Ab-surd algebra

[Rudi]

Hi all,

Trying to teach myself maths having not done any for a few years and have been working on surds. Have completed all the questions my book has to offer except two (of the algebra variety) which are currently beyond me. I have the answers in the back of the book and have tried working backwards as well as forwards but to no avail.

The problems are solving y:

y√3 = y + √3 which should give the answer y = (1/2)√3 + 3/2

and

(1/√2) + y = y/√2 which should give y = -√2 -1

If anyone knows how to get from the question to the answer in these problems I would greatly appreciate some pointers.

The presence of a y on each side of the = sign has been tripping me up.

Thanks

 

[Steven G]

Hi all,

Trying to teach myself maths having not done any for a few years and have been working on surds. Have completed all the questions my book has to offer except two (of the algebra variety) which are currently beyond me. I have the answers in the back of the book and have tried working backwards as well as forwards but to no avail.

The problems are solving y:

y√3 = y + √3 which should give the answer y = (1/2)√3 + 3/2

and

(1/√2) + y = y/√2 which should give y = -√2 -1

If anyone knows how to get from the question to the answer in these problems I would greatly appreciate some pointers.

The presence of a y on each side of the = sign has been tripping me up.

Thanks

What would you do if the sqrt(3) was say 7? Then y7 = y +7 or 7y = y +7. Then 7y - 1y =7. Then (7-1)y = 7. Then y = 7/(7-1). Now do the same exact steps with sqrt(3). Then do the same with the next equation.

 

[Ishuda]

A general way to clear square roots from the denominator is to multiply by one, i.e. multiply both denominator and numerator by their, what might call, 'opposite'. That is, for example, if the denominator had an (r-1), it's 'opposite' is (r+1) [and vice versa] and (r - 1)(r + 1) = r² -1. So if r were \(\displaystyle \sqrt{3}\) then r² -1 = 2. Oh, and I assume, for example, that you got
y = \(\displaystyle \frac{\sqrt{3}}{\sqrt{3} -1}\)
for the first answer.

BTW: This extends to \(\displaystyle \sqrt{a} \pm \sqrt{b}\) in the denominator.

 

[Rudi]

I am interested - but unfortunately I cannot spend as much time online as I would like. I'm sure it has something to do with my job - but my continuing education is hopefully my way out of that!

I am also grateful for the assistance. Thank you Jomo and Ishuda. I understand the first question now. Going from 7y - y to y(7 - 1) was the step I could not work out. I think I was also thrown because the answer in the back of the book was clearly wrong. I have been able to get to what I can now see is the correct answer.

Still struggling with the second one:

(1/√2) + y = y/√2 I multiply both sides by √2
1 + y√2 = y Then subtract y√2
1 = y - y√2
1 = y(1 - √2) Then divide (1 - √2)
y = 1 / (1 - √2) What have I done wrong? The answer claims that it should be y = - √2 - 1

 

[HallsofIvy]

I am interested - but unfortunately I cannot spend as much time online as I would like. I'm sure it has something to do with my job - but my continuing education is hopefully my way out of that!

I am also grateful for the assistance. Thank you Jomo and Ishuda. I understand the first question now. Going from 7y - y to y(7 - 1) was the step I could not work out. I think I was also thrown because the answer in the back of the book was clearly wrong. I have been able to get to what I can now see is the correct answer.

Still struggling with the second one:

(1/√2) + y = y/√2 I multiply both sides by √2
1 + y√2 = y Then subtract y√2
1 = y - y√2
1 = y(1 - √2) Then divide (1 - √2)
y = 1 / (1 - √2) What have I done wrong? The answer claims that it should be y = - √2 - 1

You haven't done anything wrong- you have the same answer in a different form. If \(\displaystyle y= \frac{1}{1- \sqrt{2}}\) then "rationalizing the denominator" by multiplying both numerator and denominator by \(\displaystyle 1+ \sqrt{2}\) gives \(\displaystyle \frac{1}{1- \sqrt{2}}\frac{1+ \sqrt{2}}{1+ \sqrt{2}}= \frac{1+ \sqrt{2}}{1- 2}= \frac{1+ \sqrt{2}}{-1}= -\sqrt{2}- 1\).

 

[Rudi]

Thanks! I get it now. The book did explain earlier how to rationalise the denominator but I, stupidly, forgot to do it.

Thank you for the useful tips everyone.

 

[Steven G]

Rationalising the denominator can be (conveniently!) skipped this way:

1/√2 + y = y/√2

multiply by √2
1 + y/√2 = y

subtract 1 from each side:
y√2 = y - 1

square both sides:
2y^2 = y^2 - 2y +1
rearrange:
y^2 + 2y - 1 = 0

Solve using quadratic formula.

Just another way; pick your poison :cool:

Sorry Mr Dennis but I do not buy this completely. The original equation is linear and does have one solution. Your quadratic will have 2 solutions and one is erroneous.

So to use the above method you must check each solution to see which ones will work

 

[Rudi]

Afraid the quadratic formula is still above my head (that looks like the next chapter).

I have been ploughing through a load more questions and while most are fine I have found one more that don't make sense to me. They are about finding y when y is an index.

In this example I think I have got it:

3^(3y+1) = 9^(y-1) First step is to recognise 9 is 3^2
3^(3y+1) = 3^2(y-1) Since the indices are both powers of 3
3y + 1 = 2y - 2 Subtract 1 and subtract 2y
y = -3

But in this example I cannot quite reach the end:

2^((y^2)+y) = 1 / 4^(y+1) I start by applying the rule a^-n = 1/a^n
2^((y^2)+y) = 4^(-y-1) Then recognise 4 is 2^2
2^((y^2)+y) = 2^(-2y-2) Since indices are both powers of 2
(y^2) + y = -2y - 2 Subtract -2y
(y^2) + 3y = -2 Rearrange
y(y+3) = -2 Divide by y+3
y = -2 / y+3

Then I am stuck. Just by throwing numbers in I can see y = -2 works and so does y=-1. Obviously I can't take the throwing numbers approach if the numbers were bigger. Is there a better way to reach a solution?

 

[Ishuda]

Afraid the quadratic formula is still above my head (that looks like the next chapter).

I have been ploughing through a load more questions and while most are fine I have found one more that don't make sense to me. They are about finding y when y is an index.

In this example I think I have got it:

3^(3y+1) = 9^(y-1) First step is to recognise 9 is 3^2
3^(3y+1) = 3^2(y-1) Since the indices are both powers of 3
3y + 1 = 2y - 2 Subtract 1 and subtract 2y
y = -3

But in this example I cannot quite reach the end:

2^((y^2)+y) = 1 / 4^(y+1) I start by applying the rule a^-n = 1/a^n
2^((y^2)+y) = 4^(-y-1) Then recognise 4 is 2^2
2^((y^2)+y) = 2^(-2y-2) Since indices are both powers of 2
(y^2) + y = -2y - 2 Subtract -2y
(y^2) + 3y = -2 Rearrange
y(y+3) = -2 Divide by y+3
y = -2 / y+3

Then I am stuck. Just by throwing numbers in I can see y = -2 works and so does y=-1. Obviously I can't take the throwing numbers approach if the numbers were bigger. Is there a better way to reach a solution?

First, if you have a new problem/question, it is generally much better to make a new thread [an entirely new post]. Sometime the new question might be ignored as 'having been answered'.

Anyway, let's go back to the
(y^2) + 3y = -2 Rearrange
This gives
y² + 3 y + 2 = 0.
In this particular case, you could recognize that
y² + 3 y + 2 = (y+2) (y+1)
thus
(y+2) (y+1) = 0
or, one of the terms must be zero or we have y=-1 or y=-2.

In the more general case
2^((y^2)+y) = 1 / 4^(a y+ b)
you can still do the same type thing but end up with
y² + (1+2a) y + 2b = 0
which might require the general solution to the quadratic equation.

 

[Steven G]

First, if you have a new problem/question, it is generally much better to make a new thread [an entirely new post]. Sometime the new question might be ignored as 'having been answered'.

Anyway, let's go back to the
(y^2) + 3y = -2 Rearrange
This gives
y² + 3 y + 2 = 0.
In this particular case, you could recognize that
y² + 3 y + 2 = (y+2) (y+1)
thus
(y+2) (y+1) = 0
or, one of the terms must be zero or we have y=-1 or y=-2. Ouch! Factor not term

In the more general case
2^((y^2)+y) = 1 / 4^(a y+ b)
you can still do the same type thing but end up with
y² + (1+2a) y + 2b = 0
which might require the general solution to the quadratic equation.

Comment above

 

[Ishuda]

Comment above

Just to repeat your comment (in red) on what I said:

...
(y+2) (y+1) = 0
or, one of the terms must be zero or we have y=-1 or y=-2. Ouch! Factor not term

Are you saying that factors are not terms of an equation? At least one definition
http://simple.wikipedia.org/wiki/Term_(mathematics)
thinks so:

Specific names for terms
...multiplication
Factor. In a multiplication of two factors, the first is called the multiplicand and the second is called the multiplier.
...

From that link, it would appear that 'term' would be a more general expression for 'factor'. So, yes it is a factor but it is also a term.

You might also look at
http://www.mathwords.com/t/term.htm
where several examples of a term are given. Especially the
5a^{[SIZE=-1]3[/SIZE]} – 2xy + 3
which is not restricted and thus, I would think, it would still be called a term in
(5a^{[SIZE=-1]3[/SIZE]} – 2xy + 3) (23 a + 7)
as would the 23 a + 7.

I'm sure more examples abound in both the virtual and physical worlds.

 

[Steven G]

Just to repeat your comment (in red) on what I said:

Are you saying that factors are not terms of an equation? At least one definition
http://simple.wikipedia.org/wiki/Term_(mathematics)
thinks so:

From that link, it would appear that 'term' would be a more general expression for 'factor'. So, yes it is a factor but it is also a term.

You might also look at
http://www.mathwords.com/t/term.htm
where several examples of a term are given. Especially the
5a^{[SIZE=-1]3[/SIZE]} – 2xy + 3
which is not restricted and thus, I would think, it would still be called a term in
(5a^{[SIZE=-1]3[/SIZE]} – 2xy + 3) (23 a + 7)
as would the 23 a + 7.

I'm sure more examples abound in both the virtual and physical worlds.

I do not use the order of operations that is usually taught (pemdas). I use terms and factors. Terms are separated by addition and subtraction symbols (not inside parenthesis). Once you find the value of each term then you add or subtract the terms. How do you compute the value of a (single) term? You identify the factors. Factors are separated by multiplication and division symbols. Ex: to evaluate 3*4^2= 3|4^2= 3*16 or 48. (note that I used | as a separator for the factors)

 

[Ishuda]

I do not use the order of operations that is usually taught (pemdas). I use terms and factors. Terms are separated by addition and subtraction symbols (not inside parenthesis). Once you find the value of each term then you add or subtract the terms. How do you compute the value of a (single) term? You identify the factors. Factors are separated by multiplication and division symbols. Ex: to evaluate 3*4^2= 3|4^2= 3*16 or 48. (note that I used | as a separator for the factors)

I sorry to hear (read) you have such a limited vocabulary. You (the generic you) could, of course, always correctly apply the most restrictive term to a specific 'thing' but that does not mean the rest of the world will always follow your example. Specifically, not following that example does not make them wrong in the eyes of those who have and use a larger and more generalized vocabulary than was used in the particular statement. Not that I necessarily have and/or use a larger and more generalized vocabulary than you (the particular you) do in general, but in this specific situation, I do believe that to be the case.

 

[Ishuda]

https://www.mathsisfun.com/definitions/term.html

Yes, I've seen the same sort of thing under a heading of "Elementary mathematics". As I said before, if one wishes to limit their vocabulary that is no reason why I should.

 

[Rudi]

Thank you for the help everyone. I will be alert in the future to use (y + 2)(y + 1) = 0.

Several hundred questions from the surds and indices chapter all successfully answered. Onwards into factorising and then quadratics! This weekend promises to be entertaining.

 

[jonah2.0]

WARNING: Beer soaked rambling/opinion/observation/reckoning ahead. Read at your own risk. Not to be taken seriously. In no event shall the wandering math knight-errant Sir jonah in his inebriated state be liable to anyone for special, collateral, incidental, or consequential damages in connection with or arising out of the use of his beer (and tequila) powered views.

I am interested - but unfortunately I cannot spend as much time online as I would like. I'm sure it has something to do with my job - but my continuing education is hopefully my way out of that!

I am also grateful for the assistance. Thank you Jomo and Ishuda. I understand the first question now. Going from 7y - y to y(7 - 1) was the step I could not work out. I think I was also thrown because the answer in the back of the book was clearly wrong. I have been able to get to what I can now see is the correct answer.

Still struggling with the second one:

(1/√2) + y = y/√2 I multiply both sides by √2
1 + y√2 = y Then subtract y√2
1 = y - y√2
1 = y(1 - √2) Then divide (1 - √2)
y = 1 / (1 - √2) What have I done wrong? The answer claims that it should be y = - √2 - 1

A bit late but here's my take.

One way to verify your answers is to use calculators. Latest models that can handle surds and fractions are especially most useful. They can easily switch from natural display of surds and fractions to display of decimal approximations. These calculators can really make maths really enjoyable. In your quoted post, you can easily verify (although Sir Hallsofivy clarified it for you already) that your answer and the book's answer are two sides of the same coin by simply evaluating their decimal approximations (it's how me and me mates did it back in 19?? when calculators were new) or by simply plugging your answer to your original equation.

Although not advisable, methinks a comprehensive book package (one where a textbook is accompanied by its solutions manual and other study aids) might reduce your relearning and online time. One possible downside is that you might not be able to remember much of what you'll learn in that way.
Results may vary of course and you might one of those lucky few who have excellent memory.
Definitely advisable is the habit of keeping your own permanent notebook solutions manual.
Ther's nothing like good old fashioned mistakes in your own handwriting to reinforce a lesson. Those same mistakes might also be a source of a good laugh or two over some beer when you're looking them over.

Good luck to you Sir Rudi in you quest.
Hopefully, after improving on you present job (whatever it may be) you'll also become a math knight-errant yourself (remember Don Quixote?) who will soon be on his way to aid the clueless, guide the lost, entertain them angry lazy ingrates, etc., in their struggle against our beloved queen. Queen to some of us, the dark Lord Mathematicus for a great unenlightened many.

Cheers!