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I have the following equations :

1) $3x-y=4$

2) $4x-2y=2$

Now on paper, I did the following :

1) at $x=0$, $y=-4$ , at $y=0$, $x=1.333$

2) at $x=0$, $y=-1$ , at $y=0$, $x=\frac{1}{2}$

Using these results, the two equations intersect at $y=3$. However, WolframAlpha and Google plot shows that they intersect at $y=5$.

Can someone tell me what am I doing wrong?

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"The intersection of two points"? I guess you mean the intersection of two lines, no? – leonbloy Jan 8 '13 at 14:37
Your method looks correct. I suppose your drawing skills need improving. ;-) The inclinations of the lines are close to each other, so it's easy to make a mistake. Or maybe you switched $x$ and $y$ -- I get $x=3$, $y=5$. I don't know what you mean by $x=-1.333$ -- that seems to have nothing to do with the problem. – Michael E2 Jan 8 '13 at 14:37
If you're the type of student who draws on an area the size of a postage stamp: make the drawing larger!. And, be careful to scale the axes correctly. (The solution is $x=3$, $y=5$, here; so your solution is close. I imagine your solution was off because your drawing was not accurate enough.) – David Mitra Jan 8 '13 at 14:41
@MichaelE2 I've edited my question to remove that part. – NLed Jan 8 '13 at 14:42
up vote 1 down vote accepted

Your approach to graph each line by finding the points at which each line intersects the x-axis and y-axis, respectively, was fine. So I suspect the problem may have been in the accuracy of your graph and the scales used. (As you can see from the graph below, the slope of the lines are close, one just a bit steeper than the other.

When estimating the point of intersection using a graph, it always helps to verify, algebraically, the precise point of intersection:

To find the point of intersection: solve the equations simultaneously -

1) $3x-y=4 \iff 6x - 2y = 8 \iff -2y = 8 - 6x$

Now substitute into equation (2):

2) $4x+ -2y=2 \iff 4x + 8 + -6x = 2 \iff -2x = - 6 \iff x = 3$

At $x = 3$, using equation (1), $3\cdot 3 - y = 4 \iff y = 9 - 4 = 5$

Hence the lines intersect at $(3, 5)$.

enter image description here

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I wonder why my plot is incorrect. I add the two points on the graph and connect the two with a straight line ... Should give me the right answer !! – NLed Jan 8 '13 at 14:44
I think you tried to compute the x and y intercepts of the two lines, which is not the points of intersection of the two lines. – amWhy Jan 8 '13 at 14:46
Nope, the problem was with the accuracy of the scales used .. I have to be careful next time, a small difference in x/y values can give a large error. – NLed Jan 8 '13 at 14:57
Yes, you're correct about accuracy of the scales used. Your approach was fine, in terms of finding two points that determine the first line, and two points that determine the second line. Then graphing the lines to see where they intersect. The technique I used helps to verify precisely the point at which the lines intersect. – amWhy Jan 8 '13 at 15:03
@amWhy: How did this not get any thumbs up (although accepted)? +1 – Amzoti May 3 '13 at 2:17

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