Math, asked by richashah299, 7 hours ago

solve this and tell me this that what solution is this in linear equations class 10​

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Answered by BrainlyTwinklingstar
1

Answer

\sf \dashrightarrow \dfrac{x + 1}{2} + \dfrac{y - 1}{3} = 8 \: \: --- (i)

\sf \dashrightarrow \dfrac{x - 1}{3} + \dfrac{y + 1}{2} = 9 \: \: --- (ii)

By first equation,

\sf \dashrightarrow \dfrac{x + 1}{2} + \dfrac{y - 1}{3} = 8

\sf \dashrightarrow \dfrac{3(x + 1) + 2(y - 1)}{6} = 8

\sf \dashrightarrow \dfrac{3x + 3 + 2y - 2}{6} = 8

\sf \dashrightarrow 3x + 3 + 2y - 2 = 8 \times 6

\sf \dashrightarrow 3x + 3 + 2y - 2 = 48

\sf \dashrightarrow 3x + 2y = 48 - 3 + 2

\sf \dashrightarrow 3x + 2y = 47 \: \: --- (iii)

By second equation,

\sf \dashrightarrow \dfrac{x - 1}{3} + \dfrac{y + 1}{2} = 9

\sf \dashrightarrow \dfrac{2(x - 1) + 3(y + 1)}{6} = 9

\sf \dashrightarrow \dfrac{2x - 2 + 3y + 3}{6} = 9

\sf \dashrightarrow 2x - 2 + 3y + 3 = 9 \times 6

\sf \dashrightarrow 2x - 2 + 3y + 3 = 54

\sf \dashrightarrow 2x + 3y = 54 + 2 - 3

\sf \dashrightarrow 2x + 3y = 53 \: \: --- (iv)

Now, by third equation,

\sf \dashrightarrow 3x + 2y = 47

\sf \dashrightarrow 3x = 47 - 2y

\sf \dashrightarrow x = \dfrac{57 - 2y}{3}

Now, we should find the value of y by fourth equation.

\sf \dashrightarrow 2x + 3y = 53

\sf \dashrightarrow 2 \bigg( \dfrac{57 - 2y}{3} \bigg) + 3y = 53

\sf \dashrightarrow \dfrac{114 - 4y}{3} + 3y = 53

\sf \dashrightarrow \dfrac{114 - 4y + 9y}{3} = 53

\sf \dashrightarrow \dfrac{114 + 5y}{3} = 53

\sf \dashrightarrow 114 + 5y = 53 \times 3

\sf \dashrightarrow 114 + 5y = 159

\sf \dashrightarrow 5y = 159 - 114

\sf \dashrightarrow 5y = 45

\sf \dashrightarrow y = \dfrac{45}{5}

\sf \dashrightarrow y = 9

Now, we can find the value of x by third equation.

\sf \dashrightarrow 3x + 2y = 47

\sf \dashrightarrow 3x + 2(9) = 47

\sf \dashrightarrow 3x + 18 = 47

\sf \dashrightarrow 3x = 47 - 18

\sf \dashrightarrow 3x = 29

\sf \dashrightarrow x = \dfrac{29}{3}

Hence, the values of x and y are \sf \dfrac{29}{3} and 9 respectively.

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