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Solve the system of equations:
x^2+y^2 = 41 ; and
xy = 20
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Answer. There are 4 solutions: (x,y) = (5,4) or/and (x,y) = (4,5) or/and (x,y) = (-5,-4) or/and (x,y) (-4,-5).
Solution
There are many ways to solve it. I will show you some.
1. Guessing:
41 = 25 + 16 ---> (x,y) = (5,4) or/and (x,y) = (4,5) or/and (x,y) = (-5,-4) or/and (x,y) (-4,-5).
Works good if the numbers are good and you have the feeling of numbers.
Otherwise is not so good.
2. Graphically:
Plots y = +/- (red+green) and y =
Works good. It is always good to make a plot.
3. By reducing the system to one equation and by introducing new variable:
= 41, (1)
xy = 20. (2)
Express x = from (2) and substitute it into (1). You will get
+ = 41, or, after multiplying both sides by
= 0.
Introduce new variable u = . The last equation becomes
= 0.
Factor the left side. You will get
(u-25)*(u-16) = 0.
The roots are u = 25 and/or u=16, which gives you
(x,y) = (+/-5,+/-4) and/or (+/-4,+/-5)
and all you need to do is to select the signs by an appropriate way.
Actually, it is powerful algebraic method. You can combine it with the quadratic formula, using it instead of factoring.
4. By reducing the system to one equation and by applying factoring to this equation:
= 41, (1)
xy = 20. (2)
Express x = from (2) and substitute it into (1). You will get
+ = 41, or, after multiplying both sides by
= 0.
Factor the left side:
= 0.
In this way you obtain the same set of solutions.
Works good, especially if you know the solution in advance.
May work not so good in more complicated cases.
I prefer the method #3 almost always.
5. Using the trick
= 41, (1)
xy = 20. (2)
Double the equation (2) and then add it to the equation (1). You ill get
{{x^2 + 2xy + y^2}}} = 41 + 2*20, or
= 81.
Take square root from both sides. You will get two equations.
a. x+y = 9. Express x = 9-y and substitute it into equation (2).
In this way you will get a quadratic equation which is easy to solve.
b. x+y = -9. Express x = -9-y and substitute it into equation (2).
Again, you will get a quadratic equation which is easy to solve.
The method works good if you know the trick and are ready to apply it.
Makes fantastic impression to your teacher.
See the lessons
- Solving systems of algebraic equations of degree 2
- Some tricks to solve systems of non-linear algebraic equations
in this site.
Also, be informed that there is this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic Systems of equations that are not linear.