2y+1 If: 3xy -y^2 = 8 then 3*(2y+1)*y -y^2 = 8 So: 6y^2 +3y -y^2 = 8 or 5y^2 +3y -8 = 0 Factorizing the above: (y-1)(5y+8) = 0 meaning y = 1 or y = -8/5 Solutions by substitution are: (3, 1) and (-11/5, -8/5)
The two equations are: 1) 3xy - y² = 8 2) x - 2y = 1 Make x the subject of (2): x - 2y = 1 → x = 2y + 1 substitute for x in (1) and solve for y: 3xy - y² = 8 → 3(2y + 1)y - y² = 8 → 6y + 3y - y² - 8 = 0 → 5y + 3y - 8 = 0 → (5y + 8)(y - 1) = 0 → 5y + 8 = 0 → y = -8/5 or y - 1 = 0 → y = 1 and substitute into 2 to find the corresponding x: y = -8/5 → x = 2(-8/5) + 1 = -11/5 y = 1 → x = 2(1) + 1 = 3 → the solutions are the ordered pairs (-11/5, -8/5) {or (-2.2, -1.6)} and (3, 1)
x + y = 10 y = x + 8 so, x + (x + 8) = 10 2x + 8 = 10 2x + 8 - 8 = 10 - 8 2x = 2 x = 1 and y = 9 because y = x + 8 y = 1 + 8 y = 9
x+y=-8 2x-y=4 y=-x-8 2x+x+8=4 3x=-4 x=-4/3 -4/3 + y= -8 y = -8 + 4/3 y= -6 2/3
-6
If you mean: y = -8 and y = -2x-12 then the solution is x = -2 and y = -8
6 + y = 8 y = 8 - 6 y = 2
If you mean: y = x+4 and y = 3x+8 then x = -2 and y = 2
x^8 - y^8 = (x - y)(x + y)(x^2 + y^2)(x^4 + y^4)
y/2 + 8 = 13 Subtract 8 from both sides: y/2 = 5 Multiply both sides by 5: y = 10
To write this algebraically: 7(y^3)|y = 2 Substitute 2 for y: 7(2^3) 2^3 = 8, so substitute 8 for (2^3): 7*8 = 56
√(2y) + 3 = 11 √(2y) = 11 - 3 = 8 √2 x √y = 8 √y = 8/√2 y = (8/√2)2 = 64/2 = 32
14
8(y - 1)(y^2 + y + 1)
It is: 8 > Y > -2
y=-2x+8