Question

In the formula \[H = \dfrac{{2ab}}{{a + b}}\]. If \[a = 5,H = 8\]. Find \[b\].

Answer 1

Hint: Here we directly substitute the values to the formula and solve using addition, multiplication, subtraction and division of numbers.
* If we have an equation in \[m\] variables and we are given values of \[(m - 1)\] variables, then we can find the value of the remaining variable by substituting the values of other variables into the equation. Example: we have an equation \[x + y + z = 7\] which is an equation having \[3\] variables, and we are given values \[x = 1,z = 3\] then we can find the value of the remaining variable \[y\]. We substitute the values \[x = 1,z = 3\] in equation \[x + y + z = 7\],
\[
  1 + y + 3 = 7 \\
  y + 4 = 7 \\
  y = 7 - 4 = 3 \\
 \]

Complete step-by-step answer:
Given, the formula \[H = \dfrac{{2ab}}{{a + b}}\]
Elaborating the given formula \[H = \dfrac{{2 \times a \times b}}{{a + b}}\]
We are given the values, \[a = 5\] and \[H = 8\]
Substituting the values \[a = 5\] and \[H = 8\] in the equation \[H = \dfrac{{2 \times a \times b}}{{a + b}}\]
\[8 = \dfrac{{2 \times 5 \times b}}{{5 + b}}\]
\[8 = \dfrac{{10 \times b}}{{(5 + b)}}\]
Cross multiplying the term in the denominator of RHS and the term of LHS of the equation.
\[8 \times (5 + b) = 10 \times b\]
Multiplying the values to bracket
\[
  8 \times 5 + 8 \times b = 10 \times b \\
  40 + 8b = 10b \\
 \]
Shifting the terms with variable to one side
\[40 = 10b - 8b\]
Taking \[b\] common on RHS of the equation
\[
  40 = b(10 - 8) \\
  40 = b \times 2 \\
 \]
Dividing both sides of the equation by \[2\]
\[\dfrac{{40}}{2} = \dfrac{{b \times 2}}{2}\]
Cancel out the same terms from numerator and denominator from both sides of the equation.
\[20 = b\]

Therefore, the value of \[b = 20\]

Note:
Students are likely to mistake while shifting the values from one side of the equation to another. Always keep in mind the sign changes from positive to negative and vice versa when shifting the values to another side of the equation. Students can also solve the formula first and then substitute the values later i.e.
\[H = \dfrac{{2ab}}{{a + b}}\]
Cross multiplying the LHS and RHS
\[H \times (a + b) = 2ab\]
\[Ha + Hb = 2ab\]
Shifting all the values to LHS of the equation
\[Ha + Hb - 2ab = 0\]
Now substitute the values \[a = 5\] and \[H = 8\]in the above equation and solve for the value of \[b\].
\[
  8 \times 5 + 8b - 2 \times 5b = 0 \\
  40 + 8b - 10b = 0 \\
  40 - 2b = 0 \\
 \]
Shifting the constant term to one side
\[ - 2b = - 40\]
Multiply both the sides of equation by \[( - 1)\]
\[
  ( - 1) \times ( - 2b) = ( - 1) \times ( - 40) \\
  2b = 40 \\
    \\
 \]
Dividing both sides of the equation by \[2\]
\[\dfrac{{b \times 2}}{2} = \dfrac{{40}}{2}\]
Cancel out the same terms from numerator and denominator from both sides of the equation.
\[b = 20\]

Therefore, the value of \[b = 20\].