An algebraic equation with an infinite number of solutions
4x2+3x+6+5x+3x+6 = 4x2+11x+12 when simplified
One.
None because without an equal sign it's not an equation
From first equation y = 3x + 5. Substitute this in second equation 2(3x + 5) = 6x + 10 This proves that 6x + 10 = 6x + 10 which is not helpful! I suppose possible solutions would be x = 0, y = 5 or x = 1, y = 8 etc etc Are you sure of your two equations, please?
There are infinitely many solutions to the equation since it simplifies to 13 = 13, which is always true.
Quadratic equation formula
None because without an equality sign it cannot be an equation
x2-5-4x2+3x = 0 -3x2+3x-5 = 0 or as 3x2-3x+5 = 0
NO,because in quadratic equation highest power of a variable is 2. e.x. 4x2-20x+10=0.
(x3 + 4x2 - 3x - 12)/(x2 - 3) = x + 4(multiply x2 - 3 by x, and subtract the product from the dividend)1. x(x2 - 3) = x3 - 3x = x3 + 0x2 - 3x2. (x3 + 4x2 - 3x - 12) - (x3 + 0x2 - 3x) = x3 + 4x2 - 3x - 12 - x3 + 3x = 4x2 - 12(multiply x2 - 3 by 4, and subtract the product from 4x2 - 12)1. 4x(x - 3) = 4x2 - 12 = 4x2 - 122. (4x2 - 12) - (4x2 - 12) = 4x2 - 12 - 4x2 + 12 = 0(remainder)
y=4x2+3x+8
An algebraic equation with an infinite number of solutions
4x2 - 3x - 9
(3x + 2)(x - 5) = 0 x = -2/3 or 5
4x2+3x+6+5x+3x+6 = 4x2+11x+12 when simplified
3x+10=12x-17