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Exercise 6.3 · Q19

Q.Find at least two equations of the straight lines in the family of the lines y=5x+by = 5x + b, for which bb and the xx-coordinate of the point of intersection of the lines with 3x−4y=63x - 4y = 6 are integers.

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Substitute y=5x+by=5x+b into 3x−4y=63x-4y=6 to get x=−6+4b17x=-\dfrac{6+4b}{17}; requiring both bb and xx integers forces b≡7(mod17)b\equiv7\pmod{17}, so e.g. b=7b=7 and b=−10b=-10 both work.

Every member of the family y=5x+by=5x+b meets 3x−4y=63x-4y=6 at exactly one point (their slopes 55 and 34\tfrac34 differ, so they are never parallel); find that intersection's xx-coordinate as a function of bb, then pin down which integer bb also make xx an integer.

Step 1. Substitute the family into 3x−4y=63x-4y=6.

3x−4(5x+b)=63x-4(5x+b)=6

3x−20x−4b=63x-20x-4b=6

−17x=6+4b-17x=6+4b

x=−6+4b17x=-\frac{6+4b}{17}

Step 2. Impose that xx is an integer.

xx is an integer exactly when 17∣(6+4b)17\mid(6+4b), i.e. 4b≡−6(mod17)4b\equiv-6\pmod{17}. Since 4⋅13=52=3(17)+14\cdot13=52=3(17)+1, the inverse of 44 modulo 1717 is 1313, so

b≡−6⋅13≡−78≡−78+85≡7(mod17)b\equiv-6\cdot13\equiv-78\equiv-78+85\equiv7\pmod{17}

So b=…,−27,−10,7,24,…b=\dots,-27,-10,7,24,\dots — infinitely many integers satisfy this, and we just need at least two.

Step 3. Take b=7b=7. …

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