Question

\( \begin{array}{l}\Rightarrow x+y=1 \text { and } x^{2}+y^{2}=1 \\ \Rightarrow 2 x y=(x+y)^{2}-\left(x^{2}+y^{2}\right)=0 \text { and } x+y= \\ \Rightarrow[x=1 \text { and } y=0] \text { or }[x=0 \text { and } y=1]\end{array} \)

          \( \begin{array}{l}\Rightarrow x+y=1 \text { and } x^{2}+y^{2}=1 \\ \Rightarrow 2 x y=(x+y)^{2}-\left(x^{2}+y^{2}\right)=0 \text { and } x+y= \\ \Rightarrow[x=1 \text { and } y=0] \text { or }[x=0 \text { and } y=1]\end{array} \)

⇒ x+y=1 and x^2+y^2=1
⇒ 2 x y=(x+y)^2-(x^2+y^2)=0 and x+y=
⇒[x=1 and y=0] or [x=0 and y=1]

Added by Mercedes V.

Beginning and Intermediate Algebra

Margaret L. Lial, John Hornsby, Terry… 7th Edition

Chapter 3

Instant Answer

Solved by Expert Ben Ault

Step 1

Step 1: Start with the given equations: \[ x + y = 1 \] \[ x^2 + y^2 = 1 \] Show more…

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\( \begin{array}{l}\Rightarrow x+y=1 \text { and } x^{2}+y^{2}=1 \\ \Rightarrow 2 x y=(x+y)^{2}-\left(x^{2}+y^{2}\right)=0 \text { and } x+y= \\ \Rightarrow[x=1 \text { and } y=0] \text { or }[x=0 \text { and } y=1]\end{array} \)