Question

Chapter = Linnear simulation equation. Please answer it fast whose answer is correct I give it brainlieast answer​

Answer 1

${\pmb{\underline{\sf{ Required \ Solution ... }}}}$

${\sf{ 2(x-y)=3 \ \ \ \ \ \cdots(1) }} \\ {\sf{ 5x+8y=14 \ \ \ \ \ \cdots(2) }}$

$\colon\implies{\sf{2(x - y) = 3}} \\ \\ \colon\implies{\sf{ 2x-2y = 3 }}$

Now, We can Equalise both Equations by Multiplying Eq.(1) with 5 and Eq.(2) with 2 as:-

${\sf{ 10x-10y=15 \ \ \ \ \ \cdots(1) }} \\ {\sf{ -10x-16y= -28 \ \ \ \ \ \cdots(2) }}$

After Calculating, we've that:-

$\colon\implies{\sf{ \cancel{-} 26y = \cancel{-} 13 }} \\ \\ \colon\implies{\sf{ 26y = 13 }} \\ \\ \colon\implies{\sf{ y = \cancel{ \dfrac{13}{26} } }} \\ \\ \colon\implies{\sf{ y = \dfrac{1}{2} }}$

Now, this time by putting value of y in any Equation to get desired value as:-

$\colon\implies{\sf{ 5x + 8y = 14 }} \\ \\ \\ \colon\implies{\sf{ 5x + \cancel{8} \times \dfrac{1}{ \cancel{2} } = 14 }} \\ \\ \\ \colon\implies{\sf{ 5x + 4 = 14 }} \\ \\ \\ \colon\implies{\sf{ 5x = 14-4}} \\ \\ \\ \colon\implies{\sf{ x = \cancel{ \dfrac{10}{5} } }} \\ \\ \\ \colon\implies{\underline{\boxed{\sf{x = 2 }}}}$

Hence,

${\pmb{\underline{\sf{ The \ value \ of \ x \ and \ y \ are \ 2 \ and \ \dfrac{1}{2} \ Respectively. }}}}$

$\\ {\pmb{\underline{\sf{More \ to \ Know ... }}}}$

Step 1: Multiply each equation by a suitable number so that the two equations have the same leading coefficient.

Step 2: Subtract the second equation from the first.

Step 3: Now, We'll get value of any variable then we can easily get the value of other variable by putting value of one variable (that we've found).

Answer 2

$\begin{gathered} \LARGE{\orange{\boxed{\pmb{\underline{\sf{ Required \: Solution↓}}}}}} \\ \end{gathered}$

$\begin{gathered} {\sf{ 2(x-y)=3 \ \ \ \ \ \cdots(1) }} \\ {\sf{ 5x+8y=14 \ \ \ \ \ \cdots(2) }} \\ \end{gathered}$

$\begin{gathered} \colon\implies{\sf{2(x - y) = 3}} \\ \\ \colon\implies{\sf{ 2x-2y = 3 }} \end{gathered}$

Now, We can Equalise both Equations by Multiplying Eq.(1) with 5 and Eq.(2) with 2 as:-

$\begin{gathered} {\sf{ 10x-10y=15 \ \ \ \ \ \cdots(1) }} \\ {\sf{ -10x-16y= -28 \ \ \ \ \ \cdots(2) }} \\ \end{gathered}$

After Calculating, we've that:-

$\begin{gathered} \colon\implies{\sf{ \cancel{-} 26y = \cancel{-} 13 }} \\ \\ \colon\implies{\sf{ 26y = 13 }} \\ \\ \colon\implies{\sf{ y = \cancel{ \dfrac{13}{26} } }} \\ \\ \colon\implies{\sf{ y = \dfrac{1}{2} }} \\ \end{gathered}$

Now, this time by putting value of y in any Equation to get desired value as:-

$\begin{gathered} \colon\implies{\sf{ 5x + 8y = 14 }} \\ \\ \\ \colon\implies{\sf{ 5x + \cancel{8} \times \dfrac{1}{ \cancel{2} } = 14 }} \\ \\ \\ \colon\implies{\sf{ 5x + 4 = 14 }} \\ \\ \\ \colon\implies{\sf{ 5x = 14-4}} \\ \\ \\ \colon\implies{\sf{ x = \cancel{ \dfrac{10}{5} } }} \\ \\ \\ \colon\implies{\underline{\boxed{\sf{x = 2 }}}} \\ \end{gathered}$

Hence,

$\begin{gathered} {\pmb{\underline{\sf{ The \: value \: of \: x \: and \: y \: are \: 2 \: and \: \dfrac{1}{2} \ Respectively. }}}} \\ \end{gathered}$

_____________________

$\begin{gathered} \\ \LARGE{\red{\boxed{\pmb{\underline{\sf{More \: Information↓}}}}}} \\ \end{gathered}$

Step 1:

Multiply each equation by a suitable number so that the two equations have the same leading coefficient.

Step 2:

Subtract the second equation from the first.

Step 3:

Now, We'll get value of any variable then we can easily get the value of other variable by putting value of one variable (that we've found).

____________________