Simple algebraic manipulation with 2 equationsMy first equations is this: d 2 = d − 30.My second equations is this: 1 d 2 = 1 12 − 1 1.066 ( d − 30 ) I am trying to solve for d 2 in the second equation and then set the two equations equal to each other to solve for d.I am having trouble solving for d 2 in the second equation. I think the fractions are giving me a hard time. Any help is much appreciated, thanks!

Question

Simple algebraic manipulation with 2 equationsMy first equations is this:       d    2    =  d  −  30.My second equations is this:             1              d        2              =            1      12        −            1              1.066        (        d        −        30        )            I am trying to solve for       d    2   in the second equation and then set the two equations equal to each other to solve for d.I am having trouble solving for       d    2   in the second equation. I think the fractions are giving me a hard time. Any help is much appreciated, thanks!

aletantas1x · Accepted Answer

1          d              2              =      1    12    −      1          1.066              d                  2                      (  1  +      1    1.066    )      1          d              2              =      1    12        d          2        =  12  (  1  +      1    1.066    )  d  =      d          2        +  30  d  =  12  (  1  +      1    1.066    )  +  30

Yahir Crane · Accepted Answer

First, I hate decimals, so let&#039;s note that 1.066 equals       533    500  .Then, since we were essentially given what       d    2   equals, let&#039;s substitute that into our second equation like so:      1          d      −      30        =      1    12    −      1                  533        500            (      d      −      30      )            1          d      −      30        =      1    12    −      500          533      (      d      −      30      )      Now we want to multiply through by the LCM (which in our case is simply   12  ⋅  533  (  d  −  3  )  6396  (  d  −  30  ) to obtain:  6396  =  533  (  d  −  30  )  −  6000Now we have a nice linear equation which we may solve for   d:  533  d  −  21990  =  6396  d  =      28386    533  With   d obtained we can easily find       d    2  :      d    2    =  d  −  30      d    2    =      28386    533    −  30      d    2    =      12396    533  And we&#039;re done.

Simple algebraic manipulation with 2 equations My first equations is this: d 2 </

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