 # Y 3x 4

## Y 3x 4

### Parallel: y = 3x 4 (2, 7)?

y = mx + b

Where M is, in this case your number is 3, so for your lines to be parallel, your new equation must also have 3.

Now enter: 7 = 3 (2) + b = 6 + b

So B is equal to one.

So your answer is y = 3x + 1

Hey, if the lines are parallel, the error is like the lines.

That is, y = 3x 4 here m = 3 [think of the common form y = mx + c]

So meter = 3

And the points are (2, 7)

So the equation is in terms of a point

(y7) = 3 (x2)

y = 3x + 1

For the linear equation, the intersection type (to the right of the moment) is y = mx + b. Dance Le Cass O س Vos Obtniz + 1 e Tout en Sutilisent pig zone du sign equivalent, Alvars IC les information vra es are: M, C East l'Mumble è M de la Ligni, ST Lee Confidence of XB East Lanterction Y, that's it . The y-axis situation where the lines intersect with the time of your query, both equations are already abnormal intersection type y = (4/3) x + (7/3) for m = 4/3 Are Multiply whatever you want and return with four / 3 so that 4/3 and = (4/3) x9 / 4 have one m = 4/3 which is 4/3 for both hits. Because your positions are the same, the tracks are parallel.

## Y 3x 4

The parallel causes (3x in this case) are the same. Draw your equation and see where the intersection point should be and it connects the point (2,7).

y = 3x + c, the same for parallel lines

Roots (2.7),

7 = 3 * 2 + c => c = 1

Answer: y = 3x + 1

Points formula:

aa1 = m (xx1)

Parallel lines have the same thing, m = 3

y1 = 7

x1 = 2

y7 = 3 (x2)

Divide to get: y7 = 3x6

Add 7 on both sides and your answer will be y = 3x + 1

Help for example!