##### How do I solve this equation?

 Algebra Tutor: None Selected Time limit: 1 Day

(30/3x-6)+(1/x-2)=11/3

Oct 20th, 2015

First eliminate the denominator on the left hand side by multiplying them on both side, to obtain

30(x-2)+1(3x-6)=11/3(3x-6)*(x-2)

open the brackets by multiplication.ythis gives you,

30x-30+3x-6=11/3(3x^2-6x-6x+12

collect the like terms together and solve them.

30x+3x-30-6=11/3(3x^2+12)

simplify the right hand equation by dividing all through by 3

33x-36=11(x^2+4)

33x-36=11x^2+44

write the equation to solve.

11x^2+44-33x+36=0

x=-b+- sqaureroot(b^2-4ac)/2a

x=33+- sqrrooot(33^2-4*11*80)/2*11

x=

Oct 20th, 2015

kindly accept my updated corrections of the question as I realized there was an error in my working  and I will be glad if you accept please.

30/3x-6 + 1/x-2=11/3

eliminate the denominator on the left hand side by multiplying them on both side, to obtain

30(x-2)+1(3x-6)=11/3(3x-6)*(x-2).Open brackets by multiplication to obtain

30x-60+3x-6=11/3(3x^2-6x-6x+12)

30x+3x-60-6=11/3(3x^2-12x+12)

33x-66=11x^2-44x+44

move the left hand side equation to the right hand side to obtain

(11x^2)-44x-33x+44+66=0

NB when a positive sign cross = it becomes negative and when negative sign cross = it becomes positive.

(11x^2) - 77x+110=0, divide by 11 to simplify the equation to

(x^2)-7x+10=0

you find two numbers which when multiplied give us 10 and when added give us -7,

the two numbers are -5 and -2. when multiplied they give us 10 and when added they give us -7.There we substitute them in our equation to obtain.

(x^2) - 2x - 5x+10=0

group the equation into two segments considering where there are terms that can be simplified further with ease as

(x^2- 2x) -(5x+10)=0, factor the like terms as follows

x(x-2)-5(x-2)=0.In our new equation you find that, the terms in brackets(x-2) are similar,Group the terms outside the brackets together to obtain

(x-5)=0 or (x-2)=0

hence value of x can be x=5 or x=2.

you can also use quadratic formula to obtain the values of X

x=(-b+ square-root (b^2 -4ac))/2  or x=(-b - square-root (b^2 -4ac))/2

where a=1, b=-7 and c=10

x=(-(-7)+square-root (-7^2 - 4*1*10))/2 or x=(-(-7) - square-root (-7^2 - 4*1*10))/2

x=(7+square-root (49 -40))/2 or x=(7 - square-root (49 -40))/2

x=(7+ square-root (9) )/2 or x=(7 - square-root (9) )/2

x=(7+3)/2 or x=(7-3)/2

x=5 or x=2

Please accept my correction and if you require more clarification on the same feel free to conduct me and I will be glad to shade more light.

kind regards

Oct 20th, 2015

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Oct 20th, 2015
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Oct 20th, 2015
Dec 7th, 2016
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