What are the solutions to the system?
Y=x^2+3x-4
Y=2x+2
To find the solutions to the system, we need to find the values of x and y that satisfy both equations. One way to do this is to substitute the expression for y in the second equation with the expression for y in the first equation, since they both equal y:
x^2 + 3x - 4 = 2x + 2
Subtracting 2x and 2 from both sides:
x^2 + x - 6 = 0
Now we can factor the left side:
(x + 3)(x - 2) = 0
So the solutions are:
x = -3 or x = 2
To find the corresponding values of y, we can substitute each value of x into either equation:
When x = -3:
y = (-3)^2 + 3(-3) - 4 = 2
When x = 2:
y = 2(2) + 2 = 6
Therefore, the solutions to the system are (-3, 2) and (2, 6).
To find the solutions to the system of equations Y = x^2 + 3x - 4 and Y = 2x + 2, we can set the two equations equal to each other and solve for x.
x^2 + 3x - 4 = 2x + 2
Rearrange the equation:
x^2 + 3x - 2x - 4 - 2 = 0
Combine like terms:
x^2 + x - 6 = 0
Now, we can factor the quadratic equation or use the quadratic formula to find the values of x. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = 1, and c = -6. Substituting these values into the quadratic formula:
x = (-1 ± √(1^2 - 4(1)(-6))) / (2(1))
Simplifying:
x = (-1 ± √(1 + 24)) / 2
x = (-1 ± √25) / 2
x = (-1 ± 5) / 2
Now we have two possible values for x:
x1 = (-1 + 5) / 2 = 4 / 2 = 2
x2 = (-1 - 5) / 2 = -6 / 2 = -3
Therefore, the solutions to the system of equations are x = 2 and x = -3. To find the corresponding values of y, substitute these values back into either of the original equations:
For x = 2:
Y = (2)^2 + 3(2) - 4
Y = 4 + 6 - 4
Y = 6
For x = -3:
Y = (-3)^2 + 3(-3) - 4
Y = 9 - 9 - 4
Y = -4
Therefore, the solutions to the system of equations are (2, 6) and (-3, -4).