Question

how do you get the last equation from the upper 3 ?​

Answer 1

★ Given :-

• $x_1 + x_2 = 2$. . . eq (i)
• $x_2 + x_3 = 0$. . . eq (ii)
• $x_3 + x_1 = 4$. . . eq (iii)

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To Derive :-

• $x_1 + x_2 + x_3 = 3$

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Solution :-

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$\longrightarrow \: \sf \: x_1 + x_2 = 2 \: \: \: \: \: \: \: \:\: \: \: \: \\ \longrightarrow \: \sf \:x_2 = 2 - x_1. . . eq (iv)$

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Now, we will put the value of $x_2$ from equation iv in equation (ii), we get,

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$\sf \longrightarrow \: 2 - x_1 + x_3 = 0 \: \: \: \: \: \: \\ \sf \longrightarrow \: x_3 = 2 + x_1 . . . eq (v)$

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Now, we will put the value of $x_3$ in eq. (iii) from eq. (v), so,

$\sf \longrightarrow \: 2 +x_1 + x_1 = 4 \\\sf \longrightarrow \: 2x_1 = 4 + 2 \: \: \: \: \: \: \: \\ \sf \longrightarrow \:x_1 = \frac{6}{2} = 3 \: \: \: \: \: \: \:$

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Now, we will put the value of $x_1$ in eq (i) to obtain the value of $x_2$,so,

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$\sf \longrightarrow \: 3 + x_2 = 2 \\ \sf \longrightarrow \:x_2 = 2 - 3 \\ \sf \longrightarrow \:x_2 = - 1 \: \: \: \:$

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Now, we will put the value of $x_2$ in eq. (ii) to obtain the value of $x_3$, so,

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$\sf \longrightarrow \: - 1 + x_3= 0 \\ \sf \longrightarrow \:x_3 = 1 \: \: \: \: \: \: \: \: \: \: \:$

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Therefore,

• $x_1$ = 3
• $x_2$ = - 1
• $x_3$ = 1

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Now,on adding these we get,

$\sf \longrightarrow \: x_1 + x_2 + x_3 \: \: \\ \sf \longrightarrow \:3 + ( - 1) + 1 \\ \sf \longrightarrow \:3 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

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Hence, we got the required equation!

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$\color{yellow} \bigstar \: \color{lavenderblush}{ \underline{\underline{ \boxed {\sf { \therefore\: x_1 + x_2 + x_3 = 3}}}}}$

Answer 2

$</p><p>x_1 + x_2 = 2x1+x2=2 . . . eq (i)</p><p></p><p>x_2 + x_3 = 0x2+x3=0 . . . eq (ii)</p><p></p><p>x_3 + x_1 = 4x3+x1=4 . . . eq (iii)</p><p></p><p>⠀</p><p>$