The equation of a straight line can be written in the form 3x + 2y - 8 = 0.
(a) Rearrange this equation to make y the subject.
(b) Write down the gradient of the line.
(c) Write down the coordinates of the point where
the line crosses the y-axis.
(pls give a full working )

see explanation

Step-by-step explanation:

Given

(a)

3x + 2y - 8 = 0 ( subtract 3x - 8 from both sides )

2y = - 3x + 8 ( divide the terms by 2 )

y = - x + 4

(b)

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = - x + 4 ← is in slope- intercept form

with slope ( gradient) m = - and

(c)

c = 4

That is crosses the y- axis at (0, 4 )

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