Question

How do you solve the system of equations $8x+y=-16$ and $-3x+y=-5$?

Answer 1

Hint: For solving the given equations, $8x+y=-16$ and $-3x+y=-5$, we can use the method of elimination. The coefficient of the variable $y$ is the same in both the equations, and is equal to $1$. Therefore, on subtracting the two equations, the variable $y$ will get removed and we will obtain a linear equation in $x$ which can be easily solved using algebraic operations to get the value of $x$. This value can be substituted in any of the two equations for getting the value of $y$.

Complete step by step solution:
The equations given in the above question are
$\begin{align}
  & \Rightarrow 8x+y=-16.......\left( i \right) \\
 & \Rightarrow -3x+y=-5.......\left( ii \right) \\
\end{align}$
In the above two equations (i) and (ii), we can see that the coefficients of the variable $y$ is the same in both the equations and is equal to $1$. Therefore, we can eliminate the variable $y$ by subtracting the equation (ii) from the equation (i) to get
\[\begin{align}
  & \Rightarrow 8x+y-\left( -3x+y \right)=-16-\left( -5 \right) \\
 & \Rightarrow 8x+y+3x-y=-16+5 \\
 & \Rightarrow 11x=-11 \\
\end{align}\]
Dividing both sides by $11$ we get
\[\begin{align}
  & \Rightarrow \dfrac{11x}{11}=\dfrac{-11}{11} \\
 & \Rightarrow x=-1 \\
\end{align}\]
Substituting this in the equation (ii) we get
$\begin{align}
  & \Rightarrow -3\left( -1 \right)+y=-5 \\
 & \Rightarrow 3+y=-5 \\
\end{align}$
Subtracting $3$ from both sides, we finally get
$\begin{align}
  & \Rightarrow 3+y-3=-5-3 \\
 & \Rightarrow y=-8 \\
\end{align}$

Hence, the solution of the given equations is $x=-1$ and $y=-8$.

Note: We have obtained the solution by substituting the value of the variable $x$ in the second equation $-3x+y=-5$. For checking whether our obtained solution is correct or not, we must substitute the final values in the first equation $8x+y=-16$ and confirm the equality of LHS and RHS. Further, we can also solve the given system using the method of substitution. For this, we can obtain the value of $y$ in terms of $x$ from the first equation and substitute this value in the second equation to get the value of $x$. On substituting this value of $x$ in any of the equations, we will obtain the value of $y$.