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Find the value of $B - A$ if the graph of $Ax + By = 3$ passes through the point $(-7,2),$ and is parallel to the graph of $x + 3y = -5.$

 May 7, 2020
 #1
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From x + 3y = -5, y = -x/3 - 5, so the slope of the line is -1/3. The slope of the new line is also -1/3, so y = -x/3 + B.  Pugging in x = 2 and y = -7, we get -7 = -2/3 + B, so B = -19/3.

 

Then the line is y = -x/3 - 19/3.  Then 3y = -x - 19, so 3y + x = -19.  We want the right-hand side to be 3, so we mutiply both sides by -3/19: -9/19*y - 3/19*x = 3.  Therefore, A + B = -9/19 - 3/19 = -12/19.

 May 7, 2020
 #2
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Why did you do A+B instead of B-A?

peepeepuupuu  May 7, 2020
 #3
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\(y=mx+c\) slope-intercept form

"Parallel to the graph of \(x+3y=-5\)" means that \(Ax+By=3\) have the same slope as it.

Both are written in standard form.

 \(x+3y=-5\) Divide both sides by 3

\(\frac{x}{3}+y=-\frac{5}{3}\) Rearrange

\(y=-\frac{1}{3}x-\frac{5}{3}\) Thus slope is \(-\frac{1}{3}\)

 

 

\(Ax+By=3\)

Rearrange to solve for y

\(\frac{A}{B}x+y=\frac{3}{B}\)

\(y=-\frac{A}{B}x+\frac{3}{B}\)

Notice \(\frac{A}{B}=\frac{1}{3}\) (Since it is the same slope as the previous line)

Thus we have

\(y=-\frac{1}{3}x+\frac{3}{B}\) 

Now, as given, \(Ax+By=3\) passes through the point (-7,2)

Substitute this into \(y=-\frac{1}{3}x+\frac{3}{B}\)

\(2=-\frac{1}{3}*(-7)+\frac{3}{B}\) , Solve for B 

\(2-\frac{7}{3}=\frac{3}{B}\) , \(B=-9\)

So the equation now becomes \(y=-\frac{1}{3}x-\frac{1}{3}\)

However, we want to find the value of A in order to answer the question "What is B-A"

\(Ax+By=3\), use (-7,2) (Also knowing the value of B= -9 to solve for A)

\(-7A+(-9)*(2)=3\)

\(A=-3\)

\(B-A=-9-(-3)=-9+3=-6\)

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 May 8, 2020

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