document.write( "Question 89993: Solve the system below using graphing. Write down the solution as an ordered pair.
\n" ); document.write( "x+2y= -4
\n" ); document.write( "x-y=5\r
\n" ); document.write( "\n" ); document.write( "Solution( , ) \r
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\n" ); document.write( "\n" ); document.write( "The perimeter of a rectangular field is 460 yards. If the length is 30 yards more than the width, find the dimensions of the field.
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Algebra.Com's Answer #65355 by checkley75(3666)\"\" \"About 
You can put this solution on YOUR website!
X+2Y=-4 OR 2Y=-X-4 OR Y=-X/2-4/2 OR Y=-X/2-2 (RED LINE)
\n" ); document.write( "X-Y=5
\n" ); document.write( "-Y=-X+5 OR Y=X-5 (GREEN LINE)
\n" ); document.write( "\"+graph%28+300%2C+200%2C+-6%2C+5%2C+-10%2C+10%2C+y+=+-x%2F2+-2%2C+y+=+x+-5%29+\" (graph 300x200 pixels, x from -6 to 5, y from -10 to 10, of TWO functions y = -x/2 -2 and y = x -5).
\n" ); document.write( "ANSWER (2,-3).
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\n" ); document.write( "L=W+30 THEN THE PERIMETERIS:
\n" ); document.write( "2(W+30)+2W=460
\n" ); document.write( "2W+60+2W=560
\n" ); document.write( "4W=560-60
\n" ); document.write( "4W=500
\n" ); document.write( "W=500/4
\n" ); document.write( "W=125 YDS.
\n" ); document.write( "L=125=30
\n" ); document.write( "L=155 YDS.
\n" ); document.write( "PROOF
\n" ); document.write( "2*125+2*155=560
\n" ); document.write( "250+310=560
\n" ); document.write( "560=560
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