Write an equation in slope-intercept form for the line that is parallel to the given line and passes through the given point.

𝑦 = 4π‘₯ + 6, (2, 7)

Y = mx+b

Y = 4x+6
7 = 4*2+b
7 = 8+b
b = -1.
Y = 4x+b.

Y = 4x-1.

Well, I could take the serious approach and give you the equation right away, but where's the fun in that, right? Let's add a touch of humor to your equation!

To find a line parallel to 𝑦 = 4π‘₯ + 6, we know the slope must be the same because parallel lines have the same slope. So the new line also has a slope of 4.

Now, let's find the equation in slope-intercept form. We already know the slope, so we just need to find the value of the y-intercept. We have the point (2, 7), which means that when x = 2, y = 7.

Since we have x, y, and the slope, we can use the good old slope-intercept form (𝑦 = π‘šπ‘₯ + 𝑏) and substitute in our values:

7 = 4(2) + 𝑏

Simplifying that, we get:

7 = 8 + 𝑏

Subtracting 8 from both sides:

-1 = 𝑏

So the equation for the line parallel to 𝑦 = 4π‘₯ + 6 and passing through the point (2, 7) is:

𝑦 = 4π‘₯ - 1.

And there you have it! A line that will make mathematicians laugh with its brilliance and humor.

To find the equation of a line that is parallel to the given line 𝑦 = 4π‘₯ + 6 and passes through the given point (2, 7), we need two pieces of information: the slope of the given line and the coordinates of the given point.

The given line is in slope-intercept form, which is 𝑦 = π‘šπ‘₯ + 𝑏, where π‘š represents the slope and 𝑏 represents the y-intercept.

We can observe that the slope of the given line is 4. Since a line parallel to the given line will have the same slope, the slope of our new line will also be 4.

Now, we can use the point-slope form of a linear equation, which is 𝑦 βˆ’ 𝑦₁ = π‘š(π‘₯ βˆ’ π‘₯₁), where (π‘₯₁, 𝑦₁) represents the coordinates of the given point and π‘š represents the slope.

Plugging in the values we have, we get:

𝑦 βˆ’ 7 = 4(π‘₯ βˆ’ 2)

Now, let's simplify the equation:

𝑦 βˆ’ 7 = 4π‘₯ βˆ’ 8

To put the equation in slope-intercept form, we isolate y:

𝑦 = 4π‘₯ βˆ’ 8 + 7

𝑦 = 4π‘₯ βˆ’ 1

So, the equation in slope-intercept form for the line that is parallel to 𝑦 = 4π‘₯ + 6 and passes through the point (2, 7) is 𝑦 = 4π‘₯ βˆ’ 1.

To find an equation of a line that is parallel to the given line 𝑦 = 4π‘₯ + 6, we need to use the fact that parallel lines have the same slope.

The given line has a slope of 4, so any line that is parallel to it will also have a slope of 4.

We also have a point that the parallel line needs to pass through, which is (2, 7).

To write an equation in slope-intercept form (𝑦 = π‘šπ‘₯ + 𝑏), we can substitute the slope (π‘š = 4) and the coordinates of the point (π‘₯ = 2, 𝑦 = 7) into the equation and solve for 𝑏, which represents the y-intercept.

Let's follow the steps:

Step 1: Substitute π‘š = 4, π‘₯ = 2, and 𝑦 = 7 into the equation 𝑦 = π‘šπ‘₯ + 𝑏:

7 = 4(2) + 𝑏

Step 2: Simplify the equation:

7 = 8 + 𝑏

Step 3: Isolate 𝑏 by subtracting 8 from both sides of the equation:

-1 = 𝑏

The value of 𝑏 is -1.

Therefore, the equation of the line parallel to 𝑦 = 4π‘₯ + 6 and passing through the point (2, 7) is:

𝑦 = 4π‘₯ - 1