It happens
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Probability
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This is another example of the age related problems I'm still trying to fully understand. I'm thinking my main problem is that I don't understand how to present the data from the questions!
Five years ago, Julie was three times as old as her son Darren was then. Julie is 47. How old is Darren?
Setting the information out seems to be a problem to me. So this is what I'm thinking;
Julie now is 47. So this must mean that Julie five years ago was 42. So in this part of the question it looks like I'm asked to find the age of Darren five years ago?
So Julie was 42 five years ago, hence;
[MATH]{42}={3}({a}-{5})[/MATH]
The right hand side of the equals is looking for the age of Julie who is three times the age of Darren?
[MATH]{42}={3}({a}-{5})[/MATH]
[MATH]{42}={3a}-{15}[/MATH]
[MATH]{57}={3a}[/MATH]
[MATH]{a}={19}[/MATH]
So now the age of Darren is 19.
This is the confusing part;
Julie is 47 now. Julie five years ago was 42, but according to the math above Julie is 57 when her son Darren is 19 at this time now!
So fifteen years ago Darren must have been 4 years old. If I take five years off Darren age now he is 14, and if I multiply this by a factor of 3 from the equation above then I end up with an age of 42.
So what have I established?
Julie is said to be 47 now. Julie five years ago was 42. Darren age now is 19. Five years ago he was 14, which made Julie 42 years old at the time, but she now is 57.
Just for clarity for anyone getting lost in all this, Julie is said to be 47 years old now, she is 57 when her son Darren is 19. Something does not quite add up just right for me at the moment!
Five years ago, Julie was three times as old as her son Darren was then. Julie is 47. How old is Darren?
Setting the information out seems to be a problem to me. So this is what I'm thinking;
Julie now is 47. So this must mean that Julie five years ago was 42. So in this part of the question it looks like I'm asked to find the age of Darren five years ago?
So Julie was 42 five years ago, hence;
[MATH]{42}={3}({a}-{5})[/MATH]
The right hand side of the equals is looking for the age of Julie who is three times the age of Darren?
[MATH]{42}={3}({a}-{5})[/MATH]
[MATH]{42}={3a}-{15}[/MATH]
[MATH]{57}={3a}[/MATH]
[MATH]{a}={19}[/MATH]
So now the age of Darren is 19.
This is the confusing part;
Julie is 47 now. Julie five years ago was 42, but according to the math above Julie is 57 when her son Darren is 19 at this time now!
So fifteen years ago Darren must have been 4 years old. If I take five years off Darren age now he is 14, and if I multiply this by a factor of 3 from the equation above then I end up with an age of 42.
So what have I established?
Julie is said to be 47 now. Julie five years ago was 42. Darren age now is 19. Five years ago he was 14, which made Julie 42 years old at the time, but she now is 57.
Just for clarity for anyone getting lost in all this, Julie is said to be 47 years old now, she is 57 when her son Darren is 19. Something does not quite add up just right for me at the moment!
HallsofIvy
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I wouldn't have done it in quite that way. Specifically, I would have just used arithmetic rather than algebra:
Julia is now 47. 5 years ago she was 42. At that time she was three times as old as her son so her son was 42/3= 14. Now her son is 14+ 5= 19, just what you got.
Yes, and fifteen years ago, Julie was 47- 15= 32, but what does "fifteen years ago" have to do with anything?
Yes, which was Julie's age 5 years ago as should be.
What?? No, she is 47, not 57!
Where in the world did you get 57??? Apparently you are adding 15 to 42 but what does "fifteen years ago" have to do with anything?
The original problem was
"Five years ago, Julie was three times as old as her son Darren was then. Julie is 47. How old is Darren?"
There is no mention of "fifteen years ago". Where did you get that?
Probability
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Looking back at post 22. The math there is confusing me. I have;
[MATH]{42}={3}({a}-{5})[/MATH]
This is referring to five years ago, but then multiplies by 3 to end up at - 15 years. It is then added to the left hand side age of 42 years to become 57 years. The math must mean something?
We start with working five years back, we then change that to 15 years back, and then we bring it forward 15 years adding to her age 42 years to become 57 years old. I'm confused!
[MATH]{42}={3}({a}-{5})[/MATH]
This is referring to five years ago, but then multiplies by 3 to end up at - 15 years. It is then added to the left hand side age of 42 years to become 57 years. The math must mean something?
We start with working five years back, we then change that to 15 years back, and then we bring it forward 15 years adding to her age 42 years to become 57 years old. I'm confused!
HallsofIvy
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And why are you multiplying "5 years ago" by 3??
You have 42= 3(a- 5) and want to solve for a so the obvious first step is to divide both sides by 3: 14= a- 5. Then a= 14+ 5= 19.
Yes, you could multiply out "3(a- 5)" to get 3a- 15 (I presume that is where you got "15") but there is no good reason to do so. If you do that you get 42= 3a- 15 so that 3a= 42+ 15= 57 and then a= 57/3= 19 as before. But that does NOT say that anything happened "15 years ago"!
You have 42= 3(a- 5) and want to solve for a so the obvious first step is to divide both sides by 3: 14= a- 5. Then a= 14+ 5= 19.
Yes, you could multiply out "3(a- 5)" to get 3a- 15 (I presume that is where you got "15") but there is no good reason to do so. If you do that you get 42= 3a- 15 so that 3a= 42+ 15= 57 and then a= 57/3= 19 as before. But that does NOT say that anything happened "15 years ago"!
For reasons I do not understand, algebra is taught starting with linear equations in a single unknown. This is weird because algebra is not needed to solve such problems: that is what arithmetic is for.
How many unknowns are there in this problem? THREE Let's LABEL them.
[MATH]x = \text {Julis's age five years ago.}\\ y = \text {her son's age five year's ago.}\\ z = \text {her son's age today.}[/MATH]Now that is easy enough to do. I count systematically the quantities I do not know and write down a label for each one so I can think and talk about them without confusion.
To solve numerically a problem with n unknowns, I need to know at least n independent numeric relations about them. That is a fundamental rule. But here two relations are self-evident just from having written down what the labels mean.
[MATH]x = 47 - 5.\\ z = y + 5.[/MATH]And the third is given in the problem itself.
[MATH]3z = x.[/MATH]
Now everything becomes completely mechanical. I start substituting and solving equations. NO THOUGHT INVOLVED.
[MATH]x = 47 - 5 = 42.\\ 3y = x \implies 3y = 42 \implies \dfrac{3y}{3} = \dfrac{42}{3} \implies y = 14.\\ z = y + 5 \implies z = 14 + 5 \implies z = 19.[/MATH]Machines can do that. In fact they do that part more reliably than humans. So last step is: check your answer.
Is 3 times 14 equal to 47 - 5? It is. You are done.
Now multiplying 19 by 5 to get 57 has nothing to do with anything. The only reason you are doing that and thereby utterly confusing yourself is because you have never been taught a systematic way to use algebra to solve numeric problems. Here is the way to solve numerically EVERY numerical problem that can be solved by algebra. (Many problems can't be solved by algebra alone.)
Identify what relevant quantities are unknown.
Write down a label (a letter) for each unknown so you remember what the letters stand for and you can think and talk about them.
Write down in mathematical form all relevant numeric relations involving the unknowns.
Solve the resulting equations by substitution and whatever techniques you know for solving equations of the type that you have.
Check your answer.
How many unknowns are there in this problem? THREE Let's LABEL them.
[MATH]x = \text {Julis's age five years ago.}\\ y = \text {her son's age five year's ago.}\\ z = \text {her son's age today.}[/MATH]Now that is easy enough to do. I count systematically the quantities I do not know and write down a label for each one so I can think and talk about them without confusion.
To solve numerically a problem with n unknowns, I need to know at least n independent numeric relations about them. That is a fundamental rule. But here two relations are self-evident just from having written down what the labels mean.
[MATH]x = 47 - 5.\\ z = y + 5.[/MATH]And the third is given in the problem itself.
[MATH]3z = x.[/MATH]
Now everything becomes completely mechanical. I start substituting and solving equations. NO THOUGHT INVOLVED.
[MATH]x = 47 - 5 = 42.\\ 3y = x \implies 3y = 42 \implies \dfrac{3y}{3} = \dfrac{42}{3} \implies y = 14.\\ z = y + 5 \implies z = 14 + 5 \implies z = 19.[/MATH]Machines can do that. In fact they do that part more reliably than humans. So last step is: check your answer.
Is 3 times 14 equal to 47 - 5? It is. You are done.
Now multiplying 19 by 5 to get 57 has nothing to do with anything. The only reason you are doing that and thereby utterly confusing yourself is because you have never been taught a systematic way to use algebra to solve numeric problems. Here is the way to solve numerically EVERY numerical problem that can be solved by algebra. (Many problems can't be solved by algebra alone.)
Identify what relevant quantities are unknown.
Write down a label (a letter) for each unknown so you remember what the letters stand for and you can think and talk about them.
Write down in mathematical form all relevant numeric relations involving the unknowns.
Solve the resulting equations by substitution and whatever techniques you know for solving equations of the type that you have.
Check your answer.
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Probability
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I've kept a copy of the work you have put some really good effort into for helping me JeffM. I really do appreciate your efforts and will use your methods here to help me now further understand the subject. I've only seen linear equations to date for solving this type of problem, and I thought where brackets are used they must be multiplied out first, which is what is taught in the book.
Probability
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The equation [MATH]{42}={3}({a}-{5})[/MATH] is not as obvious to me to divide both sides by 3. The book shows that where brackets are used in an equation that equation has the brackets multiplied out first, and then follow through. I appreciate there will be many ways to solve this problem but we must understand I don't have that depth of experience or understanding to change the techniques of the math at this time, hence wondering about 57!
OK The problem here is that your book is teaching you a way to do word problems that I have found to seriously impede many students who are learning algebra. I tutor kids who are having trouble so I have first-hand experience of the stumbling blocks for beginners. What I showed you is not some specialized method. It works every time that a problem can be solved. It is silly to ask for help and then blow it off.
Be that as it may, the point is that the mechanical techniques you have been taught are TOOLS, not BLUEPRINTS.
How does multiplying the brackets get you any closer to figuring out what a is? You are trying to "isolate" a, meaning to get a all by itself on one side of the equation. You aim to end up with something that looks like a = some number, right? That is your goal. Can you get rid of that 3? Yes you can by dividing both sides of the equation by 3. That is the tool you need at the moment. Yes, sometimes expanding brackets is helpful so they teach you how to do that, but grabbing a hammer when you need a spanner is not productive. Keep the goal in mind.
[MATH]3(a - 5) = 42 \implies \dfrac{3(a - 5)}{3} = \dfrac{42}{3} \implies a - 5 = 14.[/MATH]
You still do not have a all by itself. You need another tool, namely WHAT?
Think of techniques as tools rather than instructions. You pick your tools to achieve your goal, which is to isolate an unknown on one side of an equation with nothing but a numeric expression on the other side.
Probability
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Looking at what you have carried out I'm not sure about what tool I'd need but I'd just say in that example;
[MATH]\dfrac{42}{3}\implies{a}-{5}={14}[/MATH]
[MATH]{a}-{5}+{5}={14}+{5}={19}[/MATH]
Perfect. Yes, the tool is addition.
The old-fashioned rule on a linear equation in one unknown is to to isolate the unknown on one side of the equation using the following steps.
Gather like terms.
Simplify both sides of the equation.
Get all numeric addends onto one side of the equation using addition or subtraction.
Simplify both sides of the equation a second time.
Get all terms involving the unknown on to the other side of the equation.
Simplify both sides of the equation a third time.
Eliminate the coefficient on the unknown by multiplication or division.
It is like a computer program. Here is an example
[MATH]6x - 17 - 2x + 11 = 11x + 100 - 13x - 88.[/MATH]
Looks like a mess. So gather like terms
[MATH]6x - 17 - 2x + 11 = 11x + 100 - 13x - 88 \implies\\ 6x - 2x + 11 - 17 = 13x - 11x + 100 - 88[/MATH]Now simplify.
[MATH]6x - 2x + 11 - 17 = 13x - 11x + 100 - 88 \implies\\ 4x - 6 = 12 - 2x.[/MATH]We are getting somewhere.
Get all numeric addends onto one side of the equation using addition or subtraction.
[MATH]4x - 6 = 12 - 2x \implies\\ 4x - 6 + 6 = -2x + 12 + 6.[/MATH]Simplify again.
[MATH]4x - 6 + 6 = -2x + 12 + 6 \implies\\ 4x = 18 - 2x.[/MATH]Looking good.
Get all terms involving the unknown on to the other side of the equation.
[MATH]4x = 18 - 2x \implies\\ 4x + 2x = 18 - 2x +2x.[/MATH]Hmm. Looks a bit messy.
Simplify again.
[MATH]4x + 2x = 18 - 2x +2x \implies\\ 6x = 18.[/MATH]We have isolated the unknown.
Eliminate the coefficient on the unknown using multiplication or division..
[MATH]6x = 18 \implies\\ \dfrac{6x}{6} = \dfrac{18}{6} \implies\\ x = 3.[/MATH]If a computer followed these steps, it would get the right answer all the time, but I am human. So I check back in the original equation.
[MATH]6(3) - 17 - 2(3) + 11 = 18 - 17 - 6 + 11 = 29 - 23 = 6.\\ 11(3) + 100 - 13(3) - 88 = 33 + 100 - 39 - 88 = 133 - 127 = 6.[/MATH]And that is how you solve linear equations in one unknown. All the thinking goes into identifying unknowns and setting up equations. The rest is pure mechanics.
Probability
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I've kept a copy JeffM to study for revision and help my understanding, thanks.
By the way, before you use the blueprint I gave you, you may need to multiply out brackets.
I reiterate that learning math is not about memorizing blueprints; it is learning to use tools. By all means, study the blueprint I just gave you, but do so to see how tools are used to get to a goal. The next type of equation you will likely see is the quadratic, and the blueprint I gave you will not work. But the tools will be just as useful there except you will need another tool in addition.
Otis
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Instead of 'eliminate', I'd rather see 'change' (or something similar).
For example, "Change the coefficient on the unknown to 1 (if needed)."
?
Probability
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Yes. That is one of the few formulas I have memorized because it arises in many cases. For real x,
[MATH]ax^2 + bx + c = 0 \implies x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \text { if } b^2 \ge 4ac.[/MATH]
Another tool. Notice as well that such an equation may have no, one, or two real solutions.
By the way, otis made a good point. You do not eliminate the unknown's coefficient: you reduce it to 1.
Steven G
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Please do not write nonsense like 70-56 = 14*4 = 56. ALL equal signs must be equal.
What exactly do you mean by 'But we started with (5)?' ? Are you saying that the answer must be 5? If I am twice your age must you be 2 years old? In case it was that simple then we would know that for your example the answer is 5 and do any work.
Probability
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I agree you are correct. I should not cut corners to shorten my work but in that instance I did know what I was doing. Look back at the start of the equation and you will see that everything inside the brackets is multiplied by 5. Working through the equation 5 was reduced to 4 and I was asking about the understanding of why numbers are changing and what numbers actually mean when they have been transferred from one side to the other. The two points were understanding number 5 changing and what 56 represented.