Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

If  { ((∣x∣ + x + y = 5)),((x + ∣y∣ −y = 10)) :} then x+y ?

$$\mathrm{If}\:\begin{cases}{\mid\mathrm{x}\mid\:+\:\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{5}}\\{\mathrm{x}\:+\:\mid\mathrm{y}\mid\:−\mathrm{y}\:=\:\mathrm{10}}\end{cases}\:\mathrm{then}\:\mathrm{x}+\mathrm{y}\:? \\ $$

Answered by MJS_new last updated on 21/Jan/21

∣x∣+x+y=5  x≤0 ⇒ y=5  but then the 2^(nd)  eq. gives x=10>0  ⇒  x>0    x+∣y∣−y=10  y≥0 ⇒ x=10  but then the 1^(st)  eq. gives y=−15<0    ⇒ x>0∧y<0  now it′s easy to solve

$$\mid{x}\mid+{x}+{y}=\mathrm{5} \\ $$$${x}\leqslant\mathrm{0}\:\Rightarrow\:{y}=\mathrm{5} \\ $$$$\mathrm{but}\:\mathrm{then}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{eq}.\:\mathrm{gives}\:{x}=\mathrm{10}>\mathrm{0} \\ $$$$\Rightarrow \\ $$$${x}>\mathrm{0} \\ $$$$ \\ $$$${x}+\mid{y}\mid−{y}=\mathrm{10} \\ $$$${y}\geqslant\mathrm{0}\:\Rightarrow\:{x}=\mathrm{10} \\ $$$$\mathrm{but}\:\mathrm{then}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{eq}.\:\mathrm{gives}\:{y}=−\mathrm{15}<\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow\:{x}>\mathrm{0}\wedge{y}<\mathrm{0} \\ $$$$\mathrm{now}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve} \\ $$

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  { ((2x+y=5)),((x−2y=10)) :}  we get (x,y)=(4,−3) so x+y = 1

$$\mathrm{If}\:\mathrm{y}\:\geqslant\:\mathrm{0}\:\mathrm{then}\:\mathrm{the}\:\mathrm{second}\:\mathrm{eq}\:\mathrm{tells}\:\mathrm{us}\:\mathrm{that} \\ $$$$\:\mathrm{x}=\mathrm{10}\:,\:\mathrm{but}\:\mathrm{substituting}\:\mathrm{this}\:\mathrm{in}\:\mathrm{the}\:\mathrm{first}\:\mathrm{eq} \\ $$$$\:\mathrm{we}\:\mathrm{find}\:\mathrm{that}\:\mathrm{y}\:=\:−\mathrm{15}\:\mathrm{contracting}\:\mathrm{that}\:\mathrm{y}\geqslant\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{conclude}\:\mathrm{that}\:\mathrm{y}\:<\:\mathrm{0}.\:\mathrm{Then}\:\mathrm{it}\:\mathrm{cannot}\: \\ $$$$\mathrm{be}\:\mathrm{that}\:\mathrm{x}\leqslant\mathrm{0}\:\mathrm{because}\:\mathrm{that}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{y}=\mathrm{5} \\ $$$$\mathrm{we}\:\mathrm{conclude}\:\mathrm{that}\:\mathrm{x}\:>\mathrm{0}\: \\ $$$$\mathrm{Then}\:\mathrm{we}\:\mathrm{find}\:\mathrm{the}\:\mathrm{following}\:\mathrm{eq}\:\begin{cases}{\mathrm{2x}+\mathrm{y}=\mathrm{5}}\\{\mathrm{x}−\mathrm{2y}=\mathrm{10}}\end{cases} \\ $$$$\mathrm{we}\:\mathrm{get}\:\left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{4},−\mathrm{3}\right)\:\mathrm{so}\:\mathrm{x}+\mathrm{y}\:=\:\mathrm{1} \\ $$

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

⇒ { ((A(x,y)=5            with A(x,y)=∣x∣+x+y and B(x,y)=x+∣y∣−y)),((B(x,y) =10)) :}  case 1  x≥0 and y≥0  s⇒ { ((2x+y=5  ⇒ { ((x=10   )),((y=−15   not solution)) :})),((x=10)) :}  case2  x≥0   y≤0  s⇒ { ((2x+y=5  ⇒ { ((2x+y=5)),((2x−4y=20 ⇒  { ((5y=−15)),((x=2y+10)) :})) :})),((x−2y=10 )) :}  ⇒ { ((y=−3     )),((x=4)) :}  case3   x≤0 and y≥0  s⇒ { ((y=5    )),((x=10  not solution!)) :}  case4  x≤0 and y≤0  s⇒ { ((y=5)),((x−2y=10  ⇒ { ((y=5)),((x=20   not solution)) :})) :}

$$\Rightarrow\begin{cases}{\mathrm{A}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{with}\:\mathrm{A}\left(\mathrm{x},\mathrm{y}\right)=\mid\mathrm{x}\mid+\mathrm{x}+\mathrm{y}\:\mathrm{and}\:\mathrm{B}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}+\mid\mathrm{y}\mid−\mathrm{y}}\\{\mathrm{B}\left(\mathrm{x},\mathrm{y}\right)\:=\mathrm{10}}\end{cases} \\ $$$$\mathrm{case}\:\mathrm{1}\:\:\mathrm{x}\geqslant\mathrm{0}\:\mathrm{and}\:\mathrm{y}\geqslant\mathrm{0}\:\:\mathrm{s}\Rightarrow\begin{cases}{\mathrm{2x}+\mathrm{y}=\mathrm{5}\:\:\Rightarrow\begin{cases}{\mathrm{x}=\mathrm{10}\:\:\:}\\{\mathrm{y}=−\mathrm{15}\:\:\:\mathrm{not}\:\mathrm{solution}}\end{cases}}\\{\mathrm{x}=\mathrm{10}}\end{cases} \\ $$$$\mathrm{case2}\:\:\mathrm{x}\geqslant\mathrm{0}\:\:\:\mathrm{y}\leqslant\mathrm{0}\:\:\mathrm{s}\Rightarrow\begin{cases}{\mathrm{2x}+\mathrm{y}=\mathrm{5}\:\:\Rightarrow\begin{cases}{\mathrm{2x}+\mathrm{y}=\mathrm{5}}\\{\mathrm{2x}−\mathrm{4y}=\mathrm{20}\:\Rightarrow\:\begin{cases}{\mathrm{5y}=−\mathrm{15}}\\{\mathrm{x}=\mathrm{2y}+\mathrm{10}}\end{cases}}\end{cases}}\\{\mathrm{x}−\mathrm{2y}=\mathrm{10}\:}\end{cases} \\ $$$$\Rightarrow\begin{cases}{\mathrm{y}=−\mathrm{3}\:\:\:\:\:}\\{\mathrm{x}=\mathrm{4}}\end{cases} \\ $$$$\mathrm{case3}\:\:\:\mathrm{x}\leqslant\mathrm{0}\:\mathrm{and}\:\mathrm{y}\geqslant\mathrm{0}\:\:\mathrm{s}\Rightarrow\begin{cases}{\mathrm{y}=\mathrm{5}\:\:\:\:}\\{\mathrm{x}=\mathrm{10}\:\:\mathrm{not}\:\mathrm{solution}!}\end{cases} \\ $$$$\mathrm{case4}\:\:\mathrm{x}\leqslant\mathrm{0}\:\mathrm{and}\:\mathrm{y}\leqslant\mathrm{0}\:\:\mathrm{s}\Rightarrow\begin{cases}{\mathrm{y}=\mathrm{5}}\\{\mathrm{x}−\mathrm{2y}=\mathrm{10}\:\:\Rightarrow\begin{cases}{\mathrm{y}=\mathrm{5}}\\{\mathrm{x}=\mathrm{20}\:\:\:\mathrm{not}\:\mathrm{solution}}\end{cases}}\end{cases} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com