An algebra problem by A Former Brilliant Member
Find the value of x + y given x 2 + 2 y + 4 + x 2 + x − y + 5 = x 2 + x + 3 y + 2 + x 2 + 2 x + 3
The answer is 2.
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6 solutions
Did the same
Does this mean we can somehow rewrite this equation as x + y = 2 ? If so, how?
I observed, from the problem, that since LHS = RHS,
2y+4=2x+3 -----equation 1
And
-y+5=3y+2 -----equation 2
So from equation 2,
-y-3y=2-5
-4y=-3
y=3/4
Then, Substitute 3/4 into the first equation
2*(3/4)+4-3= 2x
6/4+1=2x
6/4+4/4=2x
10/4=2x
10/4*1/2=x
x=5/4
Therefore,
x+y=5/4+3/4
x+y=8/4
x+y=2
Take y=0, we will get x=2.
Why must y = 0 ? Why can't y = 1 ?
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You can use any value as you want.
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Who said so? Who said that we can use whatever value we want?
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@Pi Han Goh – If we put any value of x or y then we are left with a equation in other variable and we can solve it easily and will see that x+y is always equal to 2.
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@Archit Agrawal – How do you know that we can put in any value of x or y ? And the results always yields 2 as the answer?
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@Pi Han Goh – The question says that find the sum of x and y. Then there must be a fixed value for it.
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@Archit Agrawal – You can't assume that's there's a fixed value of x + y .
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@Pi Han Goh – Why I can't assume that there is fixed value for it?
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@Archit Agrawal – Did the question says that x + y is a fixed value?
That's wrong .It seems you got that luckily. :p
Take X=1 y=1 .it satisfies the equation .and I agree wirh archit
used this python program to get the answer:(this gives the values of x and y)
'' import math
for x in range(1,10): for y in range(1,10): if( math.sqrt(pow(x,2)+(2 y)+4) + math.sqrt(pow(x,2) + x + 5 -y) == math.sqrt(pow(x,2) + x + (3 y) + 2) + math.sqrt(pow(x,2)+(2*x)+3)): print(x); print(y); ''
I do not know proper solution,but
According to question (x+y) should be same for all solution.So I will use hit & trial
We can take x^2+2y+4 = x^2+x+3y+2 & x^2+x-y+5 = x^2+2x+3
After solving, x+y=2 & x+y=2
OR
x^2+2y+4 = x^2+2x+3 & x^2+x-y+5 = x^2+x+3y+2
After solving,x=5/4 & y=3/4 ; x+y=2
Now, a "legit" solution.
First, observe that x 2 + x + 3 y + 2 = x 2 + 2 y + 4 + ( x + y − 2 )
Also, observe that x 2 + 2 x + 3 = x 2 + x − y + 5 + ( x + y − 2 )
Now, let z = x + y − 2 . Then, we get x 2 + 2 y + 4 + x 2 + x − y + 5 = x 2 + 2 y + 4 + z + x 2 + x − y + 5 + z
Thus, we can split this equation into 3 cases:
Case 1: z > 0
This implies that RHS>LHS, so no solution.
Case 2: z < 0
This implies that RHS<LHS, so no solution.
Hence, we are left with the last case, z = 0 , so we get x + y = 2 .