Absolute inequalities with both on one side (1 Viewer)

pokemonlv10

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Need help solving this properly

|4x-3| + |x-1| < 11

I would take 4 cases where its positive positive, positive negative, negative positive, negative negative for each absolute value. However i think i've made a mistake in my working.

Also unsure if i am to test these values to see if they are solutions?

4x-3 + x-1 <11
x<3

-4x+3 +x-1 < 11
x>-3

4x-3-x+1 <11
x< 13/3

-4x+3 - x+1 < 11
x>-7/5
 

InteGrand

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Need help solving this properly

|4x-3| + |x-1| < 11

I would take 4 cases where its positive positive, positive negative, negative positive, negative negative for each absolute value. However i think i've made a mistake in my working.

Also unsure if i am to test these values to see if they are solutions?

4x-3 + x-1 <11
x<3

-4x+3 +x-1 < 11
x>-3

4x-3-x+1 <11
x< 13/3

-4x+3 - x+1 < 11
x>-7/5
 

kashkow

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I suck at absolute values :S.

But I think from recall of this in year 11 you don't need to split it up into four cases, but only two; ie. when they're both the same sign and when they're different signs. Someone correct me if I'm wrong?
 
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InteGrand

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I suck at absolute values :S.

But I think from recall of this in year 11 you don't need to split it up into four cases, but only two; ie. when they're both the same sign and when they're different signs. Someone correct me if I'm wrong?
Basically you should do three cases as I outlined above. I.e. Find the points where those functions inside the absolute values change sign (which happens when they cross the x-axis, i.e. when they're equal to 0), and consider the three intervals that these two points split the real line into.
 

kashkow

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Basically you should do three cases as I outlined above. I.e. Find the points where those functions inside the absolute values change sign (which happens when they cross the x-axis, i.e. when they're equal to 0), and consider the three intervals that these two points split the real line into.
Ok thanks :) That is helpful to know
 

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