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When simplified with Boolean Algebra, $(x + y)(x + z)$ simplifies to:

  1. $x$
  2. $x + x(y + z)$
  3. $x(1 + yz)$
  4. $x + yz$
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Given that $(x+y)(x+z)$

$\implies x\cdot x + x\cdot z + y\cdot x + y\cdot z$

$\implies x + xz + xy + yz$

$\implies x(1+z+y) + yz$

$\implies x\cdot 1 + yz$     $[\because1+X = 1]$

$\implies x +yz$
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