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(A) Given equations are $y=kx+1$ $(i)$ and $3x+4y=9$ (ii).Substituting $(i)$ into (ii):$3x+4(kx+1)=9$$3x+4kx+4=9$$x(3+4k)=5$$x=\frac{5}{3+4k}$Since $x

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$(y+x-1)(y+2x-1)=0$

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(A) Given equations are $y=kx+1$ $(i)$ and $3x+4y=9$ (ii).Substituting $(i)$ into (ii):$3x+4(kx+1)=9$$3x+4kx+4=9$$x(3+4k)=5$$x=\frac{5}{3+4k}$Since $x$ must be an integer,$(3+4k)$ must be a divisor of $5$. The divisors of $5$ are $\pm 1, \pm 5$.Case $1$: $3+4k=1$ $\Rightarrow 4k=-2$ $\Rightarrow k=-0.5$ (Not an integer).Case $2$: $3+4k=-1$ $\Rightarrow 4k=-4$ $\Rightarrow k=-1$.Case $3$: $3+4k=5$ $\Rightarrow 4k=2$ $\Rightarrow k=0.5$ (Not an integer).Case $4$: $3+4k=-5$ $\Rightarrow 4k=-8$ $\Rightarrow k=-2$.Thus,the possible integer values for $k$ are $-1$ and $-2$.The lines are $y=-x+1$ and $y=-2x+1$.The combined equation is $(y+x-1)(y+2x-1)=0$.

The combined equation of the straight lines of the form $y=kx+1$ (where $k$ is an integer) such that the point of intersection of each with the line $3x+4y=9$ has an integer as its $x$-coordinate is

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