Find the equation of the line that is parallel to the line 2x-3y=-4 containing the point (-5,3)

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Find the equation of the line that is parallel to the line 2x-3y=-4 containing the point (-5,3)

Mathematics
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first you want to get it in slope intercept form
ok
slope intercept form= y=mx+b

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lets put this in slope intercept form! 2x-3y=-4 subtract 2x on both sides -3y=-2x-4 divide by negative 3 on both sides y-2/3x+4/3
mak that be y=2/3x+4/3
Ok, so we will start from solving that given line First of all, solve that line in the form of y=mx+c Where "m" is the slope. Note: Parallel line means same slope. 2x-3y=-4 Solve for y \[3y = 2x +4\]\[y = \frac23x + \frac43\] Comparing this with y=mx+c Slope is "m", so slope we get is 2/3. Now insert the point given and the slope in the formula, (i.e. slope and (-5,2) \[y-y_1 = m(x-x_1)\] \[y - 2 = \frac23 (x-(-5))\] Im sure u can solve now! :)
so now lets use a formula called point slope form. do you know the formula for that?
how do you get y-2?
It's given in the question, Point (-5,2)
its (-5,3)
Oh sorry. i overread. make it 3 please.
\[y−2=2/3(x−(−5))\] y=2/3x +5.33
i prefer writing that in fractions
is that the correct answer?
Nope.
whats the answer?
\[y = \frac23 x+\frac{19}{3}\]
its not 19/3. (2/3)*5+2= 5.33
hold on..
ok?
lol, u made a mistake, (2/3)5 + 3
hahah wow. Thanks for catching that!
;]

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