Find the equation of the line and write in general form.
Line parallel to the line x + 3y = 6 through (3, 4)

- anonymous

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- anonymous

Parallel lines have the same slope. Can you determine the slope of the given line?

- anonymous

I do not know how to find the slope of these two.

- anonymous

OK. You need to rearrange the given equation of the line into the form \(y=mx+b\)So first thing is to get rid of the \(x\) on the left hand side by subtracting \(x\) from both sides. What do you get?

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## More answers

- anonymous

3y = 5x?

- anonymous

No. You are given\[x+3y=6\]Now subtract \(x\) from both sides\[x+3y-x = 6-x\]What's left after simplifying?

- anonymous

How do I simplify it?

- anonymous

Combine like terms on the left hand side. The right hand side can't be simplified any further.

- anonymous

Is it 2 + 3y = 6 - x ?

- anonymous

What is \(x - x\) ?

- anonymous

0? or just x?

- anonymous

1 - 1 = 0
5 - 5 = 0
x - x = 0
So, after simplifying you are left with\[3y = 6-x\]It will be easier to understand if we write it as\[3y = -x + 6\]

- anonymous

Now to get it into the form \(y=mx+b\) you need to get rid of the 3 on the left hand side by dividing both sides by 3. What do you get?

- anonymous

y = -x + 3 ?

- anonymous

The left hand side is OK. But you didn't divide the -x by 3 and 6/3 is not 3. Can you fix it up?

- anonymous

y = -x + 2

- anonymous

OK. But you still haven't divided the -x by 3. When you divide both sides of an equation by 3, you must divide EVERY term by 3.

- anonymous

Like this:\[3y = -x + 6\]\[\frac{ 3y }{ 3 } = \frac{ -x }{ 3 } + \frac{ 6 }{ 3 } = -\frac{ 1 }{ 3 }x + 2\]See what I mean?

- anonymous

I was confused of that turning into a decimal I see now.

- Plasmataco

Ok. So you got the slope y+a=-1/3(x+a)

- Plasmataco

The two a are different sorry about that.

- anonymous

Is it supposed to be written as y = -1/3x +2 ?
Or no y into it?

- Plasmataco

Now, you plug in the y in the coordinates given int the a next to y. If it is positive, subtract it from y, if the y in the coordinates are negative, add it to the y

- anonymous

OK. So the slope of the given line is \(-\frac{1}{3}\). So the line you're after has that same slope. The general form of the equation of a line is \[y=mx+b\]and m represents the slope, so the line you're after looks like\[y=-\frac{ 1 }{ 3 }x + b\]Now to solve for b, us the x- and y-coordinates of the given point.

- Plasmataco

Well, there are 2 ways to write it. I find my way uses less calculations

- anonymous

You are given\[\left( x, y \right) = \left( 3,4 \right)\]By substitution, then,\[y = -\frac{ 1 }{ 3 }x + b\]\[4 = -\frac{ 1 }{ 3 }\left( 3 \right) = b\]Can you solve this for b?

- anonymous

Is the -1 I get from -1/3(3)
supposed to be b?

- anonymous

* \(+ b\) sorry for the typo

- anonymous

I made a typo. It should read\[4 = -\frac{ 1 }{ 3 }\left( 3 \right) + b\]sorry bout that

- anonymous

Its no problem ! I do not get how to solve for b.

- anonymous

OK. What do you get when you multiply -1/3 x 3 ?

- anonymous

I get -1 for that.

- anonymous

Great. So now you have\[4 = -1 + b\]To solve for b, you need to get rid of the -1 on the right hand side. To do this add 1 to both sides. What do you get?

- anonymous

Do I add +1 to b as well?

- anonymous

No. It looks like this:\[4+1 = -1 + b + 1\]Now simplfy

- anonymous

Is it 5 = b ?

- anonymous

Yayyyy! Good work.

- anonymous

Now you have everything you need for the answer. The general form of the equation of a line is \[y=mx + b\]You calculated the slope (m) earlier and you just figured out b. Put it all together. What's the equation of the line?

- anonymous

y = -1/3x +5?

- anonymous

Perfect. Well done. That is the line that is parallel to \(x+3y=6\) that passes through \((3, 4)\)

- anonymous

Thank you so much for your help! Also thank you for your time and patience! Math is my worst subject.

- anonymous

You're doing a good job. With some additional practice maybe math could be your best subject :)

- anonymous

That is true indeed. Also again thank you !

- anonymous

You're welcome

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