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Question

Question: How do you put \(2x - 3y = 6\) in slope-intercept form?...

How do you put 2x3y=62x - 3y = 6 in slope-intercept form?

Explanation

Solution

Hint : The above question is based on the concept of slope-intercept form. The main approach towards solving the equation is by applying the formula of the equation of straight line. So, using this formula we need to write the equation in that form by shifting the terms on the other side.

Complete step-by-step answer :
One of the forms of the equation of a straight line is also called slope intercept form. We know the equation of straight line in slope intercept form is y=mx+cy = mx + c
Where m denotes the slope of the line and c is the y-intercept of the line.
The standard equation of first degree is Ax+By+C=0Ax + By + C = 0 can be written in the slope intercept form as :
y=(AB)x(CB)y = \left( { - \dfrac{A}{B}} \right)x - \left( {\dfrac{C}{B}} \right)
where m=(AB)m = \left( { - \dfrac{A}{B}} \right) and c=(CB)c = \left( { - \dfrac{C}{B}} \right)
and also B0B \ne 0 .
So now the given equation is 2x3y=62x - 3y = 6
So first we need to shift the terms which are on the left-hand side towards the right hand side.
Therefore, we get

2x3y=6 3y=2x+6  \Rightarrow 2x - 3y = 6 \\\ \Rightarrow - 3y = - 2x + 6 \\\

Then by multiplying it with negative sign throughout the equation we get,
3y=2x63y = 2x - 6
Then we need to isolate the term y so we shift the number 3 on the right hand side to the denominator.
y=23x2y = \dfrac{2}{3}x - 2
Hence the standard equation of first degree is written in the above slope intercept form
So, the correct answer is “y=23x2y = \dfrac{2}{3}x - 2 ”.

Note : An important thing to note is that here the c=-2 i.e., the y-intercept of the equation is -2.In the graph we plot the equation we get to know that the equation of line will cut y-axis at -2 and the slope which is 23\dfrac{2}{3} which gives the direction of the line.