Question

Solve the following pair of equations by reducing them to a pair of linear equations:
\[\dfrac{{7x - 2y}}{{xy}} = 5\]
\[\dfrac{{8x + 7y}}{{xy}} = 15\]

Answer 1

Hint: To reduce the equations, first simplify the given equations and then substitute particular terms with another variable such that after substitution the equation reduces to a linear equation. Then solve it simultaneously by the method of substitution.

Complete step by step solution:
First consider the first equation:
       \[\dfrac{{7x - 2y}}{{xy}} = 5\]
\[ \Rightarrow \dfrac{{7x}}{{xy}} - \dfrac{{2y}}{{xy}} = 5\]
\[ \Rightarrow \dfrac{7}{y} - \dfrac{2}{x} = 5\]……(I)
Similarly for the second equation:
      \[\dfrac{{8x + 7y}}{{xy}} = 15\]
\[ \Rightarrow \dfrac{{8x}}{{xy}} + \dfrac{{7y}}{{xy}} = 15\]
\[ \Rightarrow \dfrac{8}{y} + \dfrac{7}{x} = 15\]……(II)
Observe that in both the simplified equations the terms \[\dfrac{1}{x}\] and \[\dfrac{1}{y}\] are common hence we substitute these by \[a\] and \[b\] respectively, to transform the equations to linear equations:
Substituting \[\dfrac{1}{x} = a,\dfrac{1}{y} = b\] in (I) and (II) respectively,
\[7b - 2a = 5\]……(I)
\[8b + 7a = 15\]……(II)
Next solve the above two linear equations simultaneously:
For that we use the method of substitution. First multiply equation (I) with \[7\] and equation (II) with \[2\] so that coefficient of one variable that is \[a\] becomes same and thus one variable can be cancelled out:
(\[7b - 2a = 5\]) \[ \times 7\]\[ = \] \[49b - 14a = 35\]…… (III)
(\[8b + 7a = 15\]) \[ \times 2\] \[ = \] \[16b + 14a = 30\]……(IV)
Adding equations (III) and (IV) :
     \[49b - 14a = 35\]
\[ + \] \[16b + 14a = 30\]
     _____________
       \[65b + 0 = 65\]

\[ \Rightarrow 65b = 65\]
\[ \Rightarrow b = 1\]
Next substitute the value of \[b\] in equation (I):
\[7b - 2a = 5\]
\[ \Rightarrow 7(1) - 2a = 5\]
Changing sides:
\[ \Rightarrow 2a = 7 - 5\]
\[ \Rightarrow 2a = 2\]
\[ \Rightarrow a = 1\]
Hence \[a = 1,b = 1\]
Now, \[a = \dfrac{1}{x},b = \dfrac{1}{y}\]
\[ \Rightarrow 1 = \dfrac{1}{x},1 = \dfrac{1}{y}\]
\[ \Rightarrow x = 1,y = 1\]

Hence the values of both \[x\] and \[y\] are \[1\].

Note:
Students must be careful that finding the values of the substituted variables \[a\] and \[b\] is not the solution, after finding them the values of \[x\] and \[y\]needs to be substituted and found. Moreover the value of \[a\] could also have been found by substituting the value of \[b\] in any of the equation (II), (III) or (I). Moreover while changing sides the signs of the variables and constants must be taken care of.