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raketka [301]
3 years ago
7

Is (4,2) a solution to the system y-x=2 and -3x -2y= -8

Mathematics
1 answer:
mash [69]3 years ago
3 0
To figure out the answer here, all you have to do is plug the given point into your equations and see if the result is true:

(4, 2)  .... aka (x, y)

2 - 4 = 2
-2 = 2  ... this one is FALSE, as -2 doesn't equal 2, so already you can tell that the point won't be a solution, but still, check the second equation.

-3(4) - 2(2) = -8
-12 - 4 = -8
-16 = -8  .... this one is also false. (4, 2) is not a solution for this system of equations.
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As per the question,

Let a be any positive integer and b = 4.

According to Euclid division lemma , a = 4q + r

where 0 ≤ r < b.

Thus,

r = 0, 1, 2, 3

Since, a is an odd integer, and

The only valid value of r = 1 and 3

So a = 4q + 1 or 4q + 3

<u>Case 1 :-</u> When a = 4q + 1

On squaring both sides, we get

a² = (4q + 1)²

   = 16q² + 8q + 1

   = 8(2q² + q) + 1

   = 8m + 1 , where m = 2q² + q

<u>Case 2 :-</u> when a = 4q + 3

On squaring both sides, we get

a² = (4q + 3)²

   = 16q² + 24q + 9

   = 8 (2q² + 3q + 1) + 1

   = 8m +1, where m = 2q² + 3q +1

Now,

<u>We can see that at every odd values of r, square of a is in the form of 8m +1.</u>

Also we know, a = 4q +1 and 4q +3 are not divisible by 2 means these all numbers are odd numbers.

Hence , it is clear that square of an odd positive is in form of 8m +1

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