Question

How‌ ‌do‌ ‌you‌ ‌solve‌ ‌the‌ ‌system‌ of linear equations ‌$y=-x+2,2y=4-2x$‌ ‌?‌

Answer 1

Hint: We first divide both sides of the second equation by $2$ to simplify it. Now, the two equations that we get are identical. So, we cannot solve for $x,y$ and so, there are infinite solutions to the linear system.

Complete step by step solution:
The two linear equations we have are
$y=-x+2.....\left( 1 \right)$ and
$2y=4-2x.....\left( 2 \right)$
We first divide $2$ to both sides of the equation $\left( 2 \right)$ . The equation thus becomes,
$\Rightarrow \dfrac{2y}{2}=\dfrac{4-2x}{2}.....\left( 3 \right)$
Upon simplifying the above equation, the equation thus becomes,
$\Rightarrow y=2-x....\left( 4 \right)$
In the above equations we see that $y$ is written as a function of $x$ . We now take the above equation $\left( 4 \right)$ and substitute the expression of the right-hand side of the above equation in place of $y$ in equation $\left( 1 \right)$ as shown below
$\Rightarrow y=-x+2$
$\Rightarrow 2-x=-x+2$
If we observe carefully, we will see that from the above equation, we can easily cancel off $-x$ from both the sides of the above expression. This means that we cannot solve for the value of $x$ or that $x$ has no solution for the given system. Same analogy is applicable for $y$ and thus, it is also not solvable. By ‘not solvable’ we mean that there are infinite solutions that satisfy the system.
Therefore, we can conclude that there are infinite solutions to the given linear system.

Note: While simplifying the equations, we must be careful, as in this problem if we make any error in the signs, then we will end up in two different equations which lead to some different solutions of $x,y$ . If we solve it by matrix, we will get,
\[\left( \begin{matrix}
   1 & 1 \\
   2 & 2 \\
\end{matrix} \right)\left( \begin{matrix}
   x \\
   y \\
\end{matrix} \right)=\left( \begin{matrix}
   2 \\
   4 \\
\end{matrix} \right)\]
The coefficient matrix \[\left( \begin{matrix}
   1 & 1 \\
   2 & 2 \\
\end{matrix} \right)\] being singular and the augmented matrix \[\left( \begin{matrix}
   1 \\
   2 \\
\end{matrix}\begin{matrix}
   1 \\
   2 \\
\end{matrix}\begin{matrix}
   2 \\
   4 \\
\end{matrix} \right)\] having rank $1$ , we get infinite solutions.