Sum of two numbers is 24 - and sum of their reciprocals is 1-b- Find the numbers

Let the first number be x Second number be 24 x Now, 1/x + 1/(24 x) = 1/6 (24 x + x)/ x (24 x) = 1/6 24/(24 x^2) = 1/6 So, x^2 24x + 144 = 0 (x 12)^2 = ...

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Standard VIII

Mathematics

Solving Linear Equations Using Transpose of Terms

Sum of two nu...

Question

Sum of two numbers is 24, and sum of their reciprocals is $\frac{1}{b}$ . Find the numbers

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Solution

Let the first number be x Second number be 24 - x
Now, $\frac{1}{x} + \frac{1}{(24 - x)} = \frac{1}{6}$
$(24 - x + x) / x (24 - x) = \frac{1}{6}$
$\frac{24}{(24 - x^{2})} = \frac{1}{6}$
So, $x^{2} - 24 x + 144 = 0$
$(x - 12)^{2} = 0$
$x = 12$
So, first number be 12
Second number be 24 - 12 = 12

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Let the first number be x Second number be 24 - x Now, 1x+1(24−x)=16 (24−x+x)/x(24−x)=16 24(24−x2)=16 So, x2−24x+144=0 (x−12)2=0 x=12 So, first number be 12 Second number be 24 - 12 = 12

Sum of two numbers is 24, and sum of their reciprocals is $\frac{1}{b}$ . Find the numbers

Let the first number be x Second number be 24 - x
Now, $\frac{1}{x} + \frac{1}{(24 - x)} = \frac{1}{6}$
$(24 - x + x) / x (24 - x) = \frac{1}{6}$
$\frac{24}{(24 - x^{2})} = \frac{1}{6}$
So, $x^{2} - 24 x + 144 = 0$
$(x - 12)^{2} = 0$
$x = 12$
So, first number be 12
Second number be 24 - 12 = 12