How do you solve #|x-5|+|2-2x|=7 #?

Answer 1

The roots are #x=0, x=4#.

We observe that modulii in the given equation (eqn.) vanish at #x=1 and x=5.#
So, very naturally, we consider these 3 cases : (i) #x<1#, (ii) #1<#x#<5# (iii) #x>5#
Case (i) #x<1#. #:. #(x-1)#<0#, #(x-5)#<0#. #:. #|x-1|=1-x, |x-5|=5-x# #:.# given eqn. becomes, #5-x+2(1-x)=7, or, 7-3*x=7, so, x=0.# We verify that this also satisfies the given eqn.
Case (ii) #1<#x#<5#. In this case, #(x-1)#>#0# and #(x-5)#<#0#, so that #|x-1|=x-1#, #|x-5|#=#5-x#. #:.# Eqn. becomes #5-x+2*x-2=7#. #:.# #x=4#. We verify that #1#<#4#<#5#, and this satisfy the eqn.
Case (iii) #x>5# Here, #(x-5)#>#0#, #(x-1)#>4>0, so, #|x-5|=x-5, |x-1|=x-1# #:.# eqn. becomes, #x-5+2*x-2=7#, giving #x=14/3#which is non-permissible as #x>5#. Hence no root in this case!
Altogether, the roots are #x=0, x=4#.
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Answer 2

To solve the equation ( |x-5| + |2-2x| = 7 ):

  1. Split the equation into cases based on the possible values inside the absolute value expressions.
  2. Solve each case separately.
  3. Check if the solutions obtained are valid by substituting them back into the original equation.

The possible cases are:

  1. ( x - 5 \geq 0 ) and ( 2 - 2x \geq 0 )
  2. ( x - 5 \geq 0 ) and ( 2 - 2x < 0 )
  3. ( x - 5 < 0 ) and ( 2 - 2x \geq 0 )
  4. ( x - 5 < 0 ) and ( 2 - 2x < 0 )

Solve each case separately and then check the solutions obtained to ensure they satisfy the original equation.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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