find the real solutions
x= y^2 - 9
x-4y-12= 0
I got x= -3 and y= 7 and x= 40
but how do i know what x to use and whats the ordered pair? I'm confused! help please!
I hope you used substitution and replace the x of the second equation with the x value of the first, thus...
(y^2 - 9) - 4y - 12 = 0
y^2 - 4y - 21 = 0
(y-7)(y+3) = 0
then y = 7 or y = -3
if y = 7 go back to first equation for
x = 7^2-9 = 40
if y=-3, then x = (-3)^2 - 9 = 0
so your two points are (40,7) and (0,-3)
To find the real solutions for the given system of equations, we will solve the equations simultaneously. Let's start by solving the second equation for x in terms of y:
x - 4y - 12 = 0
Rearranging the equation, we get:
x = 4y + 12
Now substitute this expression for x into the first equation:
4y + 12 = y^2 - 9
Rearrange the equation to bring all terms to one side:
y^2 - 4y - 21 = 0
To simplify this equation, factor it if possible:
(y - 7)(y + 3) = 0
Setting each factor equal to zero, we have two equations:
y - 7 = 0 and y + 3 = 0
Solving these equations will give us the values for y:
For y - 7 = 0, add 7 to both sides:
y = 7
For y + 3 = 0, subtract 3 from both sides:
y = -3
Now, substitute the obtained values of y back into the second equation to find the corresponding values of x:
For y = 7:
x = 4y + 12
x = 4(7) + 12
x = 28 + 12
x = 40
So, one solution is x = 40 and y = 7.
For y = -3:
x = 4y + 12
x = 4(-3) + 12
x = -12 + 12
x = 0
So, another solution is x = 0 and y = -3.
Thus, the real solutions to the system of equations are the ordered pairs: (40, 7) and (0, -3).