document.write( "Question 65155: I'm really confused with the process that I'm supposed to follow. I am finding it difficult to even know where to start.
\n" ); document.write( "The first problem is:
\n" ); document.write( "-x+y=2
\n" ); document.write( "_______
\n" ); document.write( " x+y=4
\n" ); document.write( "The only thing I know for sure is that I'm supposed to use the substitution method but what am I substituting, with what and why.
\n" ); document.write( "I really do appreciate whatever help you can give me.
\n" ); document.write( "Thank you
\n" ); document.write( "

Algebra.Com's Answer #45642 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
-x+y=2
\n" ); document.write( "_______
\n" ); document.write( "x+y=4
\n" ); document.write( "The only thing I know for sure is that I'm supposed to use the substitution method but what am I substituting, with what and why.
\n" ); document.write( "-------
\n" ); document.write( "It's like saying \"my brother is 2 years older than I am\".
\n" ); document.write( "If the person knows how old you are they can figure out
\n" ); document.write( "how old your brother is.
\n" ); document.write( "---------
\n" ); document.write( "So you solve one of the equations for one of the variables:
\n" ); document.write( "If x+y=4 then x=4-y
\n" ); document.write( "Now we know how y is related to x.
\n" ); document.write( "So, inplace of x in the other equation we put what x is equal to, as follows:
\n" ); document.write( "-(4-y)+y=2
\n" ); document.write( "Then we can figure out what Y is, as follows:
\n" ); document.write( "-4+y+y=2
\n" ); document.write( "-4+2y=2
\n" ); document.write( "2y=6
\n" ); document.write( "y=3
\n" ); document.write( "Now we know what y equals, so substitute back and find out what x is:
\n" ); document.write( "If x=4-y then x=4-3=1
\n" ); document.write( "Now we know that x=1 and y=3
\n" ); document.write( "Tha's it; you've found the value of both variables for these two equations:
\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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