Without a second independent equation, it's not a 'system' yet.
If you mean: y = x^2+4x+3 and y = 2x+6 Then the solution is: x = 1 or x = -3
x + y = -4x - y = 2x - y = 2x + y - y = 2 + yx = 2 + yx + y = -42 + y + y = -42 + 2y = -42 - 2 + 2y = -4 - 22y = -62y/2 = -6/2y = -3x = 2 + yx = 2 - 3x = -1The solution of the system is (-1, -3)
It is (6, -1).
If you mean (1) y + 3x = 22 and (2) y + 2x = 16, then (1) y = 22 - 3x (sub this into (2)) (2) (22 - 3x) + 2x = 16 -x = -6 x = 6 y = 22 - 3x = 22 -3(6) = 4 solution: (6,4)
Without a second independent equation, it's not a 'system' yet.
What is the solution set for the equations x-y=2 and -x+y=2
If you mean: y = x^2+4x+3 and y = 2x+6 Then the solution is: x = 1 or x = -3
x + y = -4x - y = 2x - y = 2x + y - y = 2 + yx = 2 + yx + y = -42 + y + y = -42 + 2y = -42 - 2 + 2y = -4 - 22y = -62y/2 = -6/2y = -3x = 2 + yx = 2 - 3x = -1The solution of the system is (-1, -3)
It is (6, -1).
6
If you mean (1) y + 3x = 22 and (2) y + 2x = 16, then (1) y = 22 - 3x (sub this into (2)) (2) (22 - 3x) + 2x = 16 -x = -6 x = 6 y = 22 - 3x = 22 -3(6) = 4 solution: (6,4)
(3, 6)-------------------Let's see.(6) = 3(3) - 33(3) - (6) = 36 = 9 - 39 - 6 = 36 = 63 = 3========== (3, 6) is a solution to the system of equations. The only solution? I do not know.
It would be y = 6x.
I'm assuming the equation is: yx + 3y = 6 Now, you aren't going to get one answer for this, but you can factor for either x or y. In terms of y, this is y(x + 3) = 6 or y = 6/(x+3) In terms of x, this is yx = 6 - 3y or x = 6/y - 3 If you graph these, you'll get a y=1/x type function with the vertical asymptote at x=-3
Without an equality sign the given expression is not an equation and so therefore it has no solution.
[x + y = 6] has an infinite number of solutions.