document.write( "Question 665440: How do I use the substitution method to solve the following problem: 3x^2-20y^2-12x+80y-96=0 and 3x^2+20y^2=80y+48? \n" ); document.write( "

Algebra.Com's Answer #413870 by mananth(16946) $\"\"$ $\"About$

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3x^2-20y^2-12x+80y-96=0 \r
\n" ); document.write( "\n" ); document.write( "3x^2-20y^2-12x+(80y+48) -144=0 \r
\n" ); document.write( "\n" ); document.write( "Substitute 80y+48 by 3x^2+20y^2 \r
\n" ); document.write( "\n" ); document.write( "3x^2-20y^2-12x+3x^2+20y^2-144=0\r
\n" ); document.write( "\n" ); document.write( "6x^2+12x-144=0\r
\n" ); document.write( "\n" ); document.write( "/6
\n" ); document.write( "x^2+2x-24=0\r
\n" ); document.write( "\n" ); document.write( "(x+6)(x-4)=0
\n" ); document.write( "x=-6 or 4\r
\n" ); document.write( "\n" ); document.write( "Plug the values of x to get the values of y\r
\n" ); document.write( "\n" ); document.write( "and 3x^2+20y^2=80y+48?\r
\n" ); document.write( "\n" ); document.write( "x=4
\n" ); document.write( "3(16)+20y^2=80y+48
\n" ); document.write( "48+20y^2=80y+48
\n" ); document.write( "20y^2=80y
\n" ); document.write( "/y
\n" ); document.write( "20y=80
\n" ); document.write( "/20
\n" ); document.write( "y=4\r
\n" ); document.write( "\n" ); document.write( "Similarly find the other value of y \n" ); document.write( "

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