document.write( "Question 831882: Hi, \r
\n" ); document.write( "\n" ); document.write( "Any help with these questions would be gratefully received.
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\n" ); document.write( "Many thanks, \r
\n" ); document.write( "\n" ); document.write( "Hugh. \r
\n" ); document.write( "\n" ); document.write( "Question 1. \r
\n" ); document.write( "\n" ); document.write( "Find the gradient of the straight line with equation 16x + 4y = 2. \r
\n" ); document.write( "\n" ); document.write( "Question 2. \r
\n" ); document.write( "\n" ); document.write( "b = 3-a/2 \r
\n" ); document.write( "\n" ); document.write( "Rearranged the above equation to make a the subject of the equation. The options are below.\r
\n" ); document.write( "\n" ); document.write( "a = -2b + 3  \r
\n" ); document.write( "\n" ); document.write( "a = -b + 6  \r
\n" ); document.write( "\n" ); document.write( "a = -b + 3  \r
\n" ); document.write( "\n" ); document.write( "a = b + 6  \r
\n" ); document.write( "\n" ); document.write( "a = 2b + 6  \r
\n" ); document.write( "\n" ); document.write( "a = -2b + 6 \r
\n" ); document.write( "\n" ); document.write( "a = b + 3  \r
\n" ); document.write( "\n" ); document.write( "a = 2b + 3  \r
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Algebra.Com's Answer #501703 by hovuquocan1997(83)\"\" \"About 
You can put this solution on YOUR website!
QUestion 1:
\n" ); document.write( "IN question one you need to change the equation into standard form in order to find the gradient
\n" ); document.write( "\"16x+%2B+4y+=+2\"
\n" ); document.write( "\"4y+=+-16x+%2B+2\"
\n" ); document.write( "\"y+=+-4x+%2B+1%2F2\"
\n" ); document.write( "So the gradient is -4
\n" ); document.write( ":D \r
\n" ); document.write( "\n" ); document.write( "Question 2:
\n" ); document.write( "\"b+=+3+-+a%2F2\"
\n" ); document.write( "\"b+%2B+a%2F2+=+3\"
\n" ); document.write( "\"a%2F2+=+-b+%2B+3\"
\n" ); document.write( "\"a+=+-2b+%2B+6\"
\n" ); document.write( "TA-DAH
\n" ); document.write( ":D
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