What’s m<1 and m<2 to
1 answer:
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<h3>Answer: x = 7 and y = 3</h3>
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Explanation:
Apply the difference of squares rule
x² - 4y² = 13
x² - (2y)² = 13
(x - 2y)(x + 2y) = 13
Since x and y are positive integers, this means x-2y and x+2y are both integers as well.
The value 13 is prime. Its only factors are 1 and 13
Since the above equation shows 13 factoring into x-2y and x+2y, then we have two cases:
- A) x-2y = 1 and x+2y = 13
- B) x-2y = 13 and x+2y = 1
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Let's consider case A
We have this system of equations
Add the equations straight down
- x+x becomes 2x
- -2y+2y becomes 0y = 0 which goes away
- 1+13 becomes 14
Therefore we have 2x = 14 solve to x = 7
From here, plug this into either equation to solve for y
x-2y = 1
7 - 2y = 1
-2y = 1-7
-2y = -6
y = -6/(-2)
y = 3
You should get the same result if you used x+2y = 13
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Since we've found that x = 7 and y = 3, notice how case B is not possible
Example: x-2y = 13 becomes 7-2(3) = 13 which is false.
Also, x+2y = 1 would turn into 7+2(3) = 1 which is also false.
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Let's check those x and y values in the original equation
x² - 4y² = 13
7² - 4*(3)² = 13
49 - 4(9) = 13
49 - 36 = 13
13 = 13
The answer is confirmed.