Question 375101
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Given the values you determined for the slope, you came to the correct conclusion.  Whenever the slopes are NOT equal, the lines are NOT parallel.  Unfortunately, you made an error when calculating the slope for the second equation.  The slope for the first equation is indeed 2 as a simple swap of sides shows us:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 2x\ +\ 4]


And we can see that with the equation in slope-intercept form, *[tex \Large y\ =\ mx\ +\ b], that the coefficient on *[tex \Large x], and therefore the slope, is 2.


But if you solve the other equation for *[tex \Large y] by dividing both sides by 2, you get


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 2x\ -\ \frac{7}{2}]


And the coefficient on *[tex \Large x] is again 2, so the slope of the second line is also 2.  What is different is the *[tex \Large y]-intercept, that is, the constant term.  This tells us that even though the two lines have the same slope, the place they intersect the *[tex \Large y]-axis is different.  Hence, these two lines are indeed parallel (as opposed to being the same line)


{{{drawing(
500, 500, -5, 5, -5, 5,
grid(1),
graph(
500, 500, -5, 5, -5, 5,
2x+4,2x-3.5))}}}


See?


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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