Question

What multiple of equation $2$ should be subtracted from equation $1$?
$$\begin{array}{rl}& 2x-4y=6\\ & -x+5y=0\end{array}$$
After this elimination step, solve the triangular system.

If the right-hand side changes to $(-6,0)$, what is the new solution?

Answer 1

Step-1:Finding the triangular system after elimination:

Finding the multiple of equation $2$ should be subtracted from equation $1$:

Given,

$$\begin{array}{rl}& 2x-4y=6\\ & -x+5y=0\end{array}$$

We know that ,

The system of the linear equation be $AX=b$.

writing the given equation as system of the linear equation

$\left[\begin{array}{cc}2& -4\\ -1& 5\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}6\\ 0\end{array}\right]$

So,

$$\begin{gathered} \Rightarrow A=\left[\begin{array}{cc}2& -4\\ -1& 5\end{array}\right] \\ =\left[\begin{array}{cc}2& -4\\ -2& 10\end{array}\right]\left[\because R_2\to 2R_2\right] \\ =\left[\begin{array}{cc}2& -4\\ 0& 6\end{array}\right]\left[\because R_2\to R_2+R_1\right] \end{gathered}$$

$\left[\begin{array}{cc}2& -4\\ 0& 10\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}6\\ 0\end{array}\right]$

Hence, two is the the multiple of equation $2$ should be subtracted from equation $1$.

Step-2: Finding the new solution, the right-hand side changes to $(-6,0)$:

The new system of the linear equation

$\left[\begin{array}{cc}2& -4\\ 0& 10\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-6\\ 0\end{array}\right]$

Hence the equation be

$$\begin{gathered} 2x-4y=-6 \\ 10y=0 \end{gathered}$$

Solve the above equations we get,

$\Rightarrow y=0$

Substitute in $y=0$in $2x-4y=-6$

$$\begin{gathered} \Rightarrow 2x-4\times 0=-6 \\ \Rightarrow 2x=-6 \\ \Rightarrow x=-3 \end{gathered}$$

Hence, the new solutions are $\left(-3,0\right)$.