document.write( "Question 147177: what is the solution of the system?
\n" ); document.write( "4x+5y=2
\n" ); document.write( "8x+10y=4 \n" ); document.write( "

Algebra.Com's Answer #107514 by mangopeeler07(462) $\"\"$ $\"About$

You can put this solution on YOUR website!
To figure this out, use substitution. Solve for x in the first equation. You shoud end up with $\"x=%285y%2B2%29%2F4\"$ . Then plug that into the second equation and after adding like terms get $\"20y%2B4=4\"$ . So y must be zero, since 0+4=4 and 20y+4=4. 20y just takes the place of zero. So y=0. Plug y into the first original equation (4x-5y=2). Get $\"4x-0=2\"$ . So 4x=2. Divide both sides by 4 and get x=2/4 which is reduced to 1/2. So y=0 and x=1/2. Wanna check these values? plug them both back into the second original equation. Did you end up with $\"4%2B0=4\"$ ? See? It works! \n" ); document.write( "

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