Question 200359
<font size = 8 color = "red"><b>Alan didn't know what you were supposed to do. Here's Edwin's solution:</b></font>
<pre><font size = 4 color = "indigo"><b>
1. {{{5x-2y = 1}}}

Solve for x:

{{{5x-2y = 1}}}

Add {{{""+2y}}} to both sides:

{{{5x-2y+2y = 1+2y}}}

{{{5x-cross(2y)+cross(2y) = 1+2y}}}

{{{5x = 1+2y}}}

Divide both sides by {{{5}}}.  This can be
done either of two ways:

1. Divide the whole right side by {{{5}}}

{{{(5x)/5 = (1+2y)/5}}}


{{{(cross(5)x)/cross(5) = (1+2y)/5}}}

{{{x = (1+2y)/5}}}

2. Divide each term individually by {{{5}}}:

{{{(5x)/5 = 1/5+(2y)/5}}}

{{{(cross(5)x)/cross(5)= 1/5+(2y)/5}}}

{{{x=1/5+(2y)/5}}}

Both answers are correct and equivalent.

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Solve for y:

{{{5x-2y = 1}}}

Add {{{-5x}}} to both sides:

{{{5x-5x-2y = 1-5x}}}

{{{cross(5x)-cross(5x)-2y = 1-5x}}}

{{{-2y = 1-5x}}}

Divide both sides by {{{-2}}}.  This can be
done either of two ways:

1. Divide the whole right side by {{{-2}}}

{{{(-2y)/-2 = (1-5x)/-2}}}

{{{(cross(-2)y)/cross(-2) = (1-5x)/-2}}}

{{{y = (1-5x)/-2}}}

2. Divide each term individually by {{{5}}}:

{{{(-2y)/-2 = 1/-2-(5x)/-2}}}

or canceling on the left simplifying the signs
on the right:

{{{(cross(-2)y)/cross(-2)= -1/2+(5x)/2}}}

{{{y=-1/2+(2y)/5}}}

Both answers are correct and equivalent.

2. {{{2x+y = 11}}}

Same way:

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

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


3. 5x - 3y = 1

Same way:

{{{x=(1+3y)/5}}} or {{{x=1/5+(3y)/5}}}

{{{y=(1-5x)/(-3)}}} or {{{y=-1/3+(5x)/3}}}

Edwin</pre>