You can put this solution on YOUR website! 1)Solve the inequality: 5<1-3x (less than or equall to) 10
5 < 1-3x < = 10 ----(1)
Consider 5 < 1-3x
5 - 1 <-3x
4 < -3x
That is -3x > 4
Dividing by (-3) and since division by a negative quantity alters the inequality, the greater than becoms less than
x < (-4)/3 ----(*)
Consider (1-3x) < = 10
1 - 10 <= 3x
-9 <= 3x
dividing by 3(>0) and since division by a positive quantity does not alter the inequality the less than remains less than
-3 <= x ----(**)
From (**) and (*), we conclude that
Answer: -3 <= x < (-4)/3
(2) slove for y:6y^2+y=12----(1)
6y^2+y-12 =0
6y^2+(9y-8y)-12= 0
(the sum is 1 and the product is (6)X(-12) = -72
and therefore the quantities are +9 and -8)
(6y^2+9y)-8y-12= 0 (by additive associativity)
3y(2y+3)-4(2y+3) = 0
3yp-4p = 0 (where p = (2y+3)----(*))
p(3y-4) = 0
(2y+3)(3y-4) = 0
(2y+3) =0 gives y =-3/2
(3y-4) =0 gives y = 4/3
Answer: x = -3/2 and x = 4/3
Verification: x =-3/2 in 6y^2+y=12 ----(1)
LHS = 6y^2+y = 6X(9/4)-3/2 = 27/2-3/2 =24/2 = 12 = RHS
x =4/3 in 6y^2+y=12 ----(1)
LHS = 6y^2+y = 6X(16/9)+4/3 = 32/3+4/3 =36/3 = 12 = RHS
Therefore our values are correct.
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