Question 154494
Hi, I have a problem with figuring out a linear equation. The question asks me to plot the graph showing the point and slope specified in the equation. Then I must rewrite in slope-intercept form and standard form. 
Here is the equation
{{{y-3=2/5}}}{{{(x-2)}}}
<pre><font size = 4 color = "indigo"><b>
You must learn the three forms of the equation of a line:

1. Point-slope form:

{{{y-y[1]=m(x-x[1])}}}

where the point ({{{x[1]}}},{{{y[1]}}}) is on the line,
and {{{m}}} represents the slope of the line.

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2. Slope-intercept form:

{{{y=mx+b}}}

where {{{m}}} is the slope of the line and (0,{{{b}}})
are the coordinates of the y-intercept.

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3. Standard form:

{{{Ax+By=C}}}

Where A, B, and C are whole numbers, and A is positive.
[Don't confuse the capital B with the small b in the slope-
intercept form.

-----------------------------

{{{y-3=2/5}}}{{{(x-2)}}}

Multiply both sides by 5 to clear of
fractions:

{{{5(y-3)=5}}}·{{{2/5}}}{{{(x-2)}}}

{{{5(y-3)=cross(5)}}}·{{{2/cross(5)}}}{{{(x-2)}}}

{{{5(y-3)=2(x-2)}}}

{{{5y-15=2x-4}}}

{{{-2x+5y=11}}}

Multiply through by -1 to make the first 
coefficient positive:

{{{2x-5y=-11}}}

That is the standard form.

----------------------------

To get the slope-intercept form, solve
that for y:

{{{2x-5y=-11}}}
{{{-5y=-2x-11}}}
{{{(-5y)/(-5)=((-2)/(-5))x-(11/(-5))}}}

{{{y=2/5}}}{{{x+11/5}}}

Edwin</pre>