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Solve the equation

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(x^2+2)/(3x^2-5x+2)+(-3)/(x-1)=-5/(3x-2)
x can't be equal to 1 and x can't be equal to 2/3 (each will make one or more denominators 0.)
(x^2+2)/(3x-2)(x-1)+(-3)/(x-1)=-5/(3x-2)
a common denominator is (3x-2)(x-1)
(x^2+2)/(3x-2)(x-1)+(-3)(3x-2)/(3x-2)(x-1)=-5(x-1)/(3x-2)(x-1)
[x^2+2+(-3)(3x-2)]/(3x-2)(x-1)=-5(x-1)/(3x-2)(x-1)
x^2+2-9x+6=-5x+5
x^2+2-9x+6+5x-5=0
x^2-4x+3=0
(x-3)(x-1)=0
x-3=0 and x-1=0
x=3 and x=1(in the beginning we said x cannot be 1)
x=3 is the only solution
check: (3^2+2)/(3*3^2-5*3+2)+(-3)/(3-1)=-5/(3*3-2)
11/14+(-3/2)=-5/7
11/14+(-21/14)=-5/7
-10/14=-5/7
-5/7=-5/7