Solve the Simple Equation: 1/d^2 = 1/(81m^2) + 1/(25m^2) in Just a Few Steps!

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SUMMARY

The equation 1/d² = 1/(9m)² + 1/(5m)² can be solved by first rewriting it as 1/d² = 1/81m² + 1/25m². The correct approach involves multiplying both sides by the least common multiple, specifically (25)(81)d²m², to eliminate the fractions. This leads to the conclusion that d² = 106m², resulting in d = 10.3m, which is an accurate solution when calculated correctly.

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superwolf
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How can I solve this equation:

[tex] \frac{1}{d^2} = \frac{1}{(9m)^2} + \frac{1}{(5m)^2}[/tex]

?

Try:

[tex] \frac{1}{d^2} = \frac{1}{81m^2} + \frac{1}{25m^2} \Rightarrow d^2 = 81m^2 + 25m^2 = 106m^2 \Rightarrow d = 10.3m[/tex]
 
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superwolf said:
How can I solve this equation:

[tex] \frac{1}{d^2} = \frac{1}{(9m)^2} + \frac{1}{(5m)^2}[/tex]

?

Try:

[tex] \frac{1}{d^2} = \frac{1}{81m^2} + \frac{1}{25m^2}[/tex]
up to here you are fine. But
[tex]\Rightarrow d^2 = 81m^2 + 25m^2 = 106m^2 \Rightarrow d = 10.3m[/tex]
is not true. if
[tex]\frac{1}{a}= \frac{1}{b}+ \frac{1}{c}[/tex]
it is NOT true that a= b+ c.
More specifically,
[tex]\frac{1}{\frac{1}{b}+ \frac{1}{c}}\ne b+ c[/tex]

I recommend that you multiply both sides of the equation by the least common divisor of all three fractions, (25)(81)d2.
 
Last edited by a moderator:
Or even better, by (25)(81)d2m2
 

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