Question 448805
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Graph the line that goes through point (3,8) and is parallel to the line whose equation is 6y-10x=30
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The solution in the post by @mananth is incorrect.


His final equation y = 1.67x + 3 is incorrect.


The point (3,8) does not lie on this line.


Indeed, the left side 'y' at this point is  1.67*3 + 3 = 8.01, but not 8, as it should be.


See my correct solution below.



<pre>
An equation for the line parallel to 6y - 10x = 30 is

    6y - 10x = c,

where 'c' is some constant.


We find 'c' by substituting the coordinates x= 3, y= 8 into this equation

    c = 6*8 - 10*s = 18.


Thus an equation is 6y - 10x = 18.


In the slope-intercept form it is  y = {{{(5/3)x}}} + 3,  which is different from the equation found by @mananth.
</pre>

Solved correctly.


Simply IGNORE the post by @mananth, since his solution is wrong.


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Not only this particular problem is solved incorrectly by @mananth.


Many other similar problems were solved incorrectly by @mananth.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Incorrectly solved are ALL similar problem, where coefficients of linear equations 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;are / (should be) rational numbers, which can not be presented as finite decimal fractions. 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Then @mananth, by applying his incorrect algorithm of rounding, makes everything wrong.  


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Rounding in such situations IS NOT ALLOWED.