Question

If $x+y=a, x \times y=b$ and $x \cdot a=1$, then

• A. $x=\frac{a+a \times b}{a^{2}}$
• B. $y=\frac{\left(a^{2}-1\right) a-a \times b}{a^{2}}$
• C. $x=\frac{b+a \times b}{a^{2}}$
• D. $y=\frac{\left(b^{2}-1\right) b-a \times b}{a^{2}}$

Answer 1

Answer: (A), (B)

Given
$\Rightarrow y =a-x \,\,\,$(1)
$x \times y =b\,\,\,$(2)
$x \cdot a =1\,\,\,$(3)
$x+y=a$
From (1) and (2), we get
$ x \times(a-x)=b $
$\Rightarrow x \times a-x \times x=b$
$ \Rightarrow x \times a=b$
$\Rightarrow a \times(x \times a)=a \times b $
$\Rightarrow (a \cdot a) x-(a \cdot x) a=a \times b$
$\Rightarrow |a|^{2} x-1 \cdot a=a \times b \,\,\,\,$ [From (3)]
$\Rightarrow x=\frac{(a+a \times b)}{a^{2}}$
and $y=a-x=\frac{\left(a^{2}-1\right) a-a \times b}{a^{2}}$