document.write( "Question 164983: solve by substitution:
\n" ); document.write( "\"sqrt%28x%29-y=0\"
\n" ); document.write( "\"2x%2By=3\"
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Algebra.Com's Answer #121601 by Alan3354(69443)\"\" \"About 
You can put this solution on YOUR website!
solve by substitution:
\n" ); document.write( "\"sqrt%28x%29-y=0\"
\n" ); document.write( "\"2x%2By=3\"
\n" ); document.write( "-------------
\n" ); document.write( "\"sqrt%28x%29-y=0\"
\n" ); document.write( "\"sqrt%28x%29+=+y\"
\n" ); document.write( "square both sides
\n" ); document.write( "x = y^2
\n" ); document.write( "Sub into the 2nd eqn
\n" ); document.write( "\"2x%2By=3\"
\n" ); document.write( "\"2y%5E2%2By=3\"
\n" ); document.write( "\"2y%5E2%2By-3+=+0\"
\n" ); document.write( "\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation \"ax%5E2%2Bbx%2Bc=0\" (in our case \"2x%5E2%2B1x%2B-3+=+0\") has the following solutons:
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\n" ); document.write( " \"x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca\"
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\n" ); document.write( " For these solutions to exist, the discriminant \"b%5E2-4ac\" should not be a negative number.
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\n" ); document.write( " First, we need to compute the discriminant \"b%5E2-4ac\": \"b%5E2-4ac=%281%29%5E2-4%2A2%2A-3=25\".
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\n" ); document.write( " Discriminant d=25 is greater than zero. That means that there are two solutions: \"+x%5B12%5D+=+%28-1%2B-sqrt%28+25+%29%29%2F2%5Ca\".
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\n" ); document.write( " \"x%5B1%5D+=+%28-%281%29%2Bsqrt%28+25+%29%29%2F2%5C2+=+1\"
\n" ); document.write( " \"x%5B2%5D+=+%28-%281%29-sqrt%28+25+%29%29%2F2%5C2+=+-1.5\"
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\n" ); document.write( " Quadratic expression \"2x%5E2%2B1x%2B-3\" can be factored:
\n" ); document.write( " \"2x%5E2%2B1x%2B-3+=+%28x-1%29%2A%28x--1.5%29\"
\n" ); document.write( " Again, the answer is: 1, -1.5.\n" ); document.write( "Here's your graph:
\n" ); document.write( "\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B1%2Ax%2B-3+%29\"

\n" ); document.write( "\n" ); document.write( "These are the values for y - the solver always uses x as the variable.
\n" ); document.write( "The solver has a problem sometimes.
\n" ); document.write( "It's (y-1)*(2y+3), the answers shown are correct.
\n" ); document.write( "So for y = 1, x = 1
\n" ); document.write( "For y = -3/2, x = 9/4\r
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