document.write( "Question 30639: (1)solve the inequality: 5<1-3x(lesser than or equal to 10\r
\n" ); document.write( "\n" ); document.write( "(2) slove for y:6y^2+y=12
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Algebra.Com's Answer #17379 by sdmmadam@yahoo.com(530)\"\" \"About 
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1)Solve the inequality: 5<1-3x (less than or equall to) 10
\n" ); document.write( "5 < 1-3x < = 10 ----(1)
\n" ); document.write( "Consider 5 < 1-3x
\n" ); document.write( "5 - 1 <-3x
\n" ); document.write( "4 < -3x
\n" ); document.write( "That is -3x > 4
\n" ); document.write( "Dividing by (-3) and since division by a negative quantity alters the inequality, the greater than becoms less than
\n" ); document.write( "x < (-4)/3 ----(*)
\n" ); document.write( "Consider (1-3x) < = 10
\n" ); document.write( "1 - 10 <= 3x
\n" ); document.write( "-9 <= 3x
\n" ); document.write( "dividing by 3(>0) and since division by a positive quantity does not alter the inequality the less than remains less than
\n" ); document.write( "-3 <= x ----(**)
\n" ); document.write( "From (**) and (*), we conclude that
\n" ); document.write( "Answer: -3 <= x < (-4)/3 \r
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\n" ); document.write( "\n" ); document.write( "(2) slove for y:6y^2+y=12----(1)
\n" ); document.write( "6y^2+y-12 =0
\n" ); document.write( "6y^2+(9y-8y)-12= 0
\n" ); document.write( "(the sum is 1 and the product is (6)X(-12) = -72
\n" ); document.write( "and therefore the quantities are +9 and -8)
\n" ); document.write( "(6y^2+9y)-8y-12= 0 (by additive associativity)
\n" ); document.write( "3y(2y+3)-4(2y+3) = 0
\n" ); document.write( "3yp-4p = 0 (where p = (2y+3)----(*))
\n" ); document.write( "p(3y-4) = 0
\n" ); document.write( "(2y+3)(3y-4) = 0
\n" ); document.write( "(2y+3) =0 gives y =-3/2
\n" ); document.write( "(3y-4) =0 gives y = 4/3
\n" ); document.write( "Answer: x = -3/2 and x = 4/3
\n" ); document.write( "Verification: x =-3/2 in 6y^2+y=12 ----(1)
\n" ); document.write( "LHS = 6y^2+y = 6X(9/4)-3/2 = 27/2-3/2 =24/2 = 12 = RHS
\n" ); document.write( "x =4/3 in 6y^2+y=12 ----(1)
\n" ); document.write( "LHS = 6y^2+y = 6X(16/9)+4/3 = 32/3+4/3 =36/3 = 12 = RHS
\n" ); document.write( "Therefore our values are correct.\r
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