To solve this system of equations, we can use substitution or elimination method.
Let's solve by substitution method:
From the first equation, we can express y in terms of x:y = -1 - 3x
Substitute this expression for y into the second equation:x - x(-1 - 3x) = 8x + x + 3x^2 = 84x + 3x^2 = 83x^2 + 4x - 8 = 0
Now we have a quadratic equation which we can solve. Let's factor it:(3x - 2)(x + 4) = 0
Setting each factor to zero gives us two possible solutions:3x - 2 = 03x = 2x = 2/3
x + 4 = 0x = -4
Therefore, the solutions are x = 2/3 or x = -4.
Now, substitute these values back into the initial equation to find the corresponding values of y:For x = 2/3:3(2/3) + y = -12 + y = -1y = -3
For x = -4:3(-4) + y = -1-12 + y = -1y = 11
Therefore, the solutions to the system of equations are (2/3, -3) and (-4, 11).
To solve this system of equations, we can use substitution or elimination method.
Let's solve by substitution method:
From the first equation, we can express y in terms of x:
y = -1 - 3x
Substitute this expression for y into the second equation:
x - x(-1 - 3x) = 8
x + x + 3x^2 = 8
4x + 3x^2 = 8
3x^2 + 4x - 8 = 0
Now we have a quadratic equation which we can solve. Let's factor it:
(3x - 2)(x + 4) = 0
Setting each factor to zero gives us two possible solutions:
3x - 2 = 0
3x = 2
x = 2/3
x + 4 = 0
x = -4
Therefore, the solutions are x = 2/3 or x = -4.
Now, substitute these values back into the initial equation to find the corresponding values of y:
For x = 2/3:
3(2/3) + y = -1
2 + y = -1
y = -3
For x = -4:
3(-4) + y = -1
-12 + y = -1
y = 11
Therefore, the solutions to the system of equations are (2/3, -3) and (-4, 11).