How do I solve this equation?

label Algebra
account_circle Unassigned
schedule 1 Day
account_balance_wallet $5


Oct 20th, 2015

Thank you for the opportunity to help you with your question!

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


open the brackets by multiplication.ythis gives you,


collect the like terms together and solve them.


simplify the right hand equation by dividing all through by 3



write the equation to solve.


using quadratic fomula

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

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


Please let me know if you need any clarification. I'm always happy to answer your questions.
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




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


 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 


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

Did you know? You can earn $20 for every friend you invite to Studypool!
Click here to
Refer a Friend
Oct 20th, 2015
Oct 20th, 2015
Jun 28th, 2017
Mark as Final Answer
Unmark as Final Answer
Final Answer

Secure Information

Content will be erased after question is completed.

Final Answer