These Equations Are Actually Related
{ n + m = 1 0 m 2 − n 2 = 2 0
If n and m are real numbers satisfying the system of equations above, what is the value of the m 2 + n 2 + n m ?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
3 solutions
I used simitaneous equations I rearranged m+n=10, making m the subject m=10-n Substitute m for 10-n making; (10-n)^2-n^2 I then solved that to make 100+n^2-20n-n^2 = 20 I then solved for n; -20n + 100 = 20 -20n = -80 n = -80/-20 n=4 I then put my solution for n into m+n=10, and solved it m+4 = 10 m = 10-4 m = 6 Hence, m^2+n^2+nam => 6^2+4^2+24 == 76
@Arun Garg Please be respectful while talking with the community.
I've deleted your comment
n
+
m
=
1
0
m
2
−
n
2
=
2
0
So, n = (10-m) and we can insert that into the left hand side of the second equation to get:
m
2
−
n
2
=
m
2
−
(
1
0
−
m
)
2
=
m
2
−
(
m
2
−
2
0
m
+
1
0
0
)
=
m
2
−
m
2
+
2
0
m
−
1
0
0
=
2
0
m
−
1
0
0
So:
2
0
m
−
1
0
0
=
2
0
m=6 and n=4
Therefore,
m
2
+
n
2
+
n
m
=
6
2
+
4
2
+
(
4
)
(
6
)
= 36+16+24 = 76
m + n = 1 0 ( m + n ) ( m − n ) = 2 0 ⇒ m − n = 2 ⇒ m = 6 , n = 4 m 2 + n 2 + n m = 7 6