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Re: 6 < su < 8 -8 < ut < -6 [#permalink]
Expert Reply
With inequalities is always better to work conceptually

Flip the sign of the second one

\(8>-ut>6\)

\(6<-ut<8\)

Now the two are equals but u

\(6<su<8\)

\(6<-ut<8\)

u cancel out

\(6<st<8\)

s and t are in between 6 and 8 but you DO NOT know for sure if s on the number line comes first or t could be either case

st or ts

D is the answer

I hope this helps
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Re: 6 < su < 8 -8 < ut < -6 [#permalink]
Carcass wrote:
With inequalities is always better to work conceptually

Flip the sign of the second one

\(8>-ut>6\)

\(6<-ut<8\)

Now the two are equals but u

\(6<su<8\)

\(6<-ut<8\)

u cancel out

\(6<st<8\)

s and t are in between 6 and 8 but you DO NOT know for sure if s on the number line comes first or t could be either case

st or ts

D is the answer

I hope this helps


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Re: 6 < su < 8 -8 < ut < -6 [#permalink]
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