Question

x+2y=10;2x+4y=8 prove that they are in consistence​

Answer 1

Answer:

x + 2y - 10 = 0

2x + 4y - 8 = 0

a1 = 1 , b1 = 2 , c1 = -10

a2 = 2 , b2 = 4 , c2 = -8

If they should be inconsistent then

a1/a2 = b1/b2 ≠ c1/c2

1/2 = 2/4 ≠ -10/-8

1/2 = 1/2 ≠ 5/4

a1/a2 = b1/b2 = c1/c2 then it is consistent or dependent..

a1/a2 ≠ b1/b2 then it is also consistent..

so, as the same

a1/a2 = b1/b2 ≠c1/c2 then it is inconsistent pair...

So by all these information you can get your answer

Thank you

Answer 2

${\huge{\red{\sf{Given}}}}\begin{cases}\leadsto x+2y =10\\\leadsto 2x+4y =8 \end{cases}$

$\huge\red{\sf{To\:Prove}}$$\begin{cases} They\:are\:incosistent \end{cases}$

$\huge\red{\overbrace{\underbrace{Answer}}}$

Let two equations be,

$\mapsto a_{1}x+b_{1}y+c_{1}=0$

$\mapsto a_{2}x+b_{2}y+c_{2}=0$

For incosistent solutions,

$\large\green{\boxed{\sf{\orange{\dfrac{a_{1}}{a_{2}}= \dfrac{b_{1}}{b_{2}}\neq \dfrac{c_{1}}{c_{2}}}}}}$

$\implies \dfrac{1}{2}=\dfrac{2}{4}\neq\dfrac{10}{8}$

$\implies \dfrac{1}{2}=\dfrac{\cancel{2}}{\cancel{4}^{2}}\neq\dfrac{10}{8}$

$\implies \dfrac{1}{2}=\dfrac{1}{2}\neq\dfrac{5}{4}$

${\underline{\boxed{\red{\sf{\therefore \dfrac{a_{1}}{a_{2}}= \dfrac{b_{1}}{b_{2}}\neq \dfrac{c_{1}}{c_{2}}}}}}}$

Hence proved.