Find the equation of the straight line passing through 4,1 with a gradient of 1

y = x - 3

Step-by-step explanation:

If the gradient (slope) of the line is m, then the point-slope form of the appropriate equation  is y - k = m(x - h).  If the gradient is 1, then

                                         y - k =   x - h

Since this line passes through (4, 1), we have

                                          y - 1 = x - 4, or

                                            y = x - 3

8(-10x+3.5y-7)

8×-10x + 8×3.5y + 8×-7

-80x + 28y - 56

therefore

-80x+28y-56 is the correct answer

idk

step-by-step explanation: