Simplfying expression to get $b-a$











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$$frac{3}{4}(b-a)left( frac{2a}{3}+frac{b}{3} right)^0 + left( frac{b-a}{4} right)b^0 = b-a$$



I know that $$left( frac{2a}{3}+frac{b}{3} right)^0 = 1$$



and $$left( frac{b-a}{4} right)b^0=frac{b-a}{4}$$



but then how do I simplify the rest to get the solution which Is $b-a$?






algebra-precalculus







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    up vote
    0
    down vote

    favorite












    $$frac{3}{4}(b-a)left( frac{2a}{3}+frac{b}{3} right)^0 + left( frac{b-a}{4} right)b^0 = b-a$$



    I know that $$left( frac{2a}{3}+frac{b}{3} right)^0 = 1$$



    and $$left( frac{b-a}{4} right)b^0=frac{b-a}{4}$$



    but then how do I simplify the rest to get the solution which Is $b-a$?






    algebra-precalculus







    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      $$frac{3}{4}(b-a)left( frac{2a}{3}+frac{b}{3} right)^0 + left( frac{b-a}{4} right)b^0 = b-a$$



      I know that $$left( frac{2a}{3}+frac{b}{3} right)^0 = 1$$



      and $$left( frac{b-a}{4} right)b^0=frac{b-a}{4}$$



      but then how do I simplify the rest to get the solution which Is $b-a$?






      algebra-precalculus







      share|cite|improve this question















      $$frac{3}{4}(b-a)left( frac{2a}{3}+frac{b}{3} right)^0 + left( frac{b-a}{4} right)b^0 = b-a$$



      I know that $$left( frac{2a}{3}+frac{b}{3} right)^0 = 1$$



      and $$left( frac{b-a}{4} right)b^0=frac{b-a}{4}$$



      but then how do I simplify the rest to get the solution which Is $b-a$?






      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 23 at 3:26









      Joey Kilpatrick

      1,183422




      1,183422










      asked Nov 23 at 2:41









      mt12345

      958




      958






















          3 Answers
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          up vote
          1
          down vote













          Let $t=b-a$. Then don't you simply have $$frac 34 t+ frac 14 t=t$$ which is blatantly obvious?






          share|cite|improve this answer




























            up vote
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            $$frac{3}{4}(b-a)left( frac{2a}{3}+frac{b}{3} right)^0 + left( frac{b-a}{4} right)b^0 = b-a$$
            $$frac{3}{4}(b-a)cdot1+left( frac{b-a}{4} right)cdot1=b-a$$
            $$dfrac{3b-3a}{4}+dfrac{b-a}{4}=b-a$$
            $$dfrac{4b-4a}{4}=b-a$$
            $$b-a=b-a$$






            share|cite|improve this answer























            • sorry that was a typo in my question
              – mt12345
              Nov 23 at 2:45


















            up vote
            0
            down vote













            The equation becomes $$
            frac{3}{4}(b-a) + left( frac{b-a}{4} right) = b-a \
            frac{3}{4}(b-a) + frac{1}{4}(b-a) = b-a \
            b-a=b-a
            $$






            share|cite|improve this answer























            • sorry that was a typo in my question
              – mt12345
              Nov 23 at 2:46










            • I don't think the typo has been fixed.
              – Joey Kilpatrick
              Nov 23 at 2:47










            • fixed, thanks!!
              – mt12345
              Nov 23 at 2:48











            Your Answer





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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote













            Let $t=b-a$. Then don't you simply have $$frac 34 t+ frac 14 t=t$$ which is blatantly obvious?






            share|cite|improve this answer

























              up vote
              1
              down vote













              Let $t=b-a$. Then don't you simply have $$frac 34 t+ frac 14 t=t$$ which is blatantly obvious?






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                Let $t=b-a$. Then don't you simply have $$frac 34 t+ frac 14 t=t$$ which is blatantly obvious?






                share|cite|improve this answer












                Let $t=b-a$. Then don't you simply have $$frac 34 t+ frac 14 t=t$$ which is blatantly obvious?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 23 at 2:52









                Rhys Hughes

                4,6731327




                4,6731327






















                    up vote
                    0
                    down vote













                    $$frac{3}{4}(b-a)left( frac{2a}{3}+frac{b}{3} right)^0 + left( frac{b-a}{4} right)b^0 = b-a$$
                    $$frac{3}{4}(b-a)cdot1+left( frac{b-a}{4} right)cdot1=b-a$$
                    $$dfrac{3b-3a}{4}+dfrac{b-a}{4}=b-a$$
                    $$dfrac{4b-4a}{4}=b-a$$
                    $$b-a=b-a$$






                    share|cite|improve this answer























                    • sorry that was a typo in my question
                      – mt12345
                      Nov 23 at 2:45















                    up vote
                    0
                    down vote













                    $$frac{3}{4}(b-a)left( frac{2a}{3}+frac{b}{3} right)^0 + left( frac{b-a}{4} right)b^0 = b-a$$
                    $$frac{3}{4}(b-a)cdot1+left( frac{b-a}{4} right)cdot1=b-a$$
                    $$dfrac{3b-3a}{4}+dfrac{b-a}{4}=b-a$$
                    $$dfrac{4b-4a}{4}=b-a$$
                    $$b-a=b-a$$






                    share|cite|improve this answer























                    • sorry that was a typo in my question
                      – mt12345
                      Nov 23 at 2:45













                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    $$frac{3}{4}(b-a)left( frac{2a}{3}+frac{b}{3} right)^0 + left( frac{b-a}{4} right)b^0 = b-a$$
                    $$frac{3}{4}(b-a)cdot1+left( frac{b-a}{4} right)cdot1=b-a$$
                    $$dfrac{3b-3a}{4}+dfrac{b-a}{4}=b-a$$
                    $$dfrac{4b-4a}{4}=b-a$$
                    $$b-a=b-a$$






                    share|cite|improve this answer














                    $$frac{3}{4}(b-a)left( frac{2a}{3}+frac{b}{3} right)^0 + left( frac{b-a}{4} right)b^0 = b-a$$
                    $$frac{3}{4}(b-a)cdot1+left( frac{b-a}{4} right)cdot1=b-a$$
                    $$dfrac{3b-3a}{4}+dfrac{b-a}{4}=b-a$$
                    $$dfrac{4b-4a}{4}=b-a$$
                    $$b-a=b-a$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 23 at 2:46

























                    answered Nov 23 at 2:44









                    Key Flex

                    7,44941232




                    7,44941232












                    • sorry that was a typo in my question
                      – mt12345
                      Nov 23 at 2:45


















                    • sorry that was a typo in my question
                      – mt12345
                      Nov 23 at 2:45
















                    sorry that was a typo in my question
                    – mt12345
                    Nov 23 at 2:45




                    sorry that was a typo in my question
                    – mt12345
                    Nov 23 at 2:45










                    up vote
                    0
                    down vote













                    The equation becomes $$
                    frac{3}{4}(b-a) + left( frac{b-a}{4} right) = b-a \
                    frac{3}{4}(b-a) + frac{1}{4}(b-a) = b-a \
                    b-a=b-a
                    $$






                    share|cite|improve this answer























                    • sorry that was a typo in my question
                      – mt12345
                      Nov 23 at 2:46










                    • I don't think the typo has been fixed.
                      – Joey Kilpatrick
                      Nov 23 at 2:47










                    • fixed, thanks!!
                      – mt12345
                      Nov 23 at 2:48















                    up vote
                    0
                    down vote













                    The equation becomes $$
                    frac{3}{4}(b-a) + left( frac{b-a}{4} right) = b-a \
                    frac{3}{4}(b-a) + frac{1}{4}(b-a) = b-a \
                    b-a=b-a
                    $$






                    share|cite|improve this answer























                    • sorry that was a typo in my question
                      – mt12345
                      Nov 23 at 2:46










                    • I don't think the typo has been fixed.
                      – Joey Kilpatrick
                      Nov 23 at 2:47










                    • fixed, thanks!!
                      – mt12345
                      Nov 23 at 2:48













                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    The equation becomes $$
                    frac{3}{4}(b-a) + left( frac{b-a}{4} right) = b-a \
                    frac{3}{4}(b-a) + frac{1}{4}(b-a) = b-a \
                    b-a=b-a
                    $$






                    share|cite|improve this answer














                    The equation becomes $$
                    frac{3}{4}(b-a) + left( frac{b-a}{4} right) = b-a \
                    frac{3}{4}(b-a) + frac{1}{4}(b-a) = b-a \
                    b-a=b-a
                    $$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 23 at 3:29

























                    answered Nov 23 at 2:45









                    Joey Kilpatrick

                    1,183422




                    1,183422












                    • sorry that was a typo in my question
                      – mt12345
                      Nov 23 at 2:46










                    • I don't think the typo has been fixed.
                      – Joey Kilpatrick
                      Nov 23 at 2:47










                    • fixed, thanks!!
                      – mt12345
                      Nov 23 at 2:48


















                    • sorry that was a typo in my question
                      – mt12345
                      Nov 23 at 2:46










                    • I don't think the typo has been fixed.
                      – Joey Kilpatrick
                      Nov 23 at 2:47










                    • fixed, thanks!!
                      – mt12345
                      Nov 23 at 2:48
















                    sorry that was a typo in my question
                    – mt12345
                    Nov 23 at 2:46




                    sorry that was a typo in my question
                    – mt12345
                    Nov 23 at 2:46












                    I don't think the typo has been fixed.
                    – Joey Kilpatrick
                    Nov 23 at 2:47




                    I don't think the typo has been fixed.
                    – Joey Kilpatrick
                    Nov 23 at 2:47












                    fixed, thanks!!
                    – mt12345
                    Nov 23 at 2:48




                    fixed, thanks!!
                    – mt12345
                    Nov 23 at 2:48


















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