If the straight line y=mx+3 pass though the points (3,-1) find the value of gradient?

To find the gradient (m) of the straight line y=mx+3 passing through the point (3,-1), we can substitute the coordinates of the point into the equation and solve for m.

Given point: (3,-1)

Substitute the coordinates into the equation:
-1 = m(3) + 3

Simplify the equation:
-1 = 3m + 3
-1 - 3 = 3m
-4 = 3m

Divide by 3 to solve for m:
m = -4/3

Therefore, the gradient of the straight line passing through the point (3,-1) is -4/3.

Given Points:

(3, -1)

Equation of the line:

y = mx + 3

Substituting the given point into the equation:

-1 = m(3) + 3
-1 = 3m + 3
-4 = 3m
m = -4/3

Therefore, the value of the gradient (m) is -4/3.