Question 1126668
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<pre>
The standard method for solving such problems is introducing new variables


x = {{{1/A}}},   y = {{{1/B}}}.


Then your system takes the form


2x - 3y = -1      (1)
5x - 6y = -1/2    (2)


To solve it, multiply eq(1) by 2 (both sides) and then subtract from eq(2). You will get


5x - 2*(2x) = {{{-1/2}}} - 2*(-1),    or


9x - 4x = {{{-1/2}}} + 2,

x = 1.5.


Then from eq(1),  2*1.5 -3y = -1,   3 - 3y = -1,  3 + 1 = 3y  ====>  y = {{{4/3}}}.


Now you have  x = 1.5 = {{{1/A}}} ====>  A = {{{1/1.5}}} = {{{1/((3/2))}}} = {{{2/3}}}   and

              y = {{{4/3}}} = {{{1/B}}}  ====>  B = {{{3/4}}}.


<U>Answer</U>.  A = {{{2/3}}}   and   B = {{{3/4}}}.
</pre>

Solved.


Simply and elegantly.


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The way which &nbsp;@josgarithmetic &nbsp;tries to sell you - is the way to &nbsp;NOWHERE.


For your safety, &nbsp;simply ignore it.



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For your safety, &nbsp;<U>ignore &nbsp;EACH &nbsp;and &nbsp;EVERY &nbsp;HIS &nbsp;POST</U> . . . . . . . . 



I just tired to fix all this &nbsp;RUBBISH &nbsp;after him.



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Regarding the post by @greenestamps, he writes


<pre>
    You certainly CAN introduce new variables to solve the problem, as tutor @ikleyn does. But you can solve the problem without doing that.
</pre>


Truly, &nbsp;you can try solve the problem without doing that.


But then I guarantee you that you as inexperienced person in solving such problems, &nbsp;will make about 20 errors 
on the way and will not be able to complete the solution correctly.


The way I showed you in my post - is simplest, &nbsp;straightforward and prevents you of making crude errors.


It is not without reasons the way I presented you in my post is considered as the standard and canonical.