Question

If R be a relation in a set A and (x,y)= R=(y,x)e R VX,Ye A then the relation R
is called​

Answer 1

Answer:

$\huge\sf\purple{Answer:-}$

$\tt-19 + [27 - {14 + (5 – 2) X 4 ÷2}] \\ \\ \: \: \: \: \implies \tt-19 + [27 - {14 + 3 X 4 ÷2}] \\ \\ \: \: \: \: \implies \tt-19 + [27 - {14 + 3 X 2}] \\ \\ \: \: \: \: \tt \implies-19 + [27 - {14 + 6}] \\ \\ \: \: \: \: \implies \tt-19 + [27 - {14 + 6}] \\ \\ \: \: \: \: \implies \tt - 19 + 13 \\ \\ \: \: \: \: \implies \fbox{ \tt( - 6)}$

$\huge\mathfrak \red{hope \: its \: helpful}$

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Answer 2

Answer:

$\sf \: x+2y=4; 2x+y=5$

${\huge{\mathfrak{\underbrace{\purple{Given:-}}}}}$

$\star \: \sf \: x+2y=4........ \fbox{ equation \: 1}\\\star \: \sf 2x+y=5.........\fbox{ equation \:2}$

${\huge{\mathfrak{\underbrace{\purple{Solution:-}}}}}$

$\sf \: 1st \: we \: have \: to \: multiply \: equation \: 2 \\ \tt by \: 2$

$\tt \therefore \: 2(2x + y = 5) \\ \implies \tt \: 4x + 2y = 10...... \fbox{equation \: 3} \\ \\ \tt \: now \: subtract \: equation \: 1 \: and \: 3.$

$\: \: \: \: \: \: \: \: \: \: \tt4x + 2y = 10 \\ \: \: \: \: \: \: \: - \: \: \: \tt \: x - 2y = ( - 4) \\ \: \: \: \: \: \: \: \: \: \: - - - - - - - - - - - - \\ \: \: \: \: \: \: \: \: \: \: \tt2x = 6 \\ \: \: \: \: \: \: \: \longrightarrow \tt \: x = \frac{6}{2} \\ \: \: \: \: \: \: \: \longrightarrow \fbox{ \tt \: x = 3}</p><p>$

$\tt \: put \: the \: value \: of \: x \: in \: equation \\ \tt1st$

$\tt \: x + 2y = 4 \\ \tt \: \: \implies3 + 2y = 4 \\ \: \: \tt \implies2y = 4 - 3 \\ \: \: \tt \implies \: y = \frac{1}{2}$

$\huge\mathfrak \red{hope \: its \: helpful}$

$\huge\mathfrak \green{mark \: me \: as \: brainliest}$