Question

Change the equation to standard form the state from x and y intercepts

Answer 1

We will have the following:

$y=\frac{3}{2}x-\frac{1}{2}\Rightarrow(2)y=(\frac{3}{2}x-\frac{1}{2})(2)$$\Rightarrow2y=3x-1\Rightarrow3x-2y=1$

So, the equation in standard form is:

$3x-2y=1$

And now, we determine the x & y intercepts:

x-intercept:

$3x-2(0)=1\Rightarrow3x=1$$\Rightarrow x=\frac{1}{3}$

So, the x-intercept is located at:

$(\frac{1}{3},0)$

y-intercept:

$3(0)+2y=1\Rightarrow2y=1$$\Rightarrow y=\frac{1}{2}$

So, the y-intercept is located at:

$(0,\frac{1}{2})$