Question

Solve the following system of equation-
$
  2x - \dfrac{3}{y} = 9 \\
    \\
  3x + \dfrac{7}{y} = 2,y \ne 0 \\
$

Answer 1

Hint- The equations above in the question are not linear equations so we will first make them linear using the method shown in the solution below. Then we will use the method of substitution and elimination.

Complete step-by-step answer:
Let the equations given the question be equation 1 and equation 2-
$ \Rightarrow 2x - \dfrac{3}{y} = 9$ (equation 1)
$ \Rightarrow 3x + \dfrac{7}{y} = 2$ (equation 2)
Now, as we can see that the equations given to us are not linear equations because y is in the denominator in both the equations. So, in order to make the linear, we will-
Let, $\dfrac{1}{y} = v$
Now, substituting the value of v into equation 1 and equation 2-
$ \Rightarrow 2x - 3v = 9$
$ \Rightarrow 3x + 7v = 2$
Let the above equation be equation 3 and equation 4-
$ \Rightarrow 2x - 3v = 9$ (equation 3)
$ \Rightarrow 3x + 7v = 2$ (equation 4)
Now, equation 3 and equation 4 are linear equations. So, we will solve these equations with the method of elimination by equating the coefficients.
Multiply equation 3 by 3 and equation 4 by 2 in order to get the coefficient x equal, we have-
$ \Rightarrow 6x - 9v = 27$
$ \Rightarrow 6x + 14v = 4$
Subtracting the above equations, we will have the value of v-
$
   \Rightarrow 6x - 9v - 6x - 14v = 27 - 4 \\
    \\
   \Rightarrow - 23v = 23 \\
    \\
   \Rightarrow v = - 1 \\
$
Putting the value of v in equation 3 or equation 4-
Considering equation 3-
$
   \Rightarrow 2x - 3v = 9 \\
    \\
   \Rightarrow 2x + 3 = 9 \\
    \\
   \Rightarrow x = 3 \\
$
Since we already know that $\dfrac{1}{y} = v$, we will put the value of v in it we get the value of y-
$
   \Rightarrow \dfrac{1}{y} = v \\
    \\
   \Rightarrow \dfrac{1}{y} = - 1 \\
    \\
   \Rightarrow y = - 1 \\
$
Thus, the values of x and y are $x = 3,y = - 1$.

Note: When the equations given are not linear equations, remember to make them linear first or we won’t get even close to the answer. Using the basic method of substitution and elimination will give us our required answer.