Question Number 100597 by bobhans last updated on 27/Jun/20

Commented byDwaipayan Shikari last updated on 27/Jun/20

x+2∣y∣=3 First case x+2y=3 x−3y=5 ⇒5y=−2 ⇒y=−(2/5) x=((19)/5) x−y=((21)/5) { ((x=((19)/5))),((y=−(2/5))) :} Second case x−2y=3 and x−3y=5 ⇒y=−2 and x=−1 so x−y=1 { ((x=−1)),((y=−2)) :}

Answered by mathmax by abdo last updated on 27/Jun/20

{ ((x+2∣y∣=3)),((x−3y =5 ⇒2∣y∣+3y =−2 and x =5+3y)) :} case1 y≥0 ⇒5y =−2 ⇒y =−(2/5) ⇒x =5−(6/5) =−((19)/5) so x−y =−((19)/5) +(2/5) =−((17)/5) case 2 y≤0 ⇒y =−2 ⇒x =5−6 =−1 so x−y =−1+2 =1

Commented byRasheed.Sindhi last updated on 27/Jun/20

I think sir that: In case 1 if y≥0 then y=−(2/5) is discardable and for that x=−((19)/5) is also discardable.

Answered by john santu last updated on 27/Jun/20

∣y∣ = ((3−x)/2) ⇒ y = ± (((3−x)/2)) y = ((x−5)/3) ⇒± (((3−x)/2)) = ((x−5)/3) ± (9−3x) = 2x−10 { ((9−3x = 2x−10 ; x = ((19)/5))),((3x−9 = 2x−10 ; x = −1 )) :} { ((y=((((19)/5) −5)/3) = ((−2)/5))),((y=((−6)/2) = −2)) :}