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For which value of $k$ the equations $y=2x-5$, $y=x+2$ and $y=kx-12$ have common solution.

Let $G_{2x-5}$ intersects $G_{x+2}$ at point $N(x_N;y_N)$. We can get that $x_N=7$ and $y_N=9$. From here $N$ must lie on $G_{kx-12}$. Therefore, $k=\dfrac{20}{7}$.

Here are the graphs of $y=2x-5$, $y=x+2$ and $\dfrac{20}{7}x-12$. What's the problem with the drawing? Why $N$ does not lie on the third function? enter image description here

After your help: enter image description here

Math Student
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1 Answers1

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The problem is that $k = 3$ is the proper value since $3(7) - 12 = 9$. Your $k$ value of $\frac{20}{7}$ is $\frac{1}{7}$ too small, which is why the red line is slightly to the right & below where it should be.

John Omielan
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