Difference between revisions of "2015 AMC 10A Problems/Problem 16"
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Moving terms gives the equation | Moving terms gives the equation | ||
− | <math>x^2 - y^2 = 3x - 3y.</math> | + | <math>x^2 - y^2 = (x + y)(x - y) = 3x - 3y = 3(x - y).</math> |
Because <math>x \neq y,</math> we can divide both sides of the equation by <math>x - y</math> to yield the equation | Because <math>x \neq y,</math> we can divide both sides of the equation by <math>x - y</math> to yield the equation |
Revision as of 20:39, 10 March 2020
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 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 for solution 3:
This is risky, as could be a viable answer too. Do not use this method unless you're sure about the answer. In other words, this solution is less reliable than the others, so only use it if you can't do the other methods.
Solution 4 (When you can't algebraically manipulate)
This, by the way, is a really cheap way of solving the problem but as long as you get an answer, it doesn't matter.
Looking at the first given equation, begin searching for solutions. Notice how works but when plugged into the
second equation, you get . Now, if you decrease the value by to obtain , plugging it into
the second equation will yield . Now, notice how the LHS is now less than the RHS as opposed to what it had
been when we plugged into the second given equation. We then conclude that there must be a solution between
and , so calculating of and we obtain
and
Thus we know the answer to be .
This is a really bad solution, but it works. :/
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.