Asked by algebra
                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!
            
        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!
Answers
                    Answered by
            Reiny
            
    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)
    
(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)
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