How to solve this simultaneous equations which involves exponents: { 2 x − 2 y = 1 4 x − 4 y = 5 3 Simultaneously. From the second equation I factorized to get ( 2 x + 2 y ) ( 2 x − 2 y ) = 5 3 and from the first equation, since ( 2 x − 2 y ) = 1, I deduced that ( 2 x + 2 y ) ⋅ 1 = 5 3 → ( 2 x + 2 y ) = 5 3 Now I am stuck on how to proceed.

Question

How to solve this simultaneous equations which involves exponents:      {                                        2                      x                    −                      2                      y                    =          1                                                  4                      x                    −                      4                      y                    =                      5            3                                  Simultaneously. From the second equation I factorized to get  (      2    x    +      2    y    )  (      2    x    −      2    y    )  =      5    3  and from the first equation, since   (      2    x    −      2    y    )  =  1, I deduced that  (      2    x    +      2    y    )  ⋅  1  =      5    3    →  (      2    x    +      2    y    )  =      5    3  Now I am stuck on how to proceed.

EreneDreaceaw · Accepted Answer

Try adding the equations       2    x    −      2    y    =  1 and       2    x    +      2    y    =      5    3  To go further per your comment:This gives   2  ⋅      2    x    =      8    3  Subtracting the equations gives   2  ⋅      2    y    =      2    3  Now recall that             2      a              2      b        =      2          a      −      b       and divide in an appropriate way to make further progress.

veirarer · Accepted Answer

{                                        2            x                    −                      2            y                    =          1                                                  2            x                    +                      2            y                    =                      5            3                                  so      2    x    −      2    y    +      2    x    −      2    y    =  2  ⋅      2    x    =  1  +      5    3    =      8    3    ⇒      2    x    =      4    3    ⇒  x  =  l  o      g    2        4    3  Finally, since       2    x    =      4    3  ,      2    y    =      5    3    −      4    3    =      1    3    ⇒  y  =  l  o      g    2        1    3  The solution is       (    l    o          g      2                4      3        ,    l    o          g      2                1      3        )

How to solve this simultaneous equations which involves exponents: { <mtable columna

Answered question

Answer & Explanation

New Questions in Algebra I