Question

How do solve the following linear system $y=-6x-9,-9x+5y=1$ ?

Answer 1

Hint: For solving the following linear system $y=-6x-9,-9x+5y=1$ , we will use the substitution method in which we will have to rearrange one equation for $y$ in terms of $x$ . After that we will substitute $y$ in another equation and then will expand the equation to solve it. After solving the equation we will get the value for $x$ and then we will put the value of $x$ in the equation that is for $y$ in terms of $x$ .

Complete step by step solution:
Here, we have two following linear equation as:
$\Rightarrow y=-6x-9$ … $\left( i \right)$
And
$\Rightarrow -9x+5y=1$ … $\left( ii \right)$
Now, we will solve the given equation by the substitution method.
Since, we have already equation $\left( i \right)$ for $y$ in terms of $x$ . So, there is not any need to rearrange the equation $\left( i \right)$since it is already arranged for $y$ in terms of $x$ as:
$\Rightarrow y=-6x-9$ … $\left( i \right)$
Now, we will substitute $-6x-9$ for $y$ in the second equation $-9x+5y=1$ for finding the value of $x$ as:
$\Rightarrow -9x+5\left( -6x-9 \right)=1$
After opening the bracket, we can write the equation as:
$\Rightarrow -9x+5\times \left( -6x \right)-5\times 9=1$
Here, we need to solve the multiplication of the above equation as:
$\Rightarrow -9x-30x-45=1$
Let us combine the equal like terms that are $-9x$ and $-30x$ in the above equation. We will get $-39x$ by adding them as:
$\Rightarrow -39x-45=1$
Now, let us change the place of $-45$ . Since, it is also a number as $1$ is a number.
$\Rightarrow -39x=1+45$
Here, we will add $1$ and $45$ so that the above equation will be as:
$\Rightarrow -39x=46$
Here, we will again change the place coefficient to $x$ for getting the value of $x$ as:
$\Rightarrow x=\dfrac{46}{-39}$
So, we will have the value of $x$ as:
$\Rightarrow x=-\dfrac{46}{39}$
Now, we will put the value of $x$ in equation $\left( i \right)$ for getting the value of $y$ as:
$\Rightarrow y=-6\left( -\dfrac{46}{39} \right)-9$
Here, we need to open the bracket and make calculation to find the value of $y$ as:
$\Rightarrow y=\dfrac{92}{13}-9$
We will have to multiply and divide by 12 with 9 to make calculation easy as:
$\Rightarrow y=\dfrac{92}{13}-9\times \dfrac{13}{13}$
Now, we will calculate the above equation for $y$ as:
$\Rightarrow y=\dfrac{92}{13}-\dfrac{117}{13}$
$\Rightarrow y=\dfrac{92-117}{13}$
$\Rightarrow y=\dfrac{-25}{13}$
So, we will have the value of $y$ as:
$\Rightarrow y=-\dfrac{25}{13}$
Hence, we have $\left( -\dfrac{46}{39} \right)$ and $\left( -\dfrac{25}{13} \right)$ for the value of $x$ and $y$ respectively.

Note: : Let us check the value of $x$ and $y$ satisfy the linear equation or not by putting $\left( -\dfrac{46}{39} \right)$ and $\left( -\dfrac{25}{13} \right)$ for the value of $x$ and $y$ in equation $\left( ii \right)$ respectively.
Since, the equation is $-9x+5y=1$. After applying the value of $x$ and $y$ respectively equation will be:
$\Rightarrow -9x+5y=1$
L.H.S.-
$\Rightarrow -9x+5y$
$\Rightarrow -9\left( -\dfrac{46}{39} \right)+5\left( -\dfrac{25}{13} \right)$
$\Rightarrow \left( \dfrac{-3\times -46}{13} \right)+\left( \dfrac{5\times -25}{13} \right)$
Now, we will make required calculation for the above equation as:
$\Rightarrow \left( \dfrac{138}{13} \right)+\left( \dfrac{-125}{13} \right)$
The above equation can be written as:
$\Rightarrow \dfrac{138}{13}-\dfrac{125}{13}$
After subtracting the above equation we have:
$\Rightarrow \dfrac{13}{13}$
Now, we have the final result from above term as
$\Rightarrow 1$
That is equal to R.H.S.
Hence, the solution is correct for the given question.