document.write( "Question 173790: Find the solution to the system by the addition (elimination)method\r
\n" ); document.write( "\n" ); document.write( "16x-6y=-28 (1)
\n" ); document.write( "4x+12y=14 (2) \n" ); document.write( "

Algebra.Com's Answer #128666 by SanDiegoMath(2) $\"\"$ $\"About$

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(1) 16x-6y = -28
\n" ); document.write( "(2) 4x+12y = 14\r
\n" ); document.write( "\n" ); document.write( "Dividing (2)'s coefficients in half:\r
\n" ); document.write( "\n" ); document.write( "(1) 16x-6y = -28
\n" ); document.write( "(3) 2x+6y = 7\r
\n" ); document.write( "\n" ); document.write( "Adding (1) and (3) cancels the 'y' term:\r
\n" ); document.write( "\n" ); document.write( "(4) 18x = -21\r
\n" ); document.write( "\n" ); document.write( "Dividing by 18 we get the value of x:\r
\n" ); document.write( "\n" ); document.write( "(5) x = -21/18 = -7/6\r
\n" ); document.write( "\n" ); document.write( "Substituting into (3):\r
\n" ); document.write( "\n" ); document.write( "(6) 2(-7/6) + 6y = 7
\n" ); document.write( "or -7/3 + 6y = 7\r
\n" ); document.write( "\n" ); document.write( "solving for y:\r
\n" ); document.write( "\n" ); document.write( "6y = 7 + 7/3 = (21/3) + (7/3) = 28/3\r
\n" ); document.write( "\n" ); document.write( "y = 28/18 = 14/9\r
\n" ); document.write( "\n" ); document.write( "Solution: x=-7/6 and y =14/9
\n" ); document.write( "Verifying:\r
\n" ); document.write( "\n" ); document.write( "(1) 16x-6y = 16(-7/6) - 6(14/9) = -(56/3)-(28/3) = -84/3 = -28 , it works\r
\n" ); document.write( "\n" ); document.write( "(2) 4x+12y = 4(-7/6) + 12(14/9) = -(14/3) + (56/3) = (56-14)/3 = 42/3 = 14 , it works, too. \n" ); document.write( "

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