Proof of inequality solution











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I have to solve this inequality:



$$5 ≤ 4|x − 1| + |2 − 3x|$$



and prove its solution with one (or 2 or 3) of this sentences:



$$∀x∀y |xy| = |x||y|$$



$$∀x∀y(y ≤ |x| ↔ y ≤ x ∨ y ≤ −x)$$



$$∀x∀y(|x| ≤ y ↔ x ≤ y ∧ −x ≤ y)$$



The solution of inequality is:



$$(-infty, frac{1}{7}> U <frac{11}{7}, infty)$$



But I have a hard time with proving the solution with the sentence. E.g. if I choose the second one I get this:



5 ≤ 4(x-1) + (2-3x) ∨ 5 ≤ - [4(x-1) + (2-3x)] <=>



5 ≤ 4x - 4 + 2 - 3x ∨ 5 ≤ -(4x-4+2-3x) <=>



5 ≤ x-2 ∨ 5 ≤ -x + 2 <=> 7 ≤ x ∨ x ≤ 3



and that is wrong.



Can someone help me out, please? Sorry for bad English, that is not my first language.



EDIT:
Sentences should be applicable. I have another inequality which is already solved and it was done like this:



|3x| ≤ |2x − 1|



(x ∈ R | −1 ≤ x ≤1/5)



Sentences:



∀x∀y(y ≤ |x| ↔ y ≤ x ∨ y ≤ −x) (1)



∀x∀y(|x| ≤ y ↔ x ≤ y ∧ −x ≤ y) (2)



Solution:
We make another sentence which we have to prove:



∀x( |3x| ≤ |2x − 1| ↔ −1 ≤ x ≤ 1/5) (3)



|3x| ≤ |2x − 1| ⇔ |3x| ≤ 2x − 1 ∨ |3x| ≤ −(2x − 1) ⇔



3x ≤ 2x − 1 ∧ −3x ≤ 2x − 1 ∨ 3x ≤ −(2x − 1) ∧ −3x ≤ −(2x − 1) ⇔



x ≤ −1 ∧ 1 ≤ 5x ∨ 5x ≤ 1 ∧ −1 ≤ x ⇔



1/5 ≤ x ≤ −1 ∨ −1 ≤ x ≤1/5⇔ −1 ≤ x ≤1/5



So we proved sentence (3)






proof-verification inequality







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  • 1




    Not sure if you can apply your sentences there, besides, why would one make you even try do that ... there is a standard way to solve this type of problems: divide your domain into regions, where the signs of each expression under the absolute operator is same for the entire region, and then simply get rid of the absolute operator.
    – Makina
    Nov 17 at 18:53










  • also, get some graph paper (quadrille paper) and simply graph $y = 4|x-1| +|2-3x|$ for, say $0 leq x leq 2.$ Enough to divide each unit on the $x$ axis into six parts, so that one square means $1/6.$ I drew the thing yesterday, not hard.
    – Will Jagy
    Nov 17 at 19:42















up vote
-1
down vote

favorite












I have to solve this inequality:



$$5 ≤ 4|x − 1| + |2 − 3x|$$



and prove its solution with one (or 2 or 3) of this sentences:



$$∀x∀y |xy| = |x||y|$$



$$∀x∀y(y ≤ |x| ↔ y ≤ x ∨ y ≤ −x)$$



$$∀x∀y(|x| ≤ y ↔ x ≤ y ∧ −x ≤ y)$$



The solution of inequality is:



$$(-infty, frac{1}{7}> U <frac{11}{7}, infty)$$



But I have a hard time with proving the solution with the sentence. E.g. if I choose the second one I get this:



5 ≤ 4(x-1) + (2-3x) ∨ 5 ≤ - [4(x-1) + (2-3x)] <=>



5 ≤ 4x - 4 + 2 - 3x ∨ 5 ≤ -(4x-4+2-3x) <=>



5 ≤ x-2 ∨ 5 ≤ -x + 2 <=> 7 ≤ x ∨ x ≤ 3



and that is wrong.



Can someone help me out, please? Sorry for bad English, that is not my first language.



EDIT:
Sentences should be applicable. I have another inequality which is already solved and it was done like this:



|3x| ≤ |2x − 1|



(x ∈ R | −1 ≤ x ≤1/5)



Sentences:



∀x∀y(y ≤ |x| ↔ y ≤ x ∨ y ≤ −x) (1)



∀x∀y(|x| ≤ y ↔ x ≤ y ∧ −x ≤ y) (2)



Solution:
We make another sentence which we have to prove:



∀x( |3x| ≤ |2x − 1| ↔ −1 ≤ x ≤ 1/5) (3)



|3x| ≤ |2x − 1| ⇔ |3x| ≤ 2x − 1 ∨ |3x| ≤ −(2x − 1) ⇔



3x ≤ 2x − 1 ∧ −3x ≤ 2x − 1 ∨ 3x ≤ −(2x − 1) ∧ −3x ≤ −(2x − 1) ⇔



x ≤ −1 ∧ 1 ≤ 5x ∨ 5x ≤ 1 ∧ −1 ≤ x ⇔



1/5 ≤ x ≤ −1 ∨ −1 ≤ x ≤1/5⇔ −1 ≤ x ≤1/5



So we proved sentence (3)






proof-verification inequality







share|cite|improve this question




















  • 1




    Not sure if you can apply your sentences there, besides, why would one make you even try do that ... there is a standard way to solve this type of problems: divide your domain into regions, where the signs of each expression under the absolute operator is same for the entire region, and then simply get rid of the absolute operator.
    – Makina
    Nov 17 at 18:53










  • also, get some graph paper (quadrille paper) and simply graph $y = 4|x-1| +|2-3x|$ for, say $0 leq x leq 2.$ Enough to divide each unit on the $x$ axis into six parts, so that one square means $1/6.$ I drew the thing yesterday, not hard.
    – Will Jagy
    Nov 17 at 19:42













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











I have to solve this inequality:



$$5 ≤ 4|x − 1| + |2 − 3x|$$



and prove its solution with one (or 2 or 3) of this sentences:



$$∀x∀y |xy| = |x||y|$$



$$∀x∀y(y ≤ |x| ↔ y ≤ x ∨ y ≤ −x)$$



$$∀x∀y(|x| ≤ y ↔ x ≤ y ∧ −x ≤ y)$$



The solution of inequality is:



$$(-infty, frac{1}{7}> U <frac{11}{7}, infty)$$



But I have a hard time with proving the solution with the sentence. E.g. if I choose the second one I get this:



5 ≤ 4(x-1) + (2-3x) ∨ 5 ≤ - [4(x-1) + (2-3x)] <=>



5 ≤ 4x - 4 + 2 - 3x ∨ 5 ≤ -(4x-4+2-3x) <=>



5 ≤ x-2 ∨ 5 ≤ -x + 2 <=> 7 ≤ x ∨ x ≤ 3



and that is wrong.



Can someone help me out, please? Sorry for bad English, that is not my first language.



EDIT:
Sentences should be applicable. I have another inequality which is already solved and it was done like this:



|3x| ≤ |2x − 1|



(x ∈ R | −1 ≤ x ≤1/5)



Sentences:



∀x∀y(y ≤ |x| ↔ y ≤ x ∨ y ≤ −x) (1)



∀x∀y(|x| ≤ y ↔ x ≤ y ∧ −x ≤ y) (2)



Solution:
We make another sentence which we have to prove:



∀x( |3x| ≤ |2x − 1| ↔ −1 ≤ x ≤ 1/5) (3)



|3x| ≤ |2x − 1| ⇔ |3x| ≤ 2x − 1 ∨ |3x| ≤ −(2x − 1) ⇔



3x ≤ 2x − 1 ∧ −3x ≤ 2x − 1 ∨ 3x ≤ −(2x − 1) ∧ −3x ≤ −(2x − 1) ⇔



x ≤ −1 ∧ 1 ≤ 5x ∨ 5x ≤ 1 ∧ −1 ≤ x ⇔



1/5 ≤ x ≤ −1 ∨ −1 ≤ x ≤1/5⇔ −1 ≤ x ≤1/5



So we proved sentence (3)






proof-verification inequality







share|cite|improve this question















I have to solve this inequality:



$$5 ≤ 4|x − 1| + |2 − 3x|$$



and prove its solution with one (or 2 or 3) of this sentences:



$$∀x∀y |xy| = |x||y|$$



$$∀x∀y(y ≤ |x| ↔ y ≤ x ∨ y ≤ −x)$$



$$∀x∀y(|x| ≤ y ↔ x ≤ y ∧ −x ≤ y)$$



The solution of inequality is:



$$(-infty, frac{1}{7}> U <frac{11}{7}, infty)$$



But I have a hard time with proving the solution with the sentence. E.g. if I choose the second one I get this:



5 ≤ 4(x-1) + (2-3x) ∨ 5 ≤ - [4(x-1) + (2-3x)] <=>



5 ≤ 4x - 4 + 2 - 3x ∨ 5 ≤ -(4x-4+2-3x) <=>



5 ≤ x-2 ∨ 5 ≤ -x + 2 <=> 7 ≤ x ∨ x ≤ 3



and that is wrong.



Can someone help me out, please? Sorry for bad English, that is not my first language.



EDIT:
Sentences should be applicable. I have another inequality which is already solved and it was done like this:



|3x| ≤ |2x − 1|



(x ∈ R | −1 ≤ x ≤1/5)



Sentences:



∀x∀y(y ≤ |x| ↔ y ≤ x ∨ y ≤ −x) (1)



∀x∀y(|x| ≤ y ↔ x ≤ y ∧ −x ≤ y) (2)



Solution:
We make another sentence which we have to prove:



∀x( |3x| ≤ |2x − 1| ↔ −1 ≤ x ≤ 1/5) (3)



|3x| ≤ |2x − 1| ⇔ |3x| ≤ 2x − 1 ∨ |3x| ≤ −(2x − 1) ⇔



3x ≤ 2x − 1 ∧ −3x ≤ 2x − 1 ∨ 3x ≤ −(2x − 1) ∧ −3x ≤ −(2x − 1) ⇔



x ≤ −1 ∧ 1 ≤ 5x ∨ 5x ≤ 1 ∧ −1 ≤ x ⇔



1/5 ≤ x ≤ −1 ∨ −1 ≤ x ≤1/5⇔ −1 ≤ x ≤1/5



So we proved sentence (3)






proof-verification inequality




proof-verification inequality






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 18 at 11:32

























asked Nov 17 at 18:36









dfps12

11




11








  • 1




    Not sure if you can apply your sentences there, besides, why would one make you even try do that ... there is a standard way to solve this type of problems: divide your domain into regions, where the signs of each expression under the absolute operator is same for the entire region, and then simply get rid of the absolute operator.
    – Makina
    Nov 17 at 18:53










  • also, get some graph paper (quadrille paper) and simply graph $y = 4|x-1| +|2-3x|$ for, say $0 leq x leq 2.$ Enough to divide each unit on the $x$ axis into six parts, so that one square means $1/6.$ I drew the thing yesterday, not hard.
    – Will Jagy
    Nov 17 at 19:42














  • 1




    Not sure if you can apply your sentences there, besides, why would one make you even try do that ... there is a standard way to solve this type of problems: divide your domain into regions, where the signs of each expression under the absolute operator is same for the entire region, and then simply get rid of the absolute operator.
    – Makina
    Nov 17 at 18:53










  • also, get some graph paper (quadrille paper) and simply graph $y = 4|x-1| +|2-3x|$ for, say $0 leq x leq 2.$ Enough to divide each unit on the $x$ axis into six parts, so that one square means $1/6.$ I drew the thing yesterday, not hard.
    – Will Jagy
    Nov 17 at 19:42








1




1




Not sure if you can apply your sentences there, besides, why would one make you even try do that ... there is a standard way to solve this type of problems: divide your domain into regions, where the signs of each expression under the absolute operator is same for the entire region, and then simply get rid of the absolute operator.
– Makina
Nov 17 at 18:53




Not sure if you can apply your sentences there, besides, why would one make you even try do that ... there is a standard way to solve this type of problems: divide your domain into regions, where the signs of each expression under the absolute operator is same for the entire region, and then simply get rid of the absolute operator.
– Makina
Nov 17 at 18:53












also, get some graph paper (quadrille paper) and simply graph $y = 4|x-1| +|2-3x|$ for, say $0 leq x leq 2.$ Enough to divide each unit on the $x$ axis into six parts, so that one square means $1/6.$ I drew the thing yesterday, not hard.
– Will Jagy
Nov 17 at 19:42




also, get some graph paper (quadrille paper) and simply graph $y = 4|x-1| +|2-3x|$ for, say $0 leq x leq 2.$ Enough to divide each unit on the $x$ axis into six parts, so that one square means $1/6.$ I drew the thing yesterday, not hard.
– Will Jagy
Nov 17 at 19:42










2 Answers
2






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up vote
1
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The sentences (rules) you stated will not be sufficient (applicable), because the given inequality is not a single absolute value, but a sum of two.



The standard algebraic method to solve the absolute value inequality is to divide into intervals:
$$5 ≤ 4|x − 1| + |2 − 3x| Rightarrow \
begin{align}
1) &begin{cases}xle frac23\ 5le -4(x-1)+(2-3x) end{cases} Rightarrow begin{cases}xle frac23\ xle frac1{7}end{cases} Rightarrow xin (-infty,frac17] text{or} \
2) &begin{cases}frac23< x<1 \ 5le -4(x-1)-(2-3x) end{cases} Rightarrow begin{cases}frac23< x<1\ xle -3end{cases} Rightarrow xin emptyset text{or} \
3) &begin{cases}1le x \ 5le 4(x-1)-(2-3x) end{cases} Rightarrow begin{cases}1le x\ xge frac{11}7end{cases} Rightarrow xin [frac{11}7,+infty) end{align}$$
Hence, the final solution is: $$xin (-infty,frac17]cup [frac{11}7,+infty).$$



==============================================================



Let's try:
$$5 ≤ 4|x − 1| + |2 − 3x| iff \
4|x-1|ge 5-|2-3x| iff \
big[4(x-1)ge 5-|2-3x| big] lor big[-4(x-1)ge 5-|2-3x|big] iff \
big[|2-3x|ge 9-4x big] lor big[|2-3x|ge 1+4xbig] iff \
big[2-3xge 9-4x lor -(2-3x)ge 9-4xbig] lor big[2-3xge 1+4x lor -(2-3x)ge 1+4xbig] iff \
big[xge 7 lor xge frac{11}7big] lor big[xle frac17 lor xle -3big] iff \
big[xge frac{11}7big] lor big[xle frac17big] iff \
xin (-infty,frac17]cup [frac{11}7,+infty)$$






share|cite|improve this answer






























    up vote
    0
    down vote













    Part (I) fill in this table
    $$
    begin{array}{c|c|c|c|c|c|c|c}
    x & x-1&|x-1|&4|x-1| &-3x&2-3x&|2-3x|& y = 4|x-1| +|2-3x| \ hline
    frac{-1}{6} & &&&&&& \ hline
    0 & &&&&&& \ hline
    frac{1}{6} &&&&&& & \ hline
    frac{1}{3} & &&&&&& \ hline
    frac{1}{2} & &&&&&& \ hline
    frac{2}{3} & &&&&&& \ hline
    frac{5}{6} & &&&&&& \ hline
    1 & &&&&&& \ hline
    frac{7}{6} &&&&&& & \ hline
    frac{4}{3} & &&&&&& \ hline
    frac{3}{2} & &&&&&& \ hline
    frac{5}{3} & &&&&&& \ hline
    frac{11}{6} & &&&&&& \ hline
    2 & &&&&&& \ hline
    end{array}
    $$



    Part (II)
    draw on graph paper, horizontal axis called $x,$ vertical axis called $y$






    share|cite|improve this answer























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      2 Answers
      2






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      2 Answers
      2






      active

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      active

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      active

      oldest

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      up vote
      1
      down vote













      The sentences (rules) you stated will not be sufficient (applicable), because the given inequality is not a single absolute value, but a sum of two.



      The standard algebraic method to solve the absolute value inequality is to divide into intervals:
      $$5 ≤ 4|x − 1| + |2 − 3x| Rightarrow \
      begin{align}
      1) &begin{cases}xle frac23\ 5le -4(x-1)+(2-3x) end{cases} Rightarrow begin{cases}xle frac23\ xle frac1{7}end{cases} Rightarrow xin (-infty,frac17] text{or} \
      2) &begin{cases}frac23< x<1 \ 5le -4(x-1)-(2-3x) end{cases} Rightarrow begin{cases}frac23< x<1\ xle -3end{cases} Rightarrow xin emptyset text{or} \
      3) &begin{cases}1le x \ 5le 4(x-1)-(2-3x) end{cases} Rightarrow begin{cases}1le x\ xge frac{11}7end{cases} Rightarrow xin [frac{11}7,+infty) end{align}$$
      Hence, the final solution is: $$xin (-infty,frac17]cup [frac{11}7,+infty).$$



      ==============================================================



      Let's try:
      $$5 ≤ 4|x − 1| + |2 − 3x| iff \
      4|x-1|ge 5-|2-3x| iff \
      big[4(x-1)ge 5-|2-3x| big] lor big[-4(x-1)ge 5-|2-3x|big] iff \
      big[|2-3x|ge 9-4x big] lor big[|2-3x|ge 1+4xbig] iff \
      big[2-3xge 9-4x lor -(2-3x)ge 9-4xbig] lor big[2-3xge 1+4x lor -(2-3x)ge 1+4xbig] iff \
      big[xge 7 lor xge frac{11}7big] lor big[xle frac17 lor xle -3big] iff \
      big[xge frac{11}7big] lor big[xle frac17big] iff \
      xin (-infty,frac17]cup [frac{11}7,+infty)$$






      share|cite|improve this answer



























        up vote
        1
        down vote













        The sentences (rules) you stated will not be sufficient (applicable), because the given inequality is not a single absolute value, but a sum of two.



        The standard algebraic method to solve the absolute value inequality is to divide into intervals:
        $$5 ≤ 4|x − 1| + |2 − 3x| Rightarrow \
        begin{align}
        1) &begin{cases}xle frac23\ 5le -4(x-1)+(2-3x) end{cases} Rightarrow begin{cases}xle frac23\ xle frac1{7}end{cases} Rightarrow xin (-infty,frac17] text{or} \
        2) &begin{cases}frac23< x<1 \ 5le -4(x-1)-(2-3x) end{cases} Rightarrow begin{cases}frac23< x<1\ xle -3end{cases} Rightarrow xin emptyset text{or} \
        3) &begin{cases}1le x \ 5le 4(x-1)-(2-3x) end{cases} Rightarrow begin{cases}1le x\ xge frac{11}7end{cases} Rightarrow xin [frac{11}7,+infty) end{align}$$
        Hence, the final solution is: $$xin (-infty,frac17]cup [frac{11}7,+infty).$$



        ==============================================================



        Let's try:
        $$5 ≤ 4|x − 1| + |2 − 3x| iff \
        4|x-1|ge 5-|2-3x| iff \
        big[4(x-1)ge 5-|2-3x| big] lor big[-4(x-1)ge 5-|2-3x|big] iff \
        big[|2-3x|ge 9-4x big] lor big[|2-3x|ge 1+4xbig] iff \
        big[2-3xge 9-4x lor -(2-3x)ge 9-4xbig] lor big[2-3xge 1+4x lor -(2-3x)ge 1+4xbig] iff \
        big[xge 7 lor xge frac{11}7big] lor big[xle frac17 lor xle -3big] iff \
        big[xge frac{11}7big] lor big[xle frac17big] iff \
        xin (-infty,frac17]cup [frac{11}7,+infty)$$






        share|cite|improve this answer

























          up vote
          1
          down vote










          up vote
          1
          down vote









          The sentences (rules) you stated will not be sufficient (applicable), because the given inequality is not a single absolute value, but a sum of two.



          The standard algebraic method to solve the absolute value inequality is to divide into intervals:
          $$5 ≤ 4|x − 1| + |2 − 3x| Rightarrow \
          begin{align}
          1) &begin{cases}xle frac23\ 5le -4(x-1)+(2-3x) end{cases} Rightarrow begin{cases}xle frac23\ xle frac1{7}end{cases} Rightarrow xin (-infty,frac17] text{or} \
          2) &begin{cases}frac23< x<1 \ 5le -4(x-1)-(2-3x) end{cases} Rightarrow begin{cases}frac23< x<1\ xle -3end{cases} Rightarrow xin emptyset text{or} \
          3) &begin{cases}1le x \ 5le 4(x-1)-(2-3x) end{cases} Rightarrow begin{cases}1le x\ xge frac{11}7end{cases} Rightarrow xin [frac{11}7,+infty) end{align}$$
          Hence, the final solution is: $$xin (-infty,frac17]cup [frac{11}7,+infty).$$



          ==============================================================



          Let's try:
          $$5 ≤ 4|x − 1| + |2 − 3x| iff \
          4|x-1|ge 5-|2-3x| iff \
          big[4(x-1)ge 5-|2-3x| big] lor big[-4(x-1)ge 5-|2-3x|big] iff \
          big[|2-3x|ge 9-4x big] lor big[|2-3x|ge 1+4xbig] iff \
          big[2-3xge 9-4x lor -(2-3x)ge 9-4xbig] lor big[2-3xge 1+4x lor -(2-3x)ge 1+4xbig] iff \
          big[xge 7 lor xge frac{11}7big] lor big[xle frac17 lor xle -3big] iff \
          big[xge frac{11}7big] lor big[xle frac17big] iff \
          xin (-infty,frac17]cup [frac{11}7,+infty)$$






          share|cite|improve this answer














          The sentences (rules) you stated will not be sufficient (applicable), because the given inequality is not a single absolute value, but a sum of two.



          The standard algebraic method to solve the absolute value inequality is to divide into intervals:
          $$5 ≤ 4|x − 1| + |2 − 3x| Rightarrow \
          begin{align}
          1) &begin{cases}xle frac23\ 5le -4(x-1)+(2-3x) end{cases} Rightarrow begin{cases}xle frac23\ xle frac1{7}end{cases} Rightarrow xin (-infty,frac17] text{or} \
          2) &begin{cases}frac23< x<1 \ 5le -4(x-1)-(2-3x) end{cases} Rightarrow begin{cases}frac23< x<1\ xle -3end{cases} Rightarrow xin emptyset text{or} \
          3) &begin{cases}1le x \ 5le 4(x-1)-(2-3x) end{cases} Rightarrow begin{cases}1le x\ xge frac{11}7end{cases} Rightarrow xin [frac{11}7,+infty) end{align}$$
          Hence, the final solution is: $$xin (-infty,frac17]cup [frac{11}7,+infty).$$



          ==============================================================



          Let's try:
          $$5 ≤ 4|x − 1| + |2 − 3x| iff \
          4|x-1|ge 5-|2-3x| iff \
          big[4(x-1)ge 5-|2-3x| big] lor big[-4(x-1)ge 5-|2-3x|big] iff \
          big[|2-3x|ge 9-4x big] lor big[|2-3x|ge 1+4xbig] iff \
          big[2-3xge 9-4x lor -(2-3x)ge 9-4xbig] lor big[2-3xge 1+4x lor -(2-3x)ge 1+4xbig] iff \
          big[xge 7 lor xge frac{11}7big] lor big[xle frac17 lor xle -3big] iff \
          big[xge frac{11}7big] lor big[xle frac17big] iff \
          xin (-infty,frac17]cup [frac{11}7,+infty)$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 18 at 12:31

























          answered Nov 18 at 4:41









          farruhota

          18k2736




          18k2736






















              up vote
              0
              down vote













              Part (I) fill in this table
              $$
              begin{array}{c|c|c|c|c|c|c|c}
              x & x-1&|x-1|&4|x-1| &-3x&2-3x&|2-3x|& y = 4|x-1| +|2-3x| \ hline
              frac{-1}{6} & &&&&&& \ hline
              0 & &&&&&& \ hline
              frac{1}{6} &&&&&& & \ hline
              frac{1}{3} & &&&&&& \ hline
              frac{1}{2} & &&&&&& \ hline
              frac{2}{3} & &&&&&& \ hline
              frac{5}{6} & &&&&&& \ hline
              1 & &&&&&& \ hline
              frac{7}{6} &&&&&& & \ hline
              frac{4}{3} & &&&&&& \ hline
              frac{3}{2} & &&&&&& \ hline
              frac{5}{3} & &&&&&& \ hline
              frac{11}{6} & &&&&&& \ hline
              2 & &&&&&& \ hline
              end{array}
              $$



              Part (II)
              draw on graph paper, horizontal axis called $x,$ vertical axis called $y$






              share|cite|improve this answer



























                up vote
                0
                down vote













                Part (I) fill in this table
                $$
                begin{array}{c|c|c|c|c|c|c|c}
                x & x-1&|x-1|&4|x-1| &-3x&2-3x&|2-3x|& y = 4|x-1| +|2-3x| \ hline
                frac{-1}{6} & &&&&&& \ hline
                0 & &&&&&& \ hline
                frac{1}{6} &&&&&& & \ hline
                frac{1}{3} & &&&&&& \ hline
                frac{1}{2} & &&&&&& \ hline
                frac{2}{3} & &&&&&& \ hline
                frac{5}{6} & &&&&&& \ hline
                1 & &&&&&& \ hline
                frac{7}{6} &&&&&& & \ hline
                frac{4}{3} & &&&&&& \ hline
                frac{3}{2} & &&&&&& \ hline
                frac{5}{3} & &&&&&& \ hline
                frac{11}{6} & &&&&&& \ hline
                2 & &&&&&& \ hline
                end{array}
                $$



                Part (II)
                draw on graph paper, horizontal axis called $x,$ vertical axis called $y$






                share|cite|improve this answer

























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Part (I) fill in this table
                  $$
                  begin{array}{c|c|c|c|c|c|c|c}
                  x & x-1&|x-1|&4|x-1| &-3x&2-3x&|2-3x|& y = 4|x-1| +|2-3x| \ hline
                  frac{-1}{6} & &&&&&& \ hline
                  0 & &&&&&& \ hline
                  frac{1}{6} &&&&&& & \ hline
                  frac{1}{3} & &&&&&& \ hline
                  frac{1}{2} & &&&&&& \ hline
                  frac{2}{3} & &&&&&& \ hline
                  frac{5}{6} & &&&&&& \ hline
                  1 & &&&&&& \ hline
                  frac{7}{6} &&&&&& & \ hline
                  frac{4}{3} & &&&&&& \ hline
                  frac{3}{2} & &&&&&& \ hline
                  frac{5}{3} & &&&&&& \ hline
                  frac{11}{6} & &&&&&& \ hline
                  2 & &&&&&& \ hline
                  end{array}
                  $$



                  Part (II)
                  draw on graph paper, horizontal axis called $x,$ vertical axis called $y$






                  share|cite|improve this answer














                  Part (I) fill in this table
                  $$
                  begin{array}{c|c|c|c|c|c|c|c}
                  x & x-1&|x-1|&4|x-1| &-3x&2-3x&|2-3x|& y = 4|x-1| +|2-3x| \ hline
                  frac{-1}{6} & &&&&&& \ hline
                  0 & &&&&&& \ hline
                  frac{1}{6} &&&&&& & \ hline
                  frac{1}{3} & &&&&&& \ hline
                  frac{1}{2} & &&&&&& \ hline
                  frac{2}{3} & &&&&&& \ hline
                  frac{5}{6} & &&&&&& \ hline
                  1 & &&&&&& \ hline
                  frac{7}{6} &&&&&& & \ hline
                  frac{4}{3} & &&&&&& \ hline
                  frac{3}{2} & &&&&&& \ hline
                  frac{5}{3} & &&&&&& \ hline
                  frac{11}{6} & &&&&&& \ hline
                  2 & &&&&&& \ hline
                  end{array}
                  $$



                  Part (II)
                  draw on graph paper, horizontal axis called $x,$ vertical axis called $y$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 17 at 22:01

























                  answered Nov 17 at 19:50









                  Will Jagy

                  101k597198




                  101k597198






























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