The equation has infinitely many solutions.
There's an equals missing...
It is a quadratic equation and its solutions are: x = -3/2 and x = 3
Rearrange the second equation as x = 10+y and then substitute it into the first equation which will create a quadratic equation in the form of: 2y2+30y+100 = 0 and when solved y = -10 or y = -5 Therefore the solutions are: x = 0, y = -10 and x = 5, y = -5
Yes. Quite often, if you don't, you'll lose solutions. That is, the transformed equation - after taking square roots - will have less solutions than the original equation.
The equation has infinitely many solutions.
There's an equals missing...
The solutions to the quadratic equation are: x = -1 and x = 6
It is a quadratic equation and its solutions are: x = -3/2 and x = 3
Rearrange the second equation as x = 10+y and then substitute it into the first equation which will create a quadratic equation in the form of: 2y2+30y+100 = 0 and when solved y = -10 or y = -5 Therefore the solutions are: x = 0, y = -10 and x = 5, y = -5
Yes. Quite often, if you don't, you'll lose solutions. That is, the transformed equation - after taking square roots - will have less solutions than the original equation.
Using the quadratic equation formula: x = -2 plus or minus the square root of 6 which means that x = 0.4494897427831779 x = -4.4494897427831779
It often helps to take square roots on both sides of the equation. However, solutions to the original equation may be lost - it is often convenient to put a "plus or minus" sign so as not to lose solutions. Example: x2 = 25 Taking square roots: x = "plus or minus" 5
It is a quadratic equation and its solutions are: x = 4 or x = -9
Using the quadratic equation formula:- x = 1 +/- the sqrt of 13 = -2.60555 and 4.60555 (to 5 dp).
x equals negative b plus or minus the square root of b squared minus 4bc over 2a
It is a simultaneous equation and when solved its solutions are x = 71/26 and y = 50/13