Question

the sum of two natural no. is 9 and the sum of their reciprocals is 9/ 20 find the numbers​

Answer 1

${\large{\underline{\pmb{\frak{Given...}}}}}$

→ the sum of two numbers equals 9

→ sum of their reciprocals is 9/ 20

${\large{\underline{\pmb{\frak{To \; Find...}}}}}$

→ the natural numbers which make the conditions true

${\large{\underline{\pmb{\frak{Understanding \; the\; question...}}}}}$

❍ Concept : Here we two statements related to the numbers which are that

⠀⠀⠀⠀⠀⠀⠀⠀⋆ The these numbers equal 9 when they are added together

⠀⠀⠀⠀⠀⠀⠀⠀⋆ The sum of the reciprocals of the numbers equal to the 9/20

✰ Now let's frame equations according to the statement assigning suitable variables to the 1st and the 2nd numbers as they are undefined and then use substitution  method to solve them.

${\large{\underline{\pmb{\sf{RequirEd \; Solution...}}}}}$

★ The natural numbers which satisfy the statement are 4 and 5

${\large{\underline{\pmb{\frak{Full \; solution...}}}}}$

✪ Now let's assume that,

⠀⠀⠀» The first number as x

⠀⠀⠀» The second number as y

~ Let's frame an equation according to condition 1,

$\longrightarrow \tt x + y = 9$

$\longrightarrow \tt x = 9 - y ---(1)$

~ As per condition 2 let their reciprocals be

$\longrightarrow \tt \dfrac{1}{x}\; and \;\dfrac{1}{y}$

~ Framing an equation according to condition 2

$\longrightarrow \tt \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{9}{20} ---(2)$

~ Substituting the value of x in equation 2

$\longrightarrow \tt \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{9}{20}$

$\longrightarrow \tt \dfrac{1}{9 - y} + \dfrac{1}{y} = \dfrac{9}{20}$

$\longrightarrow \tt \dfrac{y + 9-y}{y(9-y)} =\dfrac{9}{20}$

$\longrightarrow \tt \dfrac{9}{9y-y^2} =\dfrac{9}{20}$

$\longrightarrow \tt \dfrac{1}{y-y^2} = \dfrac{9}{20}$

~ Now let's cross multiply and frame a quadratic equation

$\longrightarrow \tt 20(1) = 9( y - y^ 2)$

$\longrightarrow \tt 20 = 9 y-y^2$

$\longrightarrow \tt 20 - 9y + y^2 = 0$

$\longrightarrow \tt y^2 - 9y +20 = 0$

~ Now let's factorise the quadratic equation

$\longrightarrow \tt y^2 - 9y + 20=0$

$\longrightarrow \tt y^2 -5y -4y = 0$

$\longrightarrow \tt y(y - 5) -4(y-5) =0$

$\longrightarrow \tt (y - 4)(y-5)=0$

~ Now let's find the roots of the equation

$\longrightarrow \tt y - 5 = 0$

$\longrightarrow \tt y = 0+5$

$\longrightarrow \tt y= 5$

~ Now let's find the other root of the equation

$\longrightarrow \tt ( y - 4) = 0$

$\longrightarrow \tt y = 0 + 4$

$\longrightarrow \tt y = 4$

~ Now let's put the value y = 5 in equation 1 and find the value of x

$\longrightarrow \tt x = 9 - y$

$\longrightarrow \tt x = 9 - 5$

$\longrightarrow \tt x = 4$

• Henceforth the numbers are 5 and 4