if it is 2x-3=19 then you add 3 to both sides and divide by 2
but if it is 2(x-3) then you distribute 2 into x and 3: 2x-6=19 then add 6 to both sides and dividing by 2
One equation with two unknowns usually does not have a solution.
Since that isn't an equation, there is really nothing to solve.
IMPOSSIBLE
No.
It is x = -5
y2=x3+3x2
One equation with two unknowns usually does not have a solution.
(a) y = -3x + 1
2+4 (which equals 6) x3 (equals 18) +1 = 19
Since that isn't an equation, there is really nothing to solve.
IMPOSSIBLE
x^3=2x+5 x^3-2x-5=0 Graph the equation or use a calculator. x= 2.09455148
No.
no only equations with x2 and lower powers can be considered quadratic. those with x3 cannot be considered quadratic, just as x2 cannot be considered linear
No, it is not.
It is x = -5
(xn+2-1)/(x2-1)