Math, asked by khawaslaxmi24, 30 days ago

solve the simultaneously linear equations by elimation method x/2+y/3=1 x/3+y/2=1​

Answers

Answered by mathdude500
5

\large\underline{\sf{Solution-}}

Given pair of linear equations are

 \purple{\rm :\longmapsto\:\dfrac{x}{2}  + \dfrac{y}{3}  = 1}

 \purple{\rm :\longmapsto\:\dfrac{3x + 2y}{6} = 1}

 \purple{\rm :\longmapsto\:3x + 2y = 6 -  -  -  -  - (1)}

and

 \red{\rm :\longmapsto\:\dfrac{x}{3}  + \dfrac{y}{2}  = 1}

 \red{\rm :\longmapsto\:\dfrac{2x + 3y}{6}   = 1}

 \red{\rm :\longmapsto\:2x + 3y = 6 -  -  -  -  - (2)}

Now, we have two equations in simplest form as

 \purple{\rm :\longmapsto\:3x + 2y = 6 -  -  -  -  - (1)}

and

 \red{\rm :\longmapsto\:2x + 3y = 6 -  -  -  -  - (2)}

Now, multiply equation (1) by 2 and equation (2) by 3, we get

 \purple{\rm :\longmapsto\:6x + 4y = 12 -  -  -  -  - (3)}

and

 \red{\rm :\longmapsto\:6x + 9y = 18 -  -  -  -  - (4)}

On Subtracting equation (3) from (4), we get

\rm :\longmapsto\:5y = 6

\bf\implies \:y = \dfrac{6}{5}  -  -  -  - (5)

On substituting the value of x in equation (1), we get

 \purple{\rm :\longmapsto\:3x +  \dfrac{12}{5}  = 6}

 \purple{\rm :\longmapsto\:3x = 6 -   \dfrac{12}{5}}

 \purple{\rm :\longmapsto\:3x =   \dfrac{30 - 12}{5}}

 \purple{\rm :\longmapsto\:3x =   \dfrac{18}{5}}

 \purple{\bf\implies \:\:x =   \dfrac{6}{5}}

So, Solution of equations is

\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{x = \dfrac{6}{5} } \\ \\  &\sf{y = \dfrac{6}{5} } \end{cases}\end{gathered}\end{gathered}

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VERIFICATION

Consider First Equation

 \purple{\rm :\longmapsto\:\dfrac{x}{2}  + \dfrac{y}{3}  = 1}

On substituting the values of x and y, we get

 \purple{\rm :\longmapsto\:\dfrac{6}{10}  + \dfrac{6}{15}  = 1}

 \purple{\rm :\longmapsto\:\dfrac{3}{5}  + \dfrac{2}{5}  = 1}

 \purple{\rm :\longmapsto\:\dfrac{3 + 2}{5}    = 1}

 \purple{\rm :\longmapsto\:\dfrac{5}{5}    = 1}

 \purple{\rm :\longmapsto\:1    = 1}

HENCE, VERIFIED

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