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Help!!!!!

Write an equation that is parallel to 3x−2y=14 and passes through the point (-6, -11) in slope-intercept form.

User Djunehor
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2 Answers

7 votes

Answer: First, remember that parallel lines have the same slope. So we first need to find the slope of the line we are given, 3x+2y=27. The quickest way to find the slope of this line is to put it into slope-intercept form (y=mx+b)

subract 3x from both sides:

2y=-3x+27

divide each term by 2:

y=-3/2x +27/2

Now we can see that the slope is -3/2.

From here, we use the slope we found and the ordered pair given (7,-4) and find the equation of the new line. We will have to use point-slope form (y-y1)=m(x-X1) to find the new line because the ordered pair given is not the y-intercept.

(y-(-4))=-3/2(x-7)

distribute -3/2:

(y-(-4))= -3/2x + 21/2

simplify double negative on left side of equals sign:

y+4=-3/2x + 21/2

Subtract 4 from both sides:

y= -3/2x + 13/2

This is your answer in slope intercept form. Because the directions specifically ask for standard form, we have one more step. Standard form (ax+by=c) needs the terms with x and y to be on the same side of the equals sign. So, we simply have to move the term with x (-3/2x) to the other side by adding.

Finished product:

3/2x+y= 13/2

Explanation:

User Sameer Singh
by
4.8k points
4 votes

3*-6-2*-11=14

-18+22=14

22=14+18

10

User Maxxer
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4.0k points