Math help??

Since you are given the slope and a point on the line, write the equation in the "point-slope" form, y-y₀ = m(x-x₀):

y - (-5) = (2/3)(x - (-6))

y + 5 = (2/3)(x + 6)

That satisfies the requirements of the question, a linear equation with given slope and point, but since the required form of the equation is not specified, the equation can be restated in a number of different forms:

Standard Form: Ax + By = C

3y + 15 = 2x + 12

2x - 3y = 3

Slope Intercept Form:

y = (2/3)(x + 6) - 5

y = (2/3)x + 4 - 5

y = (2/3)x - 1

Slope is 2/3, y-intercept is -1

Intercept Form:

x/(3/2) + y/(-1) = 1

(intercepts are x = 3/2 and y = -1 

A line with a slope of m = 2/3 

 and passes through the point (-6, -5)

 y = mx + b 

 -5 = (2/3)(-6) + b 

 -5 = -4 + b

-1 = b

 The equation of the line:

 y = (2/3)x - 1

 y = (2x - 3)/3

Line, L, say,  with slope m = (2/3) passes through P(-6,-5). Then equation of L in

slope, y-intercept form is y = mx +b where m is the slope and b is the y-intercept.

So y = (2/3)x + b. Now P(-6,-5) is on L. Then (-5) = (2/3)(-6) + b = b-4. Then b = -1

and y = (2/3)x -1 is the slope, y-intercept form of the equation L.

y - y1 = m(x - x1)

y - (-5) = 2/3[x - (-6)]

y + 5 = 2/3(x + 6)

3(y + 5) = 2(x + 6)

3y + 15 = 2x + 12

  3y = 2x + 12 - 15

 3y = 2x - 3

  y = 2/3x - 1 answer//

Write a linear equation based on the following information: 

 A line with a slope of m = 2/3 and passes through the point (-6, -5)

 (y - -5) = (2/3)(x - -6)

 (y + 5) = (2/3)(x + 6)

 3(y + 5) = 2(x + 6)

 2x - 3y = 3

The general form of a line:

y = mx + b

You are given an m, x, and y.  Substitute and solve for the remaining unknown:

-5 = (2/3)(-6) + b

-5 = -4 + b

-1 = b

Your equation is:

y = (2/3)x - 1

Point of passage: (-6, -5)

Slope: 2/3

Equation: y = (2/3)x -1

Solution to your problem is --

Slope m of the line = 2/3

Let the line be  y = mx + c

=> y = (2/3) x + c ................ (1)

Since this line passes through (-6, -5 )

=>  - 5 = (2/3) (-6)  +  c,  multiply both sides by 3

=>  - 15  =  - 12 + 3 c 

=> - 3  =  3 c

=>  c =  - 1

Substitute this value of c in (1) above 

=>  y  =  (2/3) x  -  1  

This first form is called the point-slope form. It is a good place to start if that is what you have, a point and the slope. From there you can change it to other forms.

y + 5 = (2/3)(x + 6) ... point-slope form

3y + 15 = 2(x + 6)

3y + 15 = 2x + 12

2x - 3y - 3 = 0 ... general form

a line with a slope of 2/3 has an equation of 

y = (2/3)x + k

use the point to find k

–5 =  (2/3)(–6) + k

k = –5 + 4 = –1

equation is

y = (2/3)x – 1

or

3y = 2x – 3