Question

How do you solve the system of equations \[-8x+3y=7\] and \[13x-3y=-1\]?

Answer 1

Hint: In this problem, we have to solve and find the value of the given system of equations. We can use elimination methods to solve the given system of equations by subtracting any one of the variables to get one value of the variable and we can substitute it in any of the two equations to get the value of another variable.

Complete step by step solution:
We know that the given system of equations are,
\[-8x+3y=7\]…….. (1)
\[13x-3y=-1\]…….. (2)
we can now use the elimination method to solve for the equation.
We can see that the equations have similar terms in the coefficient of y with opposite signs, which can be cancelled and we can write the remaining terms.
We can now add the equations (1) and (2), we get
\[\Rightarrow -8x+3y-7+13x-3y+1=0\]
We can now cancel the similar terms and simplify the above step, we get
\[\begin{align}
  & \Rightarrow 5x-6=0 \\
 & \Rightarrow x=\dfrac{6}{5} \\
\end{align}\]
We can now substitute the value of x in equation (1), we get
\[\begin{align}
  & \Rightarrow -8\left( \dfrac{6}{5} \right)+3y=7 \\
 & \Rightarrow y=\dfrac{83}{15} \\
\end{align}\]
Therefore, the value of \[x=\dfrac{6}{5},y=\dfrac{83}{15}\].

Note: Students make mistakes while using the elimination method, where we will solve for the first variable in one of the equations, then substitute the result into the other. We should also concentrate while cancelling similar terms with opposite signs, if not it gives an incorrect answer. We can now check the answers by substituting it.
We can now substitute \[x=\dfrac{6}{5},y=\dfrac{83}{15}\] in (1), we get
\[\Rightarrow -8\left( \dfrac{6}{5} \right)+3\left( \dfrac{83}{15} \right)=-\dfrac{48}{5}+\dfrac{83}{5}=7\]
Therefore, the answer is correct.