<PRE><B>Question 1155473</B>
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I would never solve a system of equations like this using substitution; elimination is far easier.&lt;br&gt;
However, if I did solve it using substitution, I would not solve for y and get an expression involving fractions and then substitute in the other equation.&lt;br&gt;
Remember that, using substitution, you don&#39;t need to solve for y or x -- you can solve for 3y, or 7x, or whatever is convenient.&lt;br&gt;
So here is how I would solve this using substitution.&lt;br&gt;
{{{3x+3 = 5y}}} --&gt; {{{6x+6 = 10y}}}&lt;br&gt;
{{{4x+2y=-4}}} --&gt; {{{20x+10y = -20}}}&lt;br&gt;
Now substitute &quot;6x+6&quot; for &quot;10y&quot; in the second equation:&lt;br&gt;
{{{20x+(6x+6) = -20}}}
{{{26x+6 = -20}}}
{{{26x = 26}}}
{{{x = 1}}}&lt;br&gt;
Then finish from there by any method you choose.&lt;br&gt;

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