If-x-x-y-5-x-y-y-10-then-x-y-

Question Number 129979 by liberty last updated on 21/Jan/21

If {\begin{cases} ∣ x ∣ + x + y = 5 \\ x + ∣ y ∣ - y = 10 \end{cases} then x + y ?

Answered by MJS_new last updated on 21/Jan/21

∣ x ∣ + x + y = 5

x ⩽ 0 \Rightarrow y = 5

but then the 2^{nd} eq . gives x = 10 > 0

\Rightarrow

x > 0

x + ∣ y ∣ - y = 10

y ⩾ 0 \Rightarrow x = 10

but then the 1^{st} eq . gives y = - 15 < 0

\Rightarrow x > 0 \land y < 0

now {it}^{'} s easy to solve

Answered by EDWIN88 last updated on 21/Jan/21

If y ⩾ 0 then the second eq tells us that

x = 10, but substituting this in the first eq

we find that y = - 15 contracting that y ⩾ 0

we conclude that y < 0 . Then it cannot

be that x ⩽ 0 because that leads to y = 5

we conclude that x > 0

Then we find the following eq {\begin{cases} 2 x + y = 5 \\ x - 2 y = 10 \end{cases}

we get (x, y) = (4, - 3) so x + y = 1

Answered by mathmax by abdo last updated on 21/Jan/21

\Rightarrow {\begin{cases} A (x, y) = 5 with A (x, y) =∣ x ∣ + x + y and B (x, y) = x + ∣ y ∣ - y \\ B (x, y) = 10 \end{cases}

case 1 x ⩾ 0 and y ⩾ 0 s \Rightarrow {\begin{cases} 2 x + y = 5 \Rightarrow {\begin{cases} x = 10 \\ y = - 15 not solution \end{cases} \\ x = 10 \end{cases}

case 2 x ⩾ 0 y ⩽ 0 s \Rightarrow {\begin{cases} 2 x + y = 5 \Rightarrow {\begin{cases} 2 x + y = 5 \\ 2 x - 4 y = 20 \Rightarrow {\begin{cases} 5 y = - 15 \\ x = 2 y + 10 \end{cases} \end{cases} \\ x - 2 y = 10 \end{cases}

\Rightarrow {\begin{cases} y = - 3 \\ x = 4 \end{cases}

case 3 x ⩽ 0 and y ⩾ 0 s \Rightarrow {\begin{cases} y = 5 \\ x = 10 not solution! \end{cases}

case 4 x ⩽ 0 and y ⩽ 0 s \Rightarrow {\begin{cases} y = 5 \\ x - 2 y = 10 \Rightarrow {\begin{cases} y = 5 \\ x = 20 not solution \end{cases} \end{cases}