document.write( "Question 25543: How do you solve this equation OF Linear equations by substitution :
\n" ); document.write( "y=3-2x y=2-3x \n" ); document.write( "

Algebra.Com's Answer #13694 by rapaljer(4671) $\"\"$ $\"About$

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Since y=3-2x, and y=2-3x, you can substitute the y = of the first equation into the y = of the second equation, and you have:\r
\n" ); document.write( "\n" ); document.write( "3-2x= 2-3x\r
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\n" ); document.write( "\n" ); document.write( "Add +3x to each side:
\n" ); document.write( "3-2x + 3x = 2 -3x + 3x
\n" ); document.write( "3+x = 2\r
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\n" ); document.write( "\n" ); document.write( "Subtract 3 from each side:
\n" ); document.write( "x=-1\r
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\n" ); document.write( "\n" ); document.write( "Substitute back into the first equation:
\n" ); document.write( "y = 3-2x
\n" ); document.write( "y = 3-2(-1)
\n" ); document.write( "y = 5\r
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\n" ); document.write( "\n" ); document.write( "Check in the second equation:
\n" ); document.write( "y = 2-3x
\n" ); document.write( "5 = 2-3(-1)
\n" ); document.write( "5=5
\n" ); document.write( "It checks!\r
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\n" ); document.write( "\n" ); document.write( "The solution is the point (-1,5).\r
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\n" ); document.write( "\n" ); document.write( "R^2 at SCC \n" ); document.write( "

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