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4 solutions
Very nice!
a
+
b
=
3
⇒
a
=
3
−
b
.
b
+
c
=
7
⇒
c
=
7
−
b
.
c
+
d
=
9
⇒
(
7
−
b
)
+
d
=
9
⇒
d
=
2
+
b
.
Therefore,
a
+
d
=
(
3
−
b
)
+
(
2
+
b
)
=
5
.
I accidentally clicked 13 because i added instead of substracted the equations. I find substitution is a long way to solved this. By substracting equation 1 from 2 you find -a + c. Substracting that from equation 3 gives a+ d = 5
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Its actually one and the same.Your method is called elimination method.
First, add all the solutions.
( a + b ) + ( b + c ) + ( c + d ) = a + 2b + 2c + d => 19
Take away 2 ( b + c )
a + 2b + 2c + d - { 2 ( b + c ) } = a + b => 5
That's nice!
Since (b + c) = 7 and (c + d) = 9, you know that d = b + 2; therefore, since (a + b) = 3, it can be determined that (a + d) = 5 because you just add 2 to the sum of a + b.
( a + b ) − ( b + c ) + ( c + d ) = a + d So we have ( 3 ) − ( 7 ) + ( 9 ) = 5