How do you evaluate $| x - y | + y - 1$ $abs(x-y)+y-1$ when x=-3, y=-6?

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Answer 1 · 2016-11-23T05:17:03

$| x - y | + y - 1 = - 4$ $|x-y|+y-1=-4$ , when $x = - 3$ $x=-3$ and $y = - 6$ $y=-6$

Value of $| x - y | + y - 1$ $|x-y|+y-1$ , when $x = - 3$ $x=-3$ and $y = - 6$ $y=-6$ can be obtained by substituting these values.

$| x - y | + y - 1$ $|x-y|+y-1$

= $| (- 3) - (- 6) | + (- 6) - 1$ $|(-3)-(-6)|+(-6)-1$

= $| - 3 + 6 | + (- 6) - 1$ $|-3+6|+(-6)-1$

= $| 3 | - 6 - 1$ $|3|-6-1$

= $3 - 6 - 1 = 3 - 7 = - 4$ $3-6-1=3-7=-4$

How do you evaluate |x−y|+y−1abs(x-y)+y-1 when x=-3, y=-6?