Tag: cartesian product of two sets

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Multiple choice mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

Determine all ordered pairs that satisfy $(x - y)^{2} + x^{2} = 25$, where $x$ and $y$ are integers and $x \geq 0$. Find the number of different values of $y$ that occur

  1. $3$
  2. $4$
  3. $5$
  4. $6$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${ \left( x-y \right)  }^{ 2 }+{ x }^{ 2 }=25$
Now, ${ 3 }^{ 2 }+{ 4 }^{ 2 }={ 5 }^{ 2 }$
$9+16=25$
$\therefore $There are 2 possibilities:
$I.{ \left( x-y \right)  }^{ 2 }=9$ and ${ x }^{ 2 }=16$
$\therefore x=\pm 4$ and $x-y=\pm 3$
$\left( i \right) .x-y=3\Rightarrow \left( 4,1 \right) $ and $\left( -4,-7 \right) $
$\left( ii \right) .x-y=-3\Rightarrow \left( 4,7 \right) $ and $\left( -4,-1 \right) $
$II.{ \left( x-y \right)  }^{ 2 }=16$ and ${ x }^{ 2 }=9$
$\therefore x=\pm 3$ and $x-y=\pm 4$
$\left( i \right) .x-y=4\Rightarrow \left( 3,-1 \right) $ and $\left( -3,-7 \right) $
$\left( ii \right) .x-y=-4\Rightarrow \left( 3,7 \right) $ and $\left( -3,-1 \right) $
$\therefore $ Different values of y are $1,-1,7,-7$
$\therefore 4$ different values of y occur.