Here, it can be observed that for \(\triangle\) \(XYZ\), \(YL\) is an altitude drawn exterior to side \(XZ\) which is extended up to point \(L\).
Give first the step you will use to separate the variable and then solve the equation:
(a) x – 1 = 0
(b) x + 1 = 0
(c) x – 1 = 5
(d) x + 6 = 2
(e) y – 4 = – 7
(f) y – 4 = 4
(g) y + 4 = 4
(h) y + 4 = – 4