Algebra 1 - 1AL1.4: Absolute Value Equations ...

1 The Team1 Algebra 1 - : Absolute Value Equations & Inequalities The TeamMULTIPLE 5__ 8. Find all values ofx which make this equation {1, 413}c.{413, 1}b.{1, 1}d.{413, 413}ANS:A3x 5__ 8 means that3x 5 8or3x 5 8.( or means both answers make the equation true )Each of these Equations requires a two-step 5 83x 8 53x 3 this is the first stepx 33 this is the second step So,x=1 is one 5 83x 8 53x 13x 133 So,x= 413 is another : The solution set is: {1, 413}Geometrically,The Team2 PTS:1 DIF:Grade 8 :Students solve Equations and inequalities involving Absolute : Algebra IKEY: Absolute values | equationsMSC:DynamicThe 11__ 3. Find all values ofx which make this equation {7,4}c.{ 4, 7}b.{7, 7}d.{ 4,4}ANS:A2x 11__ 3 means that2x 11 3or2x 11 3.( or means both answers make the equation true )Each of these Equations requires a two-step 11 32x 3 112x 14 this is the first stepx 142 this is the second step So,x=7 is one 11 32x 3 112x 8x 82 So,x=4 is another : The solution set is: {7,4}Geometrically,PTS:1 DIF:Grade 8 :Students solve Equations and inequalities involving Absolute : Algebra IKEY: Absolute values | equationsMSC:DynamicThe 2x__ 14.

2 Find all values ofx which make this equation {8, 6}c.{6, 8}b.{8, 8}d.{6, 6}ANS:A2 2x__ 14 is the same as2x 2__ 14 (why?) and this Absolute Value equation is easier to work with.(Of course, you can work with the first Absolute Value equation and you MUST get the same answers.)2x 2__ 14 means that2x 2 14or2x 2 14.( or means both answers make the equation true )Each of these Equations requires a two-step 2 142x 14 22x 16 this is the first stepx 162 this is the second step So,x=8 is one 2 142x 14 22x 12x 122 So,x= 6 is another : The solution set is: {8, 6}Geometrically, we have:PTS:1 DIF:Grade 8 :Students solve Equations and inequalities involving Absolute : Algebra IKEY: Absolute values | equationsMSC:DynamicThe 3__ 12. Find all values ofx which make this equation {3, 9}c.{9, 3}b.{3, 3}d.{9, 9}ANS:A2x 3__ 12 is the same as:x 3__ x to make this true:x 3 6orx 3 6 Each of these Equations can be solved in 3 6x 6 3x 3 this is the only step So,x= 3 is one 3 6x 6 3x 9 So,x= 9 is another : The solution set is: {3, 9}Geometrically,PTS:1 DIF:Grade 8 :Students solve Equations and inequalities involving Absolute : Algebra IKEY: Absolute values | equationsMSC:DynamicThe 3__ 9.

3 Find all values ofx which make this equation {4,2}c.{ 2, 4}b.{ 2,2}d.{4, 4}ANS:A9x 3__ 9 is the same as:x 3__ to make this true:x 3 1orx 3 1 Each of these Equations can be solved in 3 1x 1 3x 4 this is the only step So,x= 4 is one 3 1x 1 3x 2 So,x= 2 is another : The solution set is: {2,4}Geometrically,PTS:1 DIF:Grade 8 :Students solve Equations and inequalities involving Absolute : Algebra IKEY: Absolute values | equationsMSC:DynamicThe 8__ 6. Find all values ofx which make this equation {10, 26}c.{26, 10}b.{10, 10}d.{26, 26}ANS:A13x 8__ 6 is the same as:x 8__ x to make this true:x 8 18orx 8 18 Each of these Equations can be solved in 8 18x 18 8x 10 this is the only step So,x= 10 is one 8 18x 18 8x 26 So,x= 26 is another : The solution set is: {10, 26}PTS:1 DIF:Grade 8 :Students solve Equations and inequalities involving Absolute : Algebra IKEY: Absolute values | equationsMSC:DynamicThe 7__ 7.

4 Find all values ofx which make this equation {70, 56}c.{56, 70}b.{56, 56}d.{70, 70}ANS:A19x 7__ 7 is the same as:x 7__ to make this true:x 7 63orx 7 63 Each of these Equations can be solved in 7 63x 63 7x 70 this is the only step So,x= 70 is one 7 63x 63 7x 56 So,x= 56 is another : The solution set is: {56, 70}PTS:1 DIF:Grade 8 :Students solve Equations and inequalities involving Absolute : Algebra IKEY: Absolute values | equationsMSC:DynamicThe 3__ 10. Find all values ofx which make this inequality {x|x ( 13,7)}c.{x|x ( 13,13)}b.{x|x ( 7, 13)}d.{x|x ( 7, 7)}ANS:Ax 3__ 10 Below on the number line are all the numbersasuch thata_ _ 10. That is, 10 a x to makex 3__ 10 true, we must have: 10 x 3 write this statement as:x 3! 10andx 3 10. That is,x must satisfy both of these inequalities can be solved in 3! 10x! 10 3x! 13 this is the only step So,x> 7 is one condition forx(green line).

5 3 10x 10 3x 7 So,x< 13 is the other condition forx(blue line).We need theintersection of these two : The solution set is: {x|x ( 13, 7)}(black line).PTS:1 DIF:Grade 8 :Students solve Equations and inequalities involving Absolute : Algebra IKEY: Absolute values | inequalityMSC:DynamicThe 4__ 7. Find all values ofx which make this inequality {x|x ( 3,11)}c.{x|x ( 11,11)}b.{x|x ( 3, 3)}d.{x|x ( 11, 3)}ANS:Ax 4__ 7 Below on the number line are all the numbersasuch thata_ _ 7. That is, 7 a means that forx 4__ 7 to be true, we must have: 7 x 4 write this statement as:x 4! 7andx 4 7. That is,x must satisfy both of these inequalities can be solved in 4! 7x! 7 4x! 3 this is the only step So,x> 3 is one condition forx(green line). 4 7x 7 4x 11 So,x< 11 is the other condition forx(blue line).We need theintersection of these two : The solution set is: {x|x ( 11, 3)}(black line).PTS:1 DIF:Grade 8 :Students solve Equations and inequalities involving Absolute : Algebra IKEY: Absolute values | inequalityMSC:DynamicThe x__ 10.

6 Find all values ofx which make this inequality {x|x ( 7,13)}c.{x|x ( 13,13)}b.{x|x ( 7, 7)}d.{x|x ( 13, 7)}ANS:A3 x__ 10 is the same asx 3__ 10, which is easier to on the number line are all the numbersasuch thata_ _ 10. That is, 10 a means that forx 3__ 10 to be true, we must have: 10 x 3 write this statement as:x 3! 10andx 3 10. That is,x must satisfy both of these inequalities can be solved in 3! 10x! 10 3x! 7 this is the only step So,x> 7 is one condition forx(green line). 3 10x 10 3x 13 So,x< 13 is the other condition forx(blue line).We need theintersection of these two : The solution set is: {x|x ( 13, 7)}(black line).PTS:1 DIF:Grade 8 :Students solve Equations and inequalities involving Absolute : Algebra IKEY: Absolute values | inequalityMSC:DynamicThe 11__t2. Find all values ofx which make this inequality {x|x ( f, 13] [ 9,f)}c.{x|x ( f, 13] [13,f)}b.{x|x ( f, 9] [13,f)}d.{x|x ( f, 9] [9,f)}ANS:Ax 11__t2 Below on the number line are all the numbersasuch thata_ _t2.

7 That is,ad x to makex 11__t2 true, we must have:x 11d 2orx is,x must satisfy at least one of these of these inequalities can be solved in 11d 2xd 2 11xd 13 this is the only s te p So, ifxd 9 the inequality in the question is true(green line). 11t2xt2 11xt 9 So, ifxt 13 the inequality in the question is also true(blue line).We need theunion of these two : The solution set is: {x|xd 13orxt 9}(black line).PTS:1 DIF:Grade 8 :Students solve Equations and inequalities involving Absolute : Algebra IKEY: Absolute values | inequalityMSC:DynamicThe 5__t4. Find all values ofx which make this equation {x|xd1orxt9}c.{x|xd 9orxt9}b.{x|xd 1 orxt1}d.{x|xd 9orxt 1}ANS:Ax 5__t4 Below on the number line are all the numbersasuch thata_ _t4. That is,ad x to makex 5__t4 true, we must have:x 5d 4orx is,x must satisfy at least one of these of these inequalities can be solved in 5d 4xd 4 5xd1 this is the only step So, ifxd 1 the inequality in the question is also true(green line).

8 5t4xt4 5xt9 So, ifxt 9 the inequality in the question is also true(blue line).We need theunion of these two : The solution set is: {x|xd1orxt9}(black line).PTS:1 DIF:Grade 8 :Students solve Equations and inequalities involving Absolute : Algebra IKEY: Absolute values | inequalityMSC:DynamicThe x__t2. Find all values ofx which make this equation {x|xd4orxt8}c.{x|xd 8orxt8}b.{x|xd 4 orxt4}d.{x|xd 8orxt 4}ANS:A6 x__t2 is the same asx 6__t2, which is easier to on the number line are all the numbersasuch thata_ _t2. That is,ad x to makex 6__t2 true, we must have:x 6d 2orx is,x must satisfy at least one of these of these inequalities can be solved in 6d 2xd 2 6xd4 this is the only step So, ifxd 4 the inequality in the question is also true(green line). 6t2xt2 6xt8 So, ifxt 8 the inequality in the question is also true(blue line).We need theunion of these two : The solution set is: {x|xd4orxt8}(black line).

9 PTS:1 DIF:Grade 8 :Students solve Equations and inequalities involving Absolute : Algebra IKEY: Absolute values | inequalityMSC:DynamicThe 10__ 4t10. Find all values ofx which make this inequality {x|xd }c.{x|xd }b.{x|xd }d.{x| }ANS:AThe first step is to reduce22x 10__ 4t10 to a simple Absolute Value 10__ 4t1022x 10__t10 422x 10__t62x 10__t622x 10__t3 Below on the number line are all the numbersasuch thata_ _t3. That is,ad x to make2x 10__t3 true, we must have:2x 10d 3or2x is,x must satisfy at least one of these of these inequalities can be solved in 10d 32xd 3 102xd 13xd 132 So, ifxd the inequality in the question is true(green line). 10t32xt3 102xt 7xt 72 So, ifxt the inequality in the question is also true(blue line).We need theunion of these two : The solution set is: {x|xd }(black line).PTS:1 DIF:Grade 8 :Students solve Equations and inequalities involving Absolute Team16 TOP: Algebra IKEY: Absolute values | inequalityMSC:DynamicThe 34x 7__d 2.

10 Find all values ofx which make this inequality {x|xd 1}c.{x|xd }b.{x|xd 1 orxt1}d.{x| }ANS:AThe first step is to reduce7 34x 7__d 2 to a simple Absolute Value inequality. Notice the change inthe inequality when we divide by the negative coefficient 34x 7__d 2 34x 7__d 2 7 34x 7__d 94x 7__t 9 34x 7__t3 Below on the number line are all the numbersasuch thata_ _t3. That is,ad x to make4x 7__t3 true, we must have:4x 7d 3or4x is,x must satisfy at least one of these of these inequalities can be solved in 7d 34xd 3 74xd 10xd 104 So, ifxd the inequality in the question is true(green line). 7t34xt3 74xt 4xt 44 So, ifxt 1 the inequality in the question is also true(blue line).We need theunion of these two : The solution set is: {x|xd 1}(black line).PTS:1 DIF:Grade 8 Team18 OBJ:Students solve Equations and inequalities involving Absolute : Algebra IKEY: Absolute values | inequalityMSC:DynamicThe 35x 6__t 9.