It works out that the solutions are: x = 3 and y = 2
3
If: 2x+y = 5 and x2-y2 = 3 Then the solutions work out as: (2, 1) and ( 14/3, -13/3)
You can easily factor this:2x2 + 3x = 0x(2x + 3) = 0Separate this into two equations, one for each solution:x = 02x + 3 = 02x = -3x = -3/2So the two solutions are 0, and -3/2.You can easily factor this:2x2 + 3x = 0x(2x + 3) = 0Separate this into two equations, one for each solution:x = 02x + 3 = 02x = -3x = -3/2So the two solutions are 0, and -3/2.You can easily factor this:2x2 + 3x = 0x(2x + 3) = 0Separate this into two equations, one for each solution:x = 02x + 3 = 02x = -3x = -3/2So the two solutions are 0, and -3/2.You can easily factor this:2x2 + 3x = 0x(2x + 3) = 0Separate this into two equations, one for each solution:x = 02x + 3 = 02x = -3x = -3/2So the two solutions are 0, and -3/2.
2x2 + 11x + 15 = (x + 3)(2x + 5).
If limited to integer solutions, the answers are: 1*117, 3*39 and 9*13
13
2x2-7x+3 = 0 (2x-1)(x-3) = 0 x = 1/2 or x = 3
2x2+5x-3 = 0 (2x-1)(x+3) = 0 x = 1/2 or x -3
2X2 = 3Xset to 02X2 - 3X = 0factorX(2X - 3) = 0zero factor rule2X = 3 = 02X = 3X = 3/2======X = 0X = 3/2
Reading this as: 4x2-6=2x2 This implies that: -6=-2x2 3=x2 x=sqrt(3) and x=-sqrt(3) So x can equal plus or minus the square root of 3
2x2-x-3 = 0= 4 - x - 3 = 0= 1 - x = 0= 1 = xTherefore x = 1