Why is $ab(frac1a)b^2cb^{-3} = c$?












0














I'm working through some textbook exercise and am unable to solve the following exercise:




Use the basic rules of algebra to simplify the following expression:
$$abfrac{1}{a}b^2cb^{-3}$$




Within the chapter of the book in question the author has covered:




  • Associative, commutative and distributive properties of an equation

  • Expanding brackets

  • Factoring

  • Quadratic factoring

  • Completing the square


The solution in the answers section is simply "$c$". The equation simplifies to $c$. I cannot for the life of me see how. The solution provided does not give the steps in between, only the final solution $c$.



How can one arrive at $c$ with all the steps in between?






algebra-precalculus







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  • 1




    My +1 for good presentation of the context.
    – Yves Daoust
    Nov 20 '18 at 17:12










  • The global exponent of $a$ is $1-1$, that of $b$, $1+2-3$ and that of $c$ is $1$.
    – Yves Daoust
    Nov 20 '18 at 17:13






  • 1




    Here's a really helpful tutorial for MathJax formatting: math.meta.stackexchange.com/questions/5020/…. It's pretty easy to pick up!
    – Robert Howard
    Nov 20 '18 at 19:24
















0














I'm working through some textbook exercise and am unable to solve the following exercise:




Use the basic rules of algebra to simplify the following expression:
$$abfrac{1}{a}b^2cb^{-3}$$




Within the chapter of the book in question the author has covered:




  • Associative, commutative and distributive properties of an equation

  • Expanding brackets

  • Factoring

  • Quadratic factoring

  • Completing the square


The solution in the answers section is simply "$c$". The equation simplifies to $c$. I cannot for the life of me see how. The solution provided does not give the steps in between, only the final solution $c$.



How can one arrive at $c$ with all the steps in between?






algebra-precalculus







share|cite|improve this question




















  • 1




    My +1 for good presentation of the context.
    – Yves Daoust
    Nov 20 '18 at 17:12










  • The global exponent of $a$ is $1-1$, that of $b$, $1+2-3$ and that of $c$ is $1$.
    – Yves Daoust
    Nov 20 '18 at 17:13






  • 1




    Here's a really helpful tutorial for MathJax formatting: math.meta.stackexchange.com/questions/5020/…. It's pretty easy to pick up!
    – Robert Howard
    Nov 20 '18 at 19:24














0












0








0







I'm working through some textbook exercise and am unable to solve the following exercise:




Use the basic rules of algebra to simplify the following expression:
$$abfrac{1}{a}b^2cb^{-3}$$




Within the chapter of the book in question the author has covered:




  • Associative, commutative and distributive properties of an equation

  • Expanding brackets

  • Factoring

  • Quadratic factoring

  • Completing the square


The solution in the answers section is simply "$c$". The equation simplifies to $c$. I cannot for the life of me see how. The solution provided does not give the steps in between, only the final solution $c$.



How can one arrive at $c$ with all the steps in between?






algebra-precalculus







share|cite|improve this question















I'm working through some textbook exercise and am unable to solve the following exercise:




Use the basic rules of algebra to simplify the following expression:
$$abfrac{1}{a}b^2cb^{-3}$$




Within the chapter of the book in question the author has covered:




  • Associative, commutative and distributive properties of an equation

  • Expanding brackets

  • Factoring

  • Quadratic factoring

  • Completing the square


The solution in the answers section is simply "$c$". The equation simplifies to $c$. I cannot for the life of me see how. The solution provided does not give the steps in between, only the final solution $c$.



How can one arrive at $c$ with all the steps in between?






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 20 '18 at 19:08









Robert Howard

1,9161822




1,9161822










asked Nov 20 '18 at 16:31









Doug Fir

2057




2057








  • 1




    My +1 for good presentation of the context.
    – Yves Daoust
    Nov 20 '18 at 17:12










  • The global exponent of $a$ is $1-1$, that of $b$, $1+2-3$ and that of $c$ is $1$.
    – Yves Daoust
    Nov 20 '18 at 17:13






  • 1




    Here's a really helpful tutorial for MathJax formatting: math.meta.stackexchange.com/questions/5020/…. It's pretty easy to pick up!
    – Robert Howard
    Nov 20 '18 at 19:24














  • 1




    My +1 for good presentation of the context.
    – Yves Daoust
    Nov 20 '18 at 17:12










  • The global exponent of $a$ is $1-1$, that of $b$, $1+2-3$ and that of $c$ is $1$.
    – Yves Daoust
    Nov 20 '18 at 17:13






  • 1




    Here's a really helpful tutorial for MathJax formatting: math.meta.stackexchange.com/questions/5020/…. It's pretty easy to pick up!
    – Robert Howard
    Nov 20 '18 at 19:24








1




1




My +1 for good presentation of the context.
– Yves Daoust
Nov 20 '18 at 17:12




My +1 for good presentation of the context.
– Yves Daoust
Nov 20 '18 at 17:12












The global exponent of $a$ is $1-1$, that of $b$, $1+2-3$ and that of $c$ is $1$.
– Yves Daoust
Nov 20 '18 at 17:13




The global exponent of $a$ is $1-1$, that of $b$, $1+2-3$ and that of $c$ is $1$.
– Yves Daoust
Nov 20 '18 at 17:13




1




1




Here's a really helpful tutorial for MathJax formatting: math.meta.stackexchange.com/questions/5020/…. It's pretty easy to pick up!
– Robert Howard
Nov 20 '18 at 19:24




Here's a really helpful tutorial for MathJax formatting: math.meta.stackexchange.com/questions/5020/…. It's pretty easy to pick up!
– Robert Howard
Nov 20 '18 at 19:24










4 Answers
4






active

oldest

votes


















4














First off, associativity means we don't need parentheses, which is good, because there are none given to us. Now, let's use the definition of exponents to get
$$
abfrac1abbcfrac1bfrac1bfrac1b
$$
Then we use commutativity to arrange things alphabetically. This gives us
$$
afrac1abbbfrac1bfrac1bfrac1bc
$$
The definition of fraction gives $afrac1a=1$ (and similarly for $b$). We get
$$
1cdot1cdot1cdot1cdot c
$$
And finally, by definition of $1$, this simplifies to $c$.






share|cite|improve this answer





















  • Thank you for the crystal clear answer. Deleting this comment in a few minutes...
    – Doug Fir
    Nov 20 '18 at 16:46






  • 2




    @DougFir, it's fine to leave your comment.
    – Decaf-Math
    Nov 20 '18 at 16:49










  • I find the use of "alphabetically" in a mathematical context weirdly funny. +1
    – DreamConspiracy
    Nov 21 '18 at 9:36



















1














First, note that $ab frac{1}{a} = b$ and then substituting that into the equation we get $bb^2cb^{-3}$ so now we collect all the $b$ terms and the cancel leaving us with $c$ as the simplified expression.






share|cite|improve this answer





























    1














    By commutativity and associativity we can rearrange the expression as
    $$left(afrac1aright)cdot(bb^2b^{-3})cdot c$$
    the first term is clearly $1$. The second term is also $1$ because of how you add powers in a product: indeed
    $$bb^2b^{-3}=b^{1+2-3}=b^0=1$$ So all you are left with is $c$






    share|cite|improve this answer





























      1














      All that is needed here from the chapter in question are the associative and commutative properties of multiplication. The expression $abfrac{1}{a}b^2cb^{-3}$ can be re-arranged with commutativity to $afrac{1}{a}bb^2b^{-3}c$, and then this expression can be grouped into $(afrac{1}{a})(bb^2b^{-3})(c)$ with associativity. Those first two groups are both equal to $1$ (assuming of course none of the variables are $0$), and so you will be left with $c$. If this symbolic manipulation is still confusing, try substituting some values in for those variables! Try $a=2$, $b=3$, and $c=5$.






      share|cite|improve this answer





















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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4














        First off, associativity means we don't need parentheses, which is good, because there are none given to us. Now, let's use the definition of exponents to get
        $$
        abfrac1abbcfrac1bfrac1bfrac1b
        $$
        Then we use commutativity to arrange things alphabetically. This gives us
        $$
        afrac1abbbfrac1bfrac1bfrac1bc
        $$
        The definition of fraction gives $afrac1a=1$ (and similarly for $b$). We get
        $$
        1cdot1cdot1cdot1cdot c
        $$
        And finally, by definition of $1$, this simplifies to $c$.






        share|cite|improve this answer





















        • Thank you for the crystal clear answer. Deleting this comment in a few minutes...
          – Doug Fir
          Nov 20 '18 at 16:46






        • 2




          @DougFir, it's fine to leave your comment.
          – Decaf-Math
          Nov 20 '18 at 16:49










        • I find the use of "alphabetically" in a mathematical context weirdly funny. +1
          – DreamConspiracy
          Nov 21 '18 at 9:36
















        4














        First off, associativity means we don't need parentheses, which is good, because there are none given to us. Now, let's use the definition of exponents to get
        $$
        abfrac1abbcfrac1bfrac1bfrac1b
        $$
        Then we use commutativity to arrange things alphabetically. This gives us
        $$
        afrac1abbbfrac1bfrac1bfrac1bc
        $$
        The definition of fraction gives $afrac1a=1$ (and similarly for $b$). We get
        $$
        1cdot1cdot1cdot1cdot c
        $$
        And finally, by definition of $1$, this simplifies to $c$.






        share|cite|improve this answer





















        • Thank you for the crystal clear answer. Deleting this comment in a few minutes...
          – Doug Fir
          Nov 20 '18 at 16:46






        • 2




          @DougFir, it's fine to leave your comment.
          – Decaf-Math
          Nov 20 '18 at 16:49










        • I find the use of "alphabetically" in a mathematical context weirdly funny. +1
          – DreamConspiracy
          Nov 21 '18 at 9:36














        4












        4








        4






        First off, associativity means we don't need parentheses, which is good, because there are none given to us. Now, let's use the definition of exponents to get
        $$
        abfrac1abbcfrac1bfrac1bfrac1b
        $$
        Then we use commutativity to arrange things alphabetically. This gives us
        $$
        afrac1abbbfrac1bfrac1bfrac1bc
        $$
        The definition of fraction gives $afrac1a=1$ (and similarly for $b$). We get
        $$
        1cdot1cdot1cdot1cdot c
        $$
        And finally, by definition of $1$, this simplifies to $c$.






        share|cite|improve this answer












        First off, associativity means we don't need parentheses, which is good, because there are none given to us. Now, let's use the definition of exponents to get
        $$
        abfrac1abbcfrac1bfrac1bfrac1b
        $$
        Then we use commutativity to arrange things alphabetically. This gives us
        $$
        afrac1abbbfrac1bfrac1bfrac1bc
        $$
        The definition of fraction gives $afrac1a=1$ (and similarly for $b$). We get
        $$
        1cdot1cdot1cdot1cdot c
        $$
        And finally, by definition of $1$, this simplifies to $c$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 20 '18 at 16:37









        Arthur

        111k7105186




        111k7105186












        • Thank you for the crystal clear answer. Deleting this comment in a few minutes...
          – Doug Fir
          Nov 20 '18 at 16:46






        • 2




          @DougFir, it's fine to leave your comment.
          – Decaf-Math
          Nov 20 '18 at 16:49










        • I find the use of "alphabetically" in a mathematical context weirdly funny. +1
          – DreamConspiracy
          Nov 21 '18 at 9:36


















        • Thank you for the crystal clear answer. Deleting this comment in a few minutes...
          – Doug Fir
          Nov 20 '18 at 16:46






        • 2




          @DougFir, it's fine to leave your comment.
          – Decaf-Math
          Nov 20 '18 at 16:49










        • I find the use of "alphabetically" in a mathematical context weirdly funny. +1
          – DreamConspiracy
          Nov 21 '18 at 9:36
















        Thank you for the crystal clear answer. Deleting this comment in a few minutes...
        – Doug Fir
        Nov 20 '18 at 16:46




        Thank you for the crystal clear answer. Deleting this comment in a few minutes...
        – Doug Fir
        Nov 20 '18 at 16:46




        2




        2




        @DougFir, it's fine to leave your comment.
        – Decaf-Math
        Nov 20 '18 at 16:49




        @DougFir, it's fine to leave your comment.
        – Decaf-Math
        Nov 20 '18 at 16:49












        I find the use of "alphabetically" in a mathematical context weirdly funny. +1
        – DreamConspiracy
        Nov 21 '18 at 9:36




        I find the use of "alphabetically" in a mathematical context weirdly funny. +1
        – DreamConspiracy
        Nov 21 '18 at 9:36











        1














        First, note that $ab frac{1}{a} = b$ and then substituting that into the equation we get $bb^2cb^{-3}$ so now we collect all the $b$ terms and the cancel leaving us with $c$ as the simplified expression.






        share|cite|improve this answer


























          1














          First, note that $ab frac{1}{a} = b$ and then substituting that into the equation we get $bb^2cb^{-3}$ so now we collect all the $b$ terms and the cancel leaving us with $c$ as the simplified expression.






          share|cite|improve this answer
























            1












            1








            1






            First, note that $ab frac{1}{a} = b$ and then substituting that into the equation we get $bb^2cb^{-3}$ so now we collect all the $b$ terms and the cancel leaving us with $c$ as the simplified expression.






            share|cite|improve this answer












            First, note that $ab frac{1}{a} = b$ and then substituting that into the equation we get $bb^2cb^{-3}$ so now we collect all the $b$ terms and the cancel leaving us with $c$ as the simplified expression.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 20 '18 at 16:34









            CyclotomicField

            2,1921313




            2,1921313























                1














                By commutativity and associativity we can rearrange the expression as
                $$left(afrac1aright)cdot(bb^2b^{-3})cdot c$$
                the first term is clearly $1$. The second term is also $1$ because of how you add powers in a product: indeed
                $$bb^2b^{-3}=b^{1+2-3}=b^0=1$$ So all you are left with is $c$






                share|cite|improve this answer


























                  1














                  By commutativity and associativity we can rearrange the expression as
                  $$left(afrac1aright)cdot(bb^2b^{-3})cdot c$$
                  the first term is clearly $1$. The second term is also $1$ because of how you add powers in a product: indeed
                  $$bb^2b^{-3}=b^{1+2-3}=b^0=1$$ So all you are left with is $c$






                  share|cite|improve this answer
























                    1












                    1








                    1






                    By commutativity and associativity we can rearrange the expression as
                    $$left(afrac1aright)cdot(bb^2b^{-3})cdot c$$
                    the first term is clearly $1$. The second term is also $1$ because of how you add powers in a product: indeed
                    $$bb^2b^{-3}=b^{1+2-3}=b^0=1$$ So all you are left with is $c$






                    share|cite|improve this answer












                    By commutativity and associativity we can rearrange the expression as
                    $$left(afrac1aright)cdot(bb^2b^{-3})cdot c$$
                    the first term is clearly $1$. The second term is also $1$ because of how you add powers in a product: indeed
                    $$bb^2b^{-3}=b^{1+2-3}=b^0=1$$ So all you are left with is $c$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 20 '18 at 16:35









                    b00n heT

                    10.2k12134




                    10.2k12134























                        1














                        All that is needed here from the chapter in question are the associative and commutative properties of multiplication. The expression $abfrac{1}{a}b^2cb^{-3}$ can be re-arranged with commutativity to $afrac{1}{a}bb^2b^{-3}c$, and then this expression can be grouped into $(afrac{1}{a})(bb^2b^{-3})(c)$ with associativity. Those first two groups are both equal to $1$ (assuming of course none of the variables are $0$), and so you will be left with $c$. If this symbolic manipulation is still confusing, try substituting some values in for those variables! Try $a=2$, $b=3$, and $c=5$.






                        share|cite|improve this answer


























                          1














                          All that is needed here from the chapter in question are the associative and commutative properties of multiplication. The expression $abfrac{1}{a}b^2cb^{-3}$ can be re-arranged with commutativity to $afrac{1}{a}bb^2b^{-3}c$, and then this expression can be grouped into $(afrac{1}{a})(bb^2b^{-3})(c)$ with associativity. Those first two groups are both equal to $1$ (assuming of course none of the variables are $0$), and so you will be left with $c$. If this symbolic manipulation is still confusing, try substituting some values in for those variables! Try $a=2$, $b=3$, and $c=5$.






                          share|cite|improve this answer
























                            1












                            1








                            1






                            All that is needed here from the chapter in question are the associative and commutative properties of multiplication. The expression $abfrac{1}{a}b^2cb^{-3}$ can be re-arranged with commutativity to $afrac{1}{a}bb^2b^{-3}c$, and then this expression can be grouped into $(afrac{1}{a})(bb^2b^{-3})(c)$ with associativity. Those first two groups are both equal to $1$ (assuming of course none of the variables are $0$), and so you will be left with $c$. If this symbolic manipulation is still confusing, try substituting some values in for those variables! Try $a=2$, $b=3$, and $c=5$.






                            share|cite|improve this answer












                            All that is needed here from the chapter in question are the associative and commutative properties of multiplication. The expression $abfrac{1}{a}b^2cb^{-3}$ can be re-arranged with commutativity to $afrac{1}{a}bb^2b^{-3}c$, and then this expression can be grouped into $(afrac{1}{a})(bb^2b^{-3})(c)$ with associativity. Those first two groups are both equal to $1$ (assuming of course none of the variables are $0$), and so you will be left with $c$. If this symbolic manipulation is still confusing, try substituting some values in for those variables! Try $a=2$, $b=3$, and $c=5$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 20 '18 at 16:36









                            Valborg

                            111




                            111






























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