Question

If a+b =4, a^2+ b^2=16, find a-b

Answer 1

The question is a bit wrong a^2 - b^2 = 16


Therefore.

a^2 - b^2 = (a+b)(a-b)

16 = 4 (a-b)
16/4 = a-b
4 = a-b

Hence the value of a- b is 4

Answer 2

$\binom{a + b = 4}{ {a}^{2} + {b}^{2} = 16} \\ \\ solve \: the \: equation \: for \: a \\ \\ \binom{a = 4 - b}{ {a}^{2} + {b}^{2} = 16} \\ \\ substitute \: a = 4 - b \: in \: {a}^{2} + {b}^{2} = 16 \\ \\ (4 - b {)}^{2} + {b}^{2} = 16$

$solve \: for \: b \\ \\ using \: (x - y) {}^{2} = {x}^{2} - 2xy+ {y}^{2} \: to \: expand \: the \: expression \\ \\ 16 - 8b + {b}^{2} + {b}^{2} = 16 \\ \\ - 8b + {b}^{2} + {b}^{2} = 16 - 16 \\ \\ - 8 {b} + {b}^{2} + {b}^{2} = 0 \\ \\ - 8b + {2b}^{2} = 0 \\ \\ factor \: out \: - 2b \: from \: the \: expression \\ \\ - 2b(4 - b) = 0$
$\\ \\ divide \: both \: sides \: by\: - 2 \\ \\ b(4 - b) = 0 \\ \\ b = \frac{0}{4 - b} \\ \\ b = 0 \\ and \: if \: we \: take \: 4 - b \: then \\ 4 - b = \frac{0}{b} \\ \\ 4 - b = 0 \\ \\ - b = - 4 \\ \\ b = 4 \\ \\ so \: two \: possible \: values \: of \: b \: are \\ \binom{b = 0}{b = 4}$

$substitute \: the \: value \: of \: b \: in \: a = 4 - b \\ \\ \binom{a = 4 - 0}{a = 4 - 4} \\ \\ \binom{a = 4}{a = 0} \\ \\ the \: possible \: solutions \: are \\ \\ (a1 ,\: b1) = (4, \: 0) \\(a2 ,\: b2 )= (0, \: 4)$

When a = 4 and b = 0

then, a-b = 4-0 = 4

When a = 0 and b = 4

then, a-b = 0-4 = -4