document.write( "Question 629004: I have three cousins on my dad's side. The product of their ages is 84. Next year the youngest will be 1/4 the age of the middle cousin. In 2 years from now, the oldest will be twice the current age of the middle. What is the age of my older cousin? \n" ); document.write( "

Algebra.Com's Answer #395995 by dfrazzetto(283) $\"\"$ $\"About$

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cousins = x,y,z\r
\n" ); document.write( "\n" ); document.write( "xyz = 84\r
\n" ); document.write( "\n" ); document.write( "x+1 = 1/4(y+1)\r
\n" ); document.write( "\n" ); document.write( "z+2 = 2y\r
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\n" ); document.write( "\n" ); document.write( "z = 2y - 2\r
\n" ); document.write( "\n" ); document.write( "x = y/4 - 3/4\r
\n" ); document.write( "\n" ); document.write( "plug into first equation;\r
\n" ); document.write( "\n" ); document.write( "(y/4 - 3/4)(y)(2y-2) = 84
\n" ); document.write( "y^2/4 - 3y/4 (2y - 2) = 84
\n" ); document.write( "((y^2 - 3y) / 4) (2y - 2) = 84
\n" ); document.write( "2y^3 - 2y^2 - 6y^2 + 6y = 84 * 4\r
\n" ); document.write( "\n" ); document.write( "2y^3 - 8y^2 + 6y = 336\r
\n" ); document.write( "\n" ); document.write( "y = 7\r
\n" ); document.write( "\n" ); document.write( "x = 1\r
\n" ); document.write( "\n" ); document.write( "z = 12\r
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