document.write( "Question 77827: I also have another question, here it is. It is on the substitution method. Here it is 3b-2a=4 and b/2-2a/3=1 \n" ); document.write( "
Algebra.Com's Answer #55800 by uma(370)\"\" \"About 
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\n" ); document.write( "3b - 2a = 4\r
\n" ); document.write( "\n" ); document.write( "==> 3b = 4 + 2a [adding 2a to both the sides]\r
\n" ); document.write( "\n" ); document.write( "==> b = (4+2a)/3\r
\n" ); document.write( "\n" ); document.write( "Plugging in this value in the second equation,\r
\n" ); document.write( "\n" ); document.write( "b/2 - 2a/3 = 1\r
\n" ); document.write( "\n" ); document.write( "==> (4+2a)/3*2 - 2a/3 = 1\r
\n" ); document.write( "\n" ); document.write( "==> 2(2+a)/6 - 2a/3 = 1 [factoring 4+2a]\r
\n" ); document.write( "\n" ); document.write( "==> 2+a/3 - 2a/3 = 1 [cancelling 2 in the first term]\r
\n" ); document.write( "\n" ); document.write( "==> 2 + a - 2a = 3 [multiply by 3]\r
\n" ); document.write( "\n" ); document.write( "==> 2 - a = 3\r
\n" ); document.write( "\n" ); document.write( "==> -a = 3 - 2\r
\n" ); document.write( "\n" ); document.write( "==> -a = 1\r
\n" ); document.write( "\n" ); document.write( "==> a = -1\r
\n" ); document.write( "\n" ); document.write( "so b = 4+2(-1)/3\r
\n" ); document.write( "\n" ); document.write( "==> b = 2/3\r
\n" ); document.write( "\n" ); document.write( "Thus a = -1 amd b= 2/3\r
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\n" ); document.write( "\n" ); document.write( "Good luck
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