Question 262455
{{{system(2x + 3y = 5,
4x - 9y = 9)}}}
<pre><font size = 4 color = "indigo"><b>
Substitution is the worst way to solve a system which
does not have any letters with coefficient of 1 or -1.
Such systems are much easier solved by elimination.

However, since you specified to do it by substitution,
I will do so, but it will be much more difficult, as
it will involve lots of work with fractions. Here goes:

Solve the first equation for x

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

Substitute {{{(5-3y)/2}}} for {{{x}}}

{{{4x - 9y = 9}}}
{{{4((5-3y)/2)-9y=9}}}

Cancel the 2 into the 4 and get a 2 where the 4 is

{{{""^2}}}{{{cross(4)((5-3y)/cross(2))-9y=9}}}

{{{2(5-3y)-9y=9}}}

Distribute to remove the parentheses:

{{{10-6y-9y=9}}}

Combine like terms:

{{{10-15y=9}}}

Add -10 to both sides:

{{{-15y=-1}}}

Divide both sides by {{{-15}}}

{{{(-15y)/(-15)=(-1)/(-15)}}}

Cancel the {{{-15}}}'s, and simplify the right side

{{{(cross(-15)y)/(cross(-15))=1/15}}}

{{{y=1/15}}}

Now we substitute {{{1/15}}} for {{{y}}} into

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

{{{x=(5-3(1/15))/2}}}

Cancel the 3 into the {{{15}}}

{{{x=(5-cross(3)(1/cross(15)[5]))/2}}}

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

Simplify the compound fraction by multiplying
the numerator and denominator by the LCD of 5

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

Distribute the 5 into the parentheses, and write 10 for the
denominator {{{5*2}}}:

{{{x=(5*5-5*(1/5))/10}}}

Write 25 for 5*5 and cancel the 5's in the second
term on top:

{{{x=(25-cross(5)*(1/cross(5)))/10}}}

{{{x=(25-1)/10}}}

{{{x=24/10}}}

We reduce the fraction:

{{{x=12/5}}}

So the solution is {{{x=12/5}}}, {{{y=1/15}}}

But that is by far the hardest way to solve a 
system of equations which has none of the terms
x, -x, y, or -y.  The easist way would be the
elimination method.  

Edwin</pre>