Question

If sum of two numbers Is 36 . If 9 is subtracted from 8 times the first numbers , we get second number . find the numbers​

Answer 1

Answer:

Missing numbers are : first number = $5$ , and second number = $31$ .

Step-by-step explanation:

To find : missing numbers .

Given :  sum of two numbers Is 36 . If 9 is subtracted from 8 times the first numbers , we get second number .

Solution :

• As per given data we know that , sum of two numbers Is $36$ . If $9$ is subtracted from $8$  times the first numbers , we get second number .
• Let  , first number be $x$ , and $(8x - 9)$ be the second number .
• According to given condition we write as , $x + ( 8x -9 )= 36$
• Using transposition method we find the value of $x$ , by transposing constants at RHS and leaving variable at LHS .

          $x + ( 8x -9 )= 36\\\\ x + 8x -9 = 36\\\\ 9x -9 = 36\\\\ 9x = 36+ 9\\\\9x = 45\\\\x = \frac{45}{9}\\\\ x= 5$

• Hence , first number $= x =5$
• Now , substituting value of $x$ in $(8x - 9)$ to find second number ,

        Second number =  $(8x - 9) = 8\times 5 - 9$

                                                      $= 40 -9 \\\\ =31$

Answer 2

Given : The Sum of two numbers is 36 .The first number is 9 less than 8 times the second number .

To Find : Find the Numbers

$\\ \qquad{\rule{200pt}{3pt}}$

SolutioN : Let us form the Equation for this question and after Solving it we can get the Numbers .Let's Solve :

$\maltese$ According to the Question :

$\longmapsto$ Let the Second Number be y .So,

$\qquad \; \; {\pmb{\sf{ Second \; No. = y }}}$

$\longmapsto$ First Number is 9 less than 8 times the Second Number .So,

$\qquad \; \; {\pmb{\sf{ First \; No. = 8y - 9 }}}$

$\maltese$ Calculating the Value of y :

$\qquad {\dashrightarrow{\qquad{\sf{ \bigg( 8y - 9 \bigg) + y = 36 }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ 9y - 9 = 36 }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ 9y = 36 + 9 }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ 9y = 45 }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ y = \dfrac{45}{9} }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ y = \cancel\dfrac{45}{9} }}}} \\ \\ \\ \ {\qquad \; \; {\dashrightarrow{\underline{\boxed{\pmb{\frak{ y = 5 }}}}}}} \; {\red{\bigstar}}$

$\maltese$ Calculating the Numbers :

• 1st No. = y = 5
• 2nd No. = 8y - 9 = 8(5) - 9 = 31

$\therefore \;$ The Two numbers are 5 and 31 .

$\\ \qquad{\rule{200pt}{3pt}}$