document.write( "Question 1116924: A) given that x+y=42, y+z=67 and x+z=55, find the value of x+y+z.
\n" ); document.write( "B) hence, find the values of x, y and z \n" ); document.write( "

Algebra.Com's Answer #731831 by Fombitz(32388) $\"\"$ $\"About$

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1. $\"x%2By=42\"$
\n" ); document.write( "2. $\"y%2Bz=67\"$
\n" ); document.write( "3. $\"x%2Bz=55\"$
\n" ); document.write( "Adding them all together,
\n" ); document.write( " $\"x%2By%2By%2Bz%2Bx%2Bz=42%2B67%2B55\"$
\n" ); document.write( " $\"2x%2B2y%2B2z=164\"$
\n" ); document.write( " $\"x%2By%2Bz=82\"$
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\n" ); document.write( "From 3,
\n" ); document.write( " $\"z=55-x\"$
\n" ); document.write( "Substituting into 2,
\n" ); document.write( " $\"y%2B%2855-x%29=67\"$
\n" ); document.write( "4. $\"-x%2By=12\"$
\n" ); document.write( "Adding 1 and 4,
\n" ); document.write( " $\"x%2By-x%2By=42%2B12\"$
\n" ); document.write( " $\"2y=54\"$
\n" ); document.write( "Solve for y then use the other equations to solve for x and z.
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