A system of Equations
If x + y + x + y = 4 2 and x − y + x − y = 2 0 , then find the sum of all possible values of x y .
The answer is 260.
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5 solutions
Correct. A common error is to accidentally include one or more of the extraneous solutions. After all, they do seem to fit in to the quadratic equation... However, a quick glance back at our original equation and we see that it is not possible.
I think there is a typo;
"The solutions for this are
b
=
1
6
,
2
5
..... "
You fixed it in the next line, but it was confusing for a second.
Either way, this is a really nice solution.
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Yes, you are correct.
By the way, we can substitute u = x + y , which will a little bit faster, since we can quickly determine the value of x + y . The same thing goes for x − y . This is how I do it.
Let u = x + y and v = x − y . We then have,
u 2 + u = 4 2
⟹ u = 6 , − 7
after some easy factorization. Similarly, by using the style above, we will have
v 2 + v = 2 0
⟹ v = 4 , − 5
But since negative numbers are not possible for the square roots, so u 2 = x + y = 3 6 and v 2 = x − y = 1 6 . Solving the simultaneous equation, we arrive x = 2 6 and y = 1 0 , implying to our desired answer which is x y = 2 6 0 .
Because I see a substitution in the first equation, let a = x + y . The equation then becomes a + a = 4 2 ⇒ a = 4 2 − a ⇒ a = a 2 − 8 4 a + 1 7 6 4 ⇒ a 2 − 8 5 a + 1 7 6 4 = 0 ⇒ ( a − 3 6 ) ( a − 4 9 ) = 0 ⇒ a = 3 6 , 4 9 . Since square roots were involved, we must check the solution.
If a = 3 6 , then 3 6 + 3 6 = 4 2 , which is creary true. If a = 4 9 , then 4 9 + 4 9 = 4 2 , which is creary not true.
Therefore, we have x + y = 3 6 .
Approaching the second equation the same way, let b = x − y . Then, b + b = 2 0 ⇒ b = 2 0 − b ⇒ b 2 − 4 1 b + 4 0 0 = 0 ⇒ ( b − 1 6 ) ( b − 2 5 ) = 0 . Checking, we see that 25 is an extraneous root, and 16 is our only root.
Therefore, x + y = 3 6 and x − y = 1 6 . Adding these equations, we have 2 x = 5 2 ⇒ x = 2 6 ⇒ y = 1 0 .
Now we must multiply 26 and 10. This is quite hard, but with a calculator, we get 2 6 ⋅ 1 0 = 2 6 0
what if 49^0.5 is taken as -7 ?
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That's not possible because you cannot take the square root of a negative number.
Minimario, you never fail to amuse me :)
You didn't have to expand a+a^0.5 =42 , you could've simply taken a^0.5= p then solved p^2 + p = 42 The same for the 2nd equation
Lets say x + y = a and x − y = b . Then the equations turns to:
A ) a + a − 4 2 = 0 → a = 4 9 or a = 3 6 .
B ) Similarly, b + b − 2 0 = 0 → b = 2 5 or b = 1 6 .
We equal:
x + y = 4 9 or x + y = 3 6
x − y = 2 5 or x − y = 1 6
We have 4 possible system of equations. Solving them, we get the folllwing possible values for x and y :
( 2 6 5 , 2 3 3 ) , ( 3 7 , 1 2 ) , ( 2 6 , 1 0 ) and ( 2 6 1 , 2 1 1 ) . Substituting them on the original equations, the only one that satisfies the system with the principal square roots is ( 2 6 , 1 0 ) . Finally, x y = 2 6 ∗ 1 0 = 2 6 0 .
(x+y)^{1/2}=a -> a^{2}+a=42 -> a=-7 or a=6... (x+y)^{1/2}=6 and not -7... -> x+y=36
(x-y)^{1/2}=b -> b^{2}+b=20 -> b=-5 or b=4... (x-y)^{1/2}=4 and not -5... -> x-y=16
x+y=36
x-y=16
so x=26 & y=10 -> xy=260
m+\sqrt(m)=42 n+\sqrt(n)=20 EXPANDING WE GET m^2+m-1764=0 solving we get the answer
Let x + y = a . From the first equation, a + a = 4 2 The solutions for this are a = 3 6 , 4 9 , but 49 is extraneous due to the square root, so x + y = 3 6 .
Let x − y = b . From the second equation, b + b = 2 0 The solutions for this are b = 1 6 , 2 5 , but 25 is extraneous due to the square root, so x − y = 1 6 .
Thus, x = 2 6 and y = 1 0 , so the answer is 2 6 ⋅ 1 0 = 2 6 0 .