Question

How do you solve this system of equations:
$2x+y=-2\And 5x+3y=-8$

Answer 1

Hint: The system of equations given in the above problem are: $2x+y=-2\And 5x+3y=-8$. Now, to solve this system of equations, we are going to multiply the first equation by 3 and then we will subtract the second equation from this modified first equation. In this way, y will be eliminated and we will get the value of x. Then, we will substitute this value of x in any of the given two equations and find the value of y.

Complete step by step answer:
The system of equations given in the above problem is given as follows:
$\begin{align}
  & 2x+y=-2.......Eq.(1) \\
 & 5x+3y=-8........Eq.(2) \\
\end{align}$
Multiplying eq. (1) by 3 we get,
$\begin{align}
  & \Rightarrow \left( 2x+y=-2 \right)\times 3 \\
 & \Rightarrow 6x+3y=-6.......Eq.(3) \\
\end{align}$
Now, subtracting eq. (2) from eq. (3) we get,
$\begin{align}
  & 6x+3y=-6 \\
 & 5x+3y=-8 \\
 & \begin{matrix}
   - & - & + \\
\end{matrix} \\
 & \overline{x+0=\left( 8-6 \right)} \\
\end{align}$
Simplifying the above equation by subtracting 6 from 8 we get,
$x=2$
Substituting the above value of x in eq. (1) we get,
$\begin{align}
  & \Rightarrow 2\left( 2 \right)+y=-2 \\
 & \Rightarrow 4+y=-2 \\
\end{align}$
Subtracting 4 on both the sides of the above equation we get,
$\begin{align}
  & \Rightarrow 4+y-4=-2-4 \\
 & \Rightarrow y=-6 \\
\end{align}$
From the above, we have solved the system of equations and the value of x and y which we are getting as:
$\begin{align}
  & x=2; \\
 & y=-6 \\
\end{align}$

Note: To check whether the values of x and y which we are getting in the above problem is correct or not we are going to substitute these values of x and y in any of the two equations and see whether they are satisfying that equation or not.
The values of x and y which we have solved above are:
$\begin{align}
  & x=2; \\
 & y=-6 \\
\end{align}$
Substituting the above values of x and y in eq. (1) we get,
$\begin{align}
  & \Rightarrow 2\left( 2 \right)+\left( -6 \right)=-2 \\
 & \Rightarrow 4-6=-2 \\
 & \Rightarrow -2=-2 \\
\end{align}$
As you can see that L.H.S = R.H.S so the values of x and y which we have solved are correct.