subtract the equations to get
3y^2 = 7
y = ±√(7/3)
so,
x^2 - 7/3 = 1
x^2 = 10/3
x = ±√(10/3)
Solve the system by substitution or elimination.
x^2 + 2y^2 = 8
x^2 - y^2 = 1
6 answers
I'm sure you'd have had no trouble if it were
x+2y=8
x-y=1
x+2y=8
x-y=1
Thank you! I tried to solve it like this:
x^2 + 2y^2 = 8
( x^2 - y^2 = 1 )2
x^2 + 2y^2 = 8
2x^2 - 2y^2 = 2
3x^2 = 10
x^2 = 10/3
x = ±√(10/3)
Any idea as to why my textbook would say that the solution is ±√(30/3)?
x^2 + 2y^2 = 8
( x^2 - y^2 = 1 )2
x^2 + 2y^2 = 8
2x^2 - 2y^2 = 2
3x^2 = 10
x^2 = 10/3
x = ±√(10/3)
Any idea as to why my textbook would say that the solution is ±√(30/3)?
I figured it out. No fraction can appear under a radical.
Check the bracket placement.
±√(10/3)
= ±√(10)/√(3)
= ±√(30)/3
= ±(⅓)√(30)
±√(10/3)
= ±√(10)/√(3)
= ±√(30)/3
= ±(⅓)√(30)
Yeah, lots of folks don't like to see radicals in the denominator. Pretty much a non-issue with me, but since it's a common practice, be sure to watch for it.
1 answer