Problems from Reduction or elimination method

Development:

In this system, operating directly between equations, it has not become possible to eliminate any unknown. But if the first equation multiplies by $$-2$$ and joins the second one it becomes possible to annul $$x$$:

$$[x+2y=0]\cdot(-2) \Rightarrow -2x-4y=0$$

This equation is equivalent to the first one, so it can be used to operate this: $$$\begin{eqnarray} & & \ \ \ 2x-5y=18 \\\\ &+ & \underline{-2x-4y= \ 0} \\\\ & & \ \ \ 0 \ \ -9y=18 \end{eqnarray}$$$

Of the resultant equation it is deduced that: $$$-9y=18 \Rightarrow y=-\dfrac{18}{9}=-2$$$ Now all that remains is to look for the value of $$x$$ substituting in the first equation: $$$x+2y=0 \Rightarrow x=-2y \Rightarrow x=-2\cdot(-2)=4$$$

Solution:

$$x=4; y=-2$$

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Development:

It is necessary to resort to the reduction method for eliminating unknowns and to simplify equations. But first, perhaps, it is possible to simplify some before starting.

The first equation of the first system can subdivided between $$2$$ to obtain a simpler equivalent equation.

$$\dfrac{2x+4y=12}{2} \Rightarrow x+2y=6$$

The exchange is realized in the system:

$$\left.\begin{array}{c} x+2y=6 \\ -x-5y=-39 \end{array} \right\}$$

It is observed that if the first equation joins the second one it becomes possible to eliminate $$x$$ from the last one: $$$\begin{eqnarray} & & -x-5y=-39 \\ &+ & \underline{ \ x \ \ +2y= \ 6} \\ & & \ 0 \ \ -3y=-33 \end{eqnarray}$$$

The resultant equation is an equivalent of the second one and it allows us to find straight away: $$$-3y=-33 \Rightarrow y=\dfrac{-33}{-3}=11$$$

Now, it is possible to obtain the value of $$x$$ substituting in the first equation: $$$x=6-2y \Rightarrow x=6-2\cdot(11) \Rightarrow x=6-22=-16$$$

Solution:

$$x=-16; y=11$$

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