Exercise 34

What value of \( y \) satisfies the following system of equations?

\[ \begin{cases} x^2 - y^2 = 4 \\[.5em] x + y = 4 \end{cases} \]

  1. \( \cfrac{5}{2} \)

  2. \( \cfrac{-5}{2} \)

  3. \( \cfrac{3}{2} \)

  4. \( \cfrac{-3}{2} \)

  5. \( 4 \)

Try to solve it before checking the answer.

  1. \( \cfrac{3}{2} \)

From the equation \( x + y = 4 \) we obtain \( x = 4 – y \).

Replacing this value of \( x \) in the first equation, we have:

\[ \begin{aligned} &(4 – y)^2 – y^2 = 4 \\[.5em] &\hspace{5em}\Rightarrow 16 – 8y + y^2 – y^2 = 4 \\[.5em] &\hspace{5em}\Rightarrow -8y = -12 \\[.5em] &\hspace{5em}\Rightarrow y = \frac{-12}{-8} \\[.5em] &\hspace{5em}\Rightarrow \boldsymbol{ y = \frac{3}{2} } \end{aligned} \]
\[ \begin{aligned} (4 – y)^2 – y^2 = 4 &\Rightarrow 16 – 8y + y^2 – y^2 = 4 \\[.5em] &\Rightarrow -8y = -12 \\[.5em] &\Rightarrow y = \frac{-12}{-8} \\[.5em] &\Rightarrow \boldsymbol{ y = \frac{3}{2} } \end{aligned} \]

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