Ok in this question, I can't draw graphs or use a calculator (I'm really wishing I could) but I need to find the intersection(s) of and y=-x+2
I keep getting (1,4) or (1,2) but the answer is actually (1,1) so I'm not sure how to get to it?
Ok in this question, I can't draw graphs or use a calculator (I'm really wishing I could) but I need to find the intersection(s) of and y=-x+2
I keep getting (1,4) or (1,2) but the answer is actually (1,1) so I'm not sure how to get to it?
In the problem, the equations are
y = sqrt(x) Which can be written as
y^2 = x.
Substitute this value in the second equation. You get
y = -y^2 + 2
Or
y^2 + y - 2 = 0
If you factorize yoy get
( y - 1 )( y + 2 ) = 0
Or y = -2 or y = 1.
Since y = sqrt(x), y cannot be negative.
So y = 1 and x = 1.
How did you get "(1, 4) or (1, 2)"? If x= 1 then y= -x+ 2= -1+ 2= 1, not either 4 or 2. Similarly, if x= 1, .
I would have done that a slightly different way. Since and y= -x+ 2, (because they are both equal to y). Now square both sides. so that . Notice that that is an equation for x where sa-ri-ga-ma got an equation for y.
gives x= 4 and x= 1. If x= 4, y= -x+ 2= -4+ 2= -2. But cannot be negative so that is NOT a solution. If x= 1, y= -1+ 2= 1. The graphs intersect at (1, 1).