# Systems of equations help?

## $- 10 x - 20 y = - 20$ $- 5 x - 10 y = 10$ With work (please!)

May 14, 2018

The systems of equns.has no solution.$\to \phi$

#### Explanation:

Here,

$- 10 x - 20 y = - 20$

Dividing each term by $\left(- 10\right)$ ,we get

color(red)(x+2y=2...to(1)

Also given that,

$- 5 x - 10 y = 10$

Dividing each term by $\left(- 5\right)$ ,we get

color(red)(x+2y=-2...to(2)

Subtracting equn.$\left(1\right)$ from $\left(2\right)$

$x + 2 y = 2$

$x + 2 y = - 2$

ul(- -color(white)(.........)+

$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots . .} 0 = 4 \to$ which is false statement.

Thus, the pair of equn. has no solution.

Let us draw the graphs of equn. $\left(1\right) \mathmr{and} \left(2\right)$

From the graph ,we can say that the lines are parallel.

i.e.two lines donot intersect anywhere.

So, the systems of equns.has no solution. Note:
We know that :If for ${a}_{1} , {b}_{1} , {c}_{1} , {a}_{2} , {b}_{2} , {c}_{2} \in \mathbb{R}$

${a}_{1} x + {b}_{1} y + {c}_{1} = 0 , w h e r e , {a}_{1}^{2} + {b}_{1}^{2} \ne 0$

${a}_{2} x + {b}_{2} y + {c}_{2} = 0 , w h e r e , {a}_{2}^{2} + {b}_{2}^{2} \ne 0 \mathmr{and}$

$\mathmr{and} {a}_{1} / {a}_{2} = {b}_{1} / {b}_{2} \ne {c}_{1} / {c}_{2} \implies N o \textcolor{w h i t e}{.} S o l u t i o n .$

In short, $\frac{1}{1} = \frac{2}{2} \ne \frac{2}{- 2} \to \phi$