The lines x + y = |a| and ax - y = 1 intersec | vedclass

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(C) Given the equations $x + y = |a|$ and $ax - y = 1$.Adding the two equations,we get $(1 + a)x = |a| + 1$,which implies $x = \frac{|a| + 1}{a + 1}$.

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$(1, \infty)$

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(C) Given the equations $x + y = |a|$ and $ax - y = 1$.Adding the two equations,we get $(1 + a)x = |a| + 1$,which implies $x = \frac{|a| + 1}{a + 1}$.Substituting $x$ into the first equation,$y = |a| - x = |a| - \frac{|a| + 1}{a + 1} = \frac{|a|(a + 1) - (|a| + 1)}{a + 1} = \frac{a|a| + |a| - |a| - 1}{a + 1} = \frac{a|a| - 1}{a + 1}$.For the intersection point to be in the first quadrant,both $x > 0$ and $y > 0$ must hold.Since $x = \frac{|a| + 1}{a + 1} > 0$ and $|a| + 1$ is always positive,we must have $a + 1 > 0$,which means $a > -1$.For $y = \frac{a|a| - 1}{a + 1} > 0$,since $a + 1 > 0$,we must have $a|a| - 1 > 0$,which means $a|a| > 1$.If $a \in (-1, 0]$,then $a|a| = -a^2$,and $-a^2 > 1$ is impossible.If $a > 0$,then $a^2 > 1$,which implies $a > 1$.Thus,the possible values of $a$ are in the interval $(1, \infty)$.

The lines $x + y = |a|$ and $ax - y = 1$ intersect each other in the first quadrant. Then,in which interval do all possible values of $a$ lie?

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