0 = 2x^2 + 2x - 4 = 2(x^2 + x - 2) = 2(x-1)(x+2) Thus, there are two real solutions: x = 1 and x = -2
2x - y = 8 x + y = 1 These are your two equations. They will have two solutions since you have two variables. The solutions are x=3 and y=-2
(2x + 1)(2x + 1) or (2x + 1)2 so x = -0.5
Only one: (3,-2)
I messed up the question I ment what are the solutions to 2x^2 + 5x + 8 = 6
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0 = 2x^2 + 2x - 4 = 2(x^2 + x - 2) = 2(x-1)(x+2) Thus, there are two real solutions: x = 1 and x = -2
There are infinitely many solutions. These are coordinates of all points on the line given by the equation 2 - y = 2x - 1 or 2x + y = 3.
2(2x-1)+2x=6(x-1)
How many solutions are there to the following system of equations?2x - y = 2-x + 5y = 3if this is your question,there is ONLY 1 way to solve it.
2x - y = 8 x + y = 1 These are your two equations. They will have two solutions since you have two variables. The solutions are x=3 and y=-2
x<-2 or x>1
It has 2 solutions and they are x = 2 and y = 1 which are applicable to both equations
3(4x*-2)=6(2x-1)
(2x + 1)(2x + 1) or (2x + 1)2 so x = -0.5
It is when a variable has an infinite number of solutions Example: 2x + 2 = 4x divided by 2 +2
Only one: (3,-2)