Question

Solve the following linear equation:

$m minus fraction numerator m minus 1 over denominator 2 end fraction equals 1 minus fraction numerator m minus 2 over denominator 3 end fraction$

Solution

straight m minus fraction numerator straight m minus 1 over denominator 2 end fraction equals 1 minus fraction numerator straight m minus 2 over denominator 3 end fraction

Since, LCM of 2 and 3 is 6.

∴ Multiplying both sides by 6, we have
 6 $cross times$ m - 6 $cross times$ $fraction numerator space space straight m minus 1 over denominator 3 end fraction space equals space 6 cross times 1 minus 6 cross times fraction numerator straight m minus 2 over denominator 3 end fraction$
or 6m -3(m-1) = 6 - 2(m-2)
or 6m - 3m + 3 = 6 -2m + 4
or (6-3)m + 3 = (6+4) - 2m
or 3m + 3 = 10 - 2m
Transposing 3 to RHS and -2m to LHS, we have
 3m + 2m = 10 - 3
or 5m =7
or $straight m equals 7 over 5$ (Dividing both sides by 5)
Check: LHS = $straight m minus fraction numerator straight m minus 1 over denominator 2 end fraction$
 = $7 over 5 minus fraction numerator begin display style 7 over 5 end style minus 1 over denominator 2 end fraction equals 7 over 5 minus fraction numerator begin display style 2 over 5 end style over denominator 2 end fraction$
 = $7 over 5 minus 2 over 5 cross times 1 half$
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 RHS = $1 minus fraction numerator straight m minus 2 over denominator 3 end fraction$
 = $1 minus fraction numerator begin display style 7 over 5 end style minus 2 over denominator 3 end fraction equals 1 minus fraction numerator begin display style fraction numerator negative 3 over denominator 5 end fraction end style over denominator 3 end fraction$

 = $space space 1 plus 3 over 5 cross times 1 third$
 = $1 plus 1 fifth equals 6 over 5$