# 4.  Find (a)  $\inline (-7)-8-(-25)$ (b)  $(-13)+32-8-1$ (c)  $\inline (-7)+(-8)+(-90)$ (d)  $\inline 50-(-40)-(-2)$

M manish

We already know that the sum of the additive inverse is Zero ($a+ (-a) = 0$

(a)  $\inline (-7)-8-(-25)$
$\\= -7 -8 +25\\ =-(7+8)+(15)+10\\ =-15+(-15)+10\\ =0+10 =10$

(b)  $(-13)+32-8-1 = (-13)+ 32 -(8+1)$
$\\= (-13)+ 32 -(9)\\ =-(13+9)+32\\ =(-22)+22+10\\ =0+10=10$

(c)  $\inline (-7)+(-8)+(-90)$
$\\= (-7)-(8)-(90)\\ =-(7+8+90)\\ =-(105)$

(d)  $\inline 50-(-40)-(-2)$
$= 50 +40 +2$
$=92$

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