\(–35\times 107\) is not same as
\(–35\times(100+7)\)
\((–35)\times 7 + (–35)\times 100\)
\(–35\times 7 + 100\)
\((–30–5)\times 107\)
Why (c) is NOT the same as ((-35) imes 107)
Key idea: Use the distributive property:
(a imes(b+c)=a imes b+a imes c)
Check (a): ((-35) imes(100+7))
First, note that:
(100+7=107)
So,
((-35) imes(100+7)=(-35) imes107)
Also, expanding by distributive law:
((-35) imes(100+7)=(-35) imes100+(-35) imes7)
So (a) matches the original.
Check (b): (((-35) imes 7)+((-35) imes 100))
This is exactly the expanded form from (a), just the two parts are written in swapped order. Addition order doesn’t change the sum:
((-35) imes100+(-35) imes7 = (-35) imes7+(-35) imes100)
So (b) also matches the original.
Check (c): ((-35) imes 7 + 100)
Carefully compare with the correct expansion:
( ext{Correct: }(-35) imes100+(-35) imes7)
But option (c) has:
((-35) imes7+100)
Notice the problem: the 100 is not multiplied by ((-35)). It should be ((-35) imes100), not just (+,100).
To be extra sure, check the actual values:
((-35) imes107=-3745)
((-35) imes7+100=-245+100=-145)
(-3745 eq -145). So (c) is not the same.
Check (d): (( -30 - 5) imes 107)
Here, (-35) is rewritten as (-30-5):
(-35=-30-5)
Apply distributive property:
(( -30 - 5) imes 107=(-30) imes107+(-5) imes107)
Combine:
((-30-5) imes107=(-35) imes107)
So (d) also matches the original.
Conclusion: Only option (c) is not the same as ((-35) imes107).