solving equation and making subject

[mathismaths]

Q1.So im practicing and find this question to make "n" the subject
i take the LCM as a & √ n but im still getting the wrong answer

Q2.in the question below you have to solve the equation
i m not able to do this. im taking LCM as j+4 and j+3


would appreciate steps to understand how u derive answer

 

[Steven G]

Can we see your work so that we can help you get the correct results.

For the record I do not think that the lcm will be a and sqrt(n). Either you mean that the lcm is a and sqrt(n) and I have no idea how to use and as part of the lcm OR you mean that the lcm can be a or sqrt(n). This can't be as there is just one lcm. This is why we ask to see your work so we know what you are trying.

 

[mathismaths]

Can we see your wok so that we can help you get the correct results.

For the record I do not think that the lcm will be a and sqrt(n). Either you mean that the lcm is a and sqrt(n) and I have no idea how to use and as part of the lcm OR you mean that the lcm can be a or sqrt(n). This can't be as there is just one lcm. This is why we ask to see your work so we know what you are trying.

LCM that im taking is both denominators ie, a (multiply by) sqrt(n) .
this is most im able to do.im going further but the answer im getting is not right so didnt post more work.


also for second question am i taking the correct LCM?

 

[topsquark]

If you had the equation x + 3 = (2)(3)x how would you solve it?

For the second problem, yes.

-Dan

 

[mathismaths]

If you had the equation x + 3 = (2)(3)x how would you solve it?

x=3/5

For the second problem, yes.

thn how to remove the denominator of first fraction 1/(j+2): if the LCM is (j+4) (j+3) thn for first fraction the denominator in 1/j+2 gets 1/j+2 leaving the remaining as j+2 j+3? correct, if not then how because only when denominators are eliminated the equation can be formed. im sorry if im being dumb

 

[Steven G]

LCM that im taking is both denominators ie, a (multiply by) sqrt(n) .
View attachment 18026this is most im able to do.im going further but the answer im getting is not right so didnt post more work.


also for second question am i taking the correct LCM?

Now bring all the terms that have sqrt(n) to one side of the equal sign and bring everything else to the other side. Do just what you did in solving x + 3 = (2)(3)x for x. Please post back.

 

[Steven G]

x=3/5



thn how to remove the denominator of first fraction 1/(j+2): if the LCM is (j+4) (j+3) thn for first fraction the denominator in 1/j+2 gets 1/j+2 leaving the remaining as j+2 j+3? correct, if not then how because only when denominators are eliminated the equation can be formed. im sorry if im being dumb
View attachment 18029

You never said what the goal is for this problem. Are you trying to solve for j or j+7???

 

[mathismaths]

Now bring all the terms that have sqrt(n) to one side of the equal sign and bring everything else to the other side. Do just what you did in solving x + 3 = (2)(3)x for x. Please post back.

pretty sure doing it wrong i was able to do uptil now even before bt how to proceed

You never said what the goal is for this problem. Are you trying to solve for j or j+7???

to solve the equation for j

 

[mathismaths]

You never said what the goal is for this problem. Are you trying to solve for j or j+7???

thats wat i have done pretty sure equation is coming up wrong

 

[Deleted member 4993]

thats wat i have done pretty sure equation is coming up wrong
View attachment 18032

What you have is a quadratic equation? Does that ring a bell?

If you do not remember, please use google to review. After that - please tell us if you are still stuck.

 

[mathismaths]

What you have is a quadratic equation? Does that ring a bell?

If you do not remember, please use google to review. After that - please tell us if you are still stuck.

i do know that but the answer to this is -1 which im not getting, also the equation seems wrong is it correct?

 

[Cubist]

thats wat i have done pretty sure equation is coming up wrong
View attachment 18032

There's a mistake on your first line...

[math]\frac{j+4}{j+2} ≠ j+2[/math]
Try putting j=10 in

[math]\frac{10+4}{10+2} = \frac{14}{12} = \frac{7}{6} ≠ 10+2[/math]

 

[Cubist]

Can you think of something else to multiply through by? Perhaps even longer than (j+4)(j+3)

 

[Steven G]

View attachment 18031 pretty sure doing it wrong i was able to do uptil now even before bt how to proceed


to solve the equation for j

Any reason why you divided 2 of the 3 terms by y. Any reason you did not use my hint to just bring all the terms that had sqrt(n) to one side (before dividing). I again suggest that you go back to the last equation you had and bring the sqrt(n) to one side. Then see what you can do from there.

 

[mathismaths]

Can you think of something else to multiply through by? Perhaps even longer than (j+4)(j+3)

nope, i know im doing the first fraction [MATH]1/j+2[/MATH] wrong although the other 2 fractions were done correct. right?

Any reason why you divided 2 of the 3 terms by y. Any reason you did not use my hint to just bring all the terms that had sqrt(n) to one side (before dividing). I again suggest that you go back to the last equation you had and bring the sqrt(n) to one side. Then see what you can do from there.

alright i think i got the first question only 2nd remains

 

[Cubist]

although the other 2 fractions were done correct. right?

(Regarding the 2^nd question)

Your other two fractions were done perfectly!

FYI I think the answer to this will become a bit "messy" in the middle. There may be a few lines where the equation takes up a lot of room on a long line before things start to cancel out and you end up with a simple result. You just need to start by multiplying through by "(j+4) * (j+3) * <something else>"

 

[mathismaths]

(Regarding the 2^nd question)
FYI I think the answer to this will become a bit "messy" in the middle. There may be a few lines where the equation takes up a lot of room on a long line before things start to cancel out and you end up with a simple result. You just need to start by multiplying through by "(j+4) * (j+3) * <something else>"

explain to me how to solve first fraction thn? [MATH]1/j+2[/MATH] * (j+4)(j+3) how to cancel the denominator ? if you multiply all three it will still be a fraction unless the denominator is cancelled entirely.so how to go about solving that?

 

[Cubist]

explain to me how to solve first fraction thn? [MATH]1/j+2[/MATH] * (j+4)(j+3) how to cancel the denominator ? if you multiply all three it will still be a fraction unless the denominator is cancelled entirely.so how to go about solving that?

"(j+4) * (j+3)" is not the correct thing to multiply by (if your goal is to remove all denominators in the equation in one step)

You need to multiplying through by "(j+4) * (j+3) * <something else>".

Can you think of a "<something else>" that will cancel out the denominator of 1/(j+2)?

 

[mathismaths]

"(j+4) * (j+3)" is not the correct thing to multiply by (if your goal is to remove all denominators in the equation in one step)

You need to multiplying through by "(j+4) * (j+3) * <something else>".

Can you think of a "<something else>" that will cancel out the denominator of 1/(j+2)?

no im hitting the wall here.not sure how to proceed, thats y im asking how to do it

 

[mathismaths]

"(j+4) * (j+3)" is not the correct thing to multiply by (if your goal is to remove all denominators in the equation in one step)

You need to multiplying through by "(j+4) * (j+3) * <something else>".

Can you think of a "<something else>" that will cancel out the denominator of 1/(j+2)?

well i think j+2 can cancel out J+2, am i right?