Answer:
c(x-3)-4=b(x+7)
(cx-c•3)-4=b(x+7)
cx-3c-4=b(x+7)
cx-3c-4=bx+b•7
-3c-7b-4=bx-cx
x(b-c)=-3c-7b-4
x(b-c)/b-c = -3c-7b-4/b-c
x=-3c+7b+4/b-c
A straight line passing through the point (-2,1) and having a gradient of -3 yields the equation y = -3x - 5.
We know that a straight line is an infinitely long line with no curves on it. A straight line's equation is
y = mx + c...(1), where m is the gradient of the straight line and c is a constant.
Given that the gradient of the given straight line = m = -3.
Putting this value in (1), we get
y = -3x + c ...(2)
Again, the given straight line passes through the point (-2,1). So, we can put x = -2 and y = 1 to get the value of the constant c.
So, 1 = (-3)(-2) + c
i.e. 6 + c = 1
i.e. c = 1 - 6 = -5
(2) can be written as
y = -3x - 5
Therefore the equation of the given straight line is
y = -3x - 5
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Answer:
THe y intercept is 30 and it represents the registration fee
Step-by-step explanation:
