Question

Solve$2x - \dfrac{3}{y} = 9,{\kern 1pt} {\kern 1pt} 3x + \dfrac{7}{y} = 2$. Hence find the value of$k$, if $x = ky + 5$

Answer 1

We are asked to solve the equations and find the value of $k$ . To solve the equations, we are going to use the elimination method by equating the coefficients of $x$ and $y$ . Eliminate either of the variables and substitute it in one of the equation to get the other variable. Using this method, we can find the value of $x$ and $y$ .

Complete answer:
We are given the equations$2x - \dfrac{3}{y} = 9,{\kern 1pt} {\kern 1pt} 3x + \dfrac{7}{y} = 2$ to find the value of $k$from$x = ky + 5$
Now let us take the LCM of the coefficients of $x$ from both the given equations,
$LCM(2,3) = 6$then,
$3 \times (2x - \dfrac{3}{y} - 9 = 0)$
$\Rightarrow 6x - \dfrac{9}{y} - 27 = 0$--------$(1)$
$2 \times ({\kern 1pt} 3x + \dfrac{7}{y} - 2 = 0)$
$\Rightarrow 6x + \dfrac{{14}}{y} - 4 = 0$--------$(2)$
Now subtracting equation $(1)$ from equation $(2)$we get,
$= 6x - \dfrac{9}{y} - 27 - (6x + \dfrac{{14}}{y} - 4) = 0$
${\kern 1pt} \Rightarrow 6x - \dfrac{9}{y} - 27 - 6x - \dfrac{{14}}{y} + 4 = 0{\kern 1pt}$
$\Rightarrow \dfrac{{ - 23}}{y} - 23 = 0$
$\Rightarrow \dfrac{{ - 23}}{y} = 23$
$\Rightarrow y = \dfrac{{ - 23}}{{23}}$
$\Rightarrow y = - 1$
Now substituting $y = - 1$in equation $6x + \dfrac{{14}}{y} - 4 = 0$we get,
$6x + \dfrac{{14}}{{ - 1}} - 4 = 0$
$\Rightarrow 6x - 18 = 0$
$\Rightarrow 6x = 18$
$\Rightarrow x = 3$
Therefore our required solution for the above pair of equations is $(3, - 1)$
Now let us substitute $(3, - 1)$in $x = ky + 5$to get,
$x = ky + 5$
$\Rightarrow 3 = k.( - 1) + 5$
$\Rightarrow 3 = - k + 5$
$\Rightarrow k = 5 - 3$
$\Rightarrow k = 2$
Hence the value of $k$is$2$.

Additional information:
A linear equation is an equation that is written for two exceptional variables. This equation may be a linear aggregate of those variables, and a steady can be present. Surprisingly, while any linear equation is plotted on a graph, it will necessarily produce an instantly line - as a result they are called: Linear equations. Linear Equations are an extensive sort of equations altogether. There may be linear equations in one variable, linear equations in two variables, and so on. In every equation, one issue stays regular: The highest and the best diploma of all variables in the equation should be$1$. Other than that, constants $0$diploma variables can be there.

Note:
It is very important that we know how to calculate the LCM of the numbers if needed and then take one variable from either of the variables to calculate the LCM of the coefficients and then multiply each of them to eliminate either of the variables and then calculate the value of the other variable using substitution method.