Whenever a problem says "one number" and "another number" we can substitute x and y. for this we have x=2y+1 and xy=1. since we know the value of x (2y+1) we can substitute it for the other equation to get (2y+1)y=10. simplify to get 2y^2+y=10. from here you can do a few methods to solve this, but the simplest in my opinion is by factoring.
In order to factor it must be equal to 0, so we have 2y^2+y-10=0. We factor this and get y=5 (we also get y=-4 but it is an extraneous root). now we can plug that into either equation and find that x=2.
this means our two numbers are 5 and 2
Answer: Just get closer
Step-by-step explanation:
Walk next to someone to hear them left foot right foot towards them.
Answer:
1 real solution
Step-by-step explanation:
y=2x^2−8x+8
We can use the discriminant to determine the number of real solutions
b^2 -4ac
a =2 b = -8 c=8
(-8)^2 - 4(2)(8)
64 - 64
0
Since the discriminant is 0 there is 1 real solution
>0 there are 2 real solutions
< 0 two complex solutions
