2015 AMC 10A Problems/Problem 16
Contents
Problem
If , and
, what is the value of
?
Solutions
Solution 1
Note that we can add the two equations to yield the equation
Moving terms gives the equation
We can also subtract the two equations to yield the equation
Moving terms gives the equation
Because we can divide both sides of the equation by
to yield the equation
Substituting this into the equation for that we derived earlier gives
Solution 1
Note that we can add the two equations to yield the equation
Moving terms gives the equation
We can also subtract the two equations to yield the equation
Moving terms gives the equation
Because we can divide both sides of the equation by
to yield the equation
Substituting this into the equation for that we derived earlier gives
Solution 2 (Algebraic)
Subtract from the left hand side of both equations, and use difference of squares to yield the equations
and
.
It may save some time to find two solutions, and
, at this point. However,
in these solutions.
Substitute into
.
This gives the equation
which can be simplified to
.
Knowing and
are solutions is now helpful, as you divide both sides by
. This can also be done using polynomial division to find
as a factor. This gives
.
Because the two equations and
are symmetric, the
and
values are the roots of the equation, which are
and
.
Squaring these and adding them together gives
.
Solution 3
By graphing the two equations on a piece of graph paper, we can see that the point where they intersect that is not on the line is close to the point
(or
).
, and the closest answer choice to
is
.
Note: This is risky, as could be a viable answer too. Do not use this method unless sure about the answer. Also, very time consuming to graph the equations, do not attempt with time limit.
See Also
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.