Practice · Match interval endpoints, splitting on the sign · Sensim Math

Variant 1 easy

The values of $x$ that satisfy the inequality $|ax+1|\le b$ are exactly $-1\le x\le 2$ . Find the integers $a$ and $b$ . (Assume $a\neq 0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Undo the absolute value to get a double inequality for $ax$ , divide by $a$ , and make the resulting interval equal $-1\le x\le 2$ . Because dividing by $a$ may flip the inequalities, split into the cases $a>0$ and $a<0$ .

Givens

The solutions of $|ax+1|\le b$ are exactly $-1\le x\le 2$ .
$a$ and $b$ are integers and $a\neq 0$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Dividing by a negative $a$ reverses both inequality signs.
Since the inequality has solutions, $b\ge 0$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Split on the sign of $a$ (two subproblems). In each case match the two endpoints, which gives two linear equations in $a$ and $b$ ; then keep only the case consistent with its sign assumption.

3 · Execute4 carry out the plan

1Remove the absolute value

#13 Convert to Algebra 6.NS.C.7

|ax+1|\le b

means

ax+1

lies between

-b

and

b

. Subtract

1

throughout.

-b\le ax+1\le b \;\Rightarrow\; -b-1\le ax\le b-1

An absolute value at most

b

means the distance from

0

is at most

b

.

2Case $a>0$

#3 Eliminate Possibilities 8.EE.C.8

Dividing by a positive

a

keeps the signs:

\frac{-b-1}{a}\le x\le\frac{b-1}{a}

. Matching endpoints gives

-b-1=-a

and

b-1=2a

. Adding the two equations gives

-2=a

, so

a=-2

, which contradicts

a>0

.

-b-1=-a,\; b-1=2a \;\Rightarrow\; -2=a \;\Rightarrow\; a=-2

The only candidate breaks its own assumption, so this case is eliminated.

3Case $a<0$

#7 Identify Subproblems 8.EE.C.8

Dividing by a negative

a

flips both signs:

\frac{b-1}{a}\le x\le\frac{-b-1}{a}

. Matching endpoints gives

b-1=-a

and

-b-1=2a

. Adding gives

-2=a

, so

a=-2

, and then

b=1-a=3

.

b-1=-a,\; -b-1=2a \;\Rightarrow\; a=-2,\; b=1-1(-2)=3

Here

a=-2<0

and

b=3\ge 0

, so the case is consistent.

4Check

#13 Convert to Algebra 7.EE.B.4

Substitute

a=-2

,

b=3

back into the original inequality.

|-2x+1|\le 3 \;\Rightarrow\; -3\le -2x+1\le 3 \;\Rightarrow\; -1\le x\le 2

The solution set is exactly the given interval.

Answer:

a=-2

,

b=3

4 · Reviewdoes it hold up?

With $a=-2$ , $b=3$ : $x=-1$ gives $|3|=3\le 3$ and $x=2$ gives $|-3|=3\le 3$ , both on the boundary, while $x=3$ gives $|-5|=5>3$ . So the endpoints are right.

Another way: Think of distance: $|ax+1|=|a|\cdot|x+\frac{1}{a}|$ , so the solutions form an interval centered at $-\frac{1}{a}$ with half-width $\frac{b}{|a|}$ . The interval $-1\le x\le 2$ has center $\frac{1}{2}$ and half-width $1\frac{1}{2}$ , so $-\frac{1}{a}=\frac{1}{2}$ gives $a=-2$ and $\frac{b}{2}=1\frac{1}{2}$ gives $b=3$ .

Standardsmin grade 8

6.NS.C.7 Order and absolute value of rational numbers — Reading $|ax+1|\le b$ as a distance condition and rewriting it as a double inequality.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving and checking the linear inequalities in $x$.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the pair of linear equations for $a$ and $b$ obtained from the endpoints.

💡Takeaway. When you divide by a letter, ask first whether it is positive or negative.

Variant 2 easy

The values of $x$ that satisfy the inequality $|ax-1|\le b$ are exactly $-1\le x\le 2$ . Find the integers $a$ and $b$ . (Assume $a\neq 0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Undo the absolute value to get a double inequality for $ax$ , divide by $a$ , and make the resulting interval equal $-1\le x\le 2$ . Because dividing by $a$ may flip the inequalities, split into the cases $a>0$ and $a<0$ .

Givens

The solutions of $|ax-1|\le b$ are exactly $-1\le x\le 2$ .
$a$ and $b$ are integers and $a\neq 0$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Dividing by a negative $a$ reverses both inequality signs.
Since the inequality has solutions, $b\ge 0$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Split on the sign of $a$ (two subproblems). In each case match the two endpoints, which gives two linear equations in $a$ and $b$ ; then keep only the case consistent with its sign assumption.

3 · Execute4 carry out the plan

1Remove the absolute value

#13 Convert to Algebra 6.NS.C.7

|ax-1|\le b

means

ax-1

lies between

-b

and

b

. Add

1

throughout.

-b\le ax-1\le b \;\Rightarrow\; -b+1\le ax\le b+1

An absolute value at most

b

means the distance from

0

is at most

b

.

2Case $a>0$

#7 Identify Subproblems 8.EE.C.8

Dividing by a positive

a

keeps the signs:

\frac{-b+1}{a}\le x\le\frac{b+1}{a}

. Matching endpoints gives

-b+1=-a

and

b+1=2a

. Adding gives

2=a

, so

a=2

, and then

b=-1+2a=3

.

-b+1=-a,\; b+1=2a \;\Rightarrow\; a=2,\; b=-1+2\cdot 2=3

Here

a=2>0

and

b=3\ge 0

, so the case is consistent.

3Case $a<0$

#3 Eliminate Possibilities 8.EE.C.8

Dividing by a negative

a

flips both signs:

\frac{b+1}{a}\le x\le\frac{-b+1}{a}

. Matching endpoints gives

b+1=-a

and

-b+1=2a

. Adding the two equations gives

2=a

, so

a=2

, which contradicts

a<0

.

b+1=-a,\; -b+1=2a \;\Rightarrow\; 2=a \;\Rightarrow\; a=2

The only candidate breaks its own assumption, so this case is eliminated.

4Check

#13 Convert to Algebra 7.EE.B.4

Substitute

a=2

,

b=3

back into the original inequality.

|2x-1|\le 3 \;\Rightarrow\; -3\le 2x-1\le 3 \;\Rightarrow\; -1\le x\le 2

The solution set is exactly the given interval.

Answer:

a=2

,

b=3

4 · Reviewdoes it hold up?

With $a=2$ , $b=3$ : $x=-1$ gives $|-3|=3\le 3$ and $x=2$ gives $|3|=3\le 3$ , both on the boundary, while $x=3$ gives $|5|=5>3$ . So the endpoints are right.

Another way: Think of distance: $|ax-1|=|a|\cdot|x+\frac{-1}{a}|$ , so the solutions form an interval centered at $-\frac{-1}{a}$ with half-width $\frac{b}{|a|}$ . The interval $-1\le x\le 2$ has center $\frac{1}{2}$ and half-width $1\frac{1}{2}$ , so $-\frac{-1}{a}=\frac{1}{2}$ gives $a=2$ and $\frac{b}{2}=1\frac{1}{2}$ gives $b=3$ .

Standardsmin grade 8

6.NS.C.7 Order and absolute value of rational numbers — Reading $|ax-1|\le b$ as a distance condition and rewriting it as a double inequality.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving and checking the linear inequalities in $x$.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the pair of linear equations for $a$ and $b$ obtained from the endpoints.

💡Takeaway. When you divide by a letter, ask first whether it is positive or negative.

Variant 3 easy

The values of $x$ that satisfy the inequality $|ax+2|\le b$ are exactly $-3\le x\le 1$ . Find the integers $a$ and $b$ . (Assume $a\neq 0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Undo the absolute value to get a double inequality for $ax$ , divide by $a$ , and make the resulting interval equal $-3\le x\le 1$ . Because dividing by $a$ may flip the inequalities, split into the cases $a>0$ and $a<0$ .

Givens

The solutions of $|ax+2|\le b$ are exactly $-3\le x\le 1$ .
$a$ and $b$ are integers and $a\neq 0$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Dividing by a negative $a$ reverses both inequality signs.
Since the inequality has solutions, $b\ge 0$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Split on the sign of $a$ (two subproblems). In each case match the two endpoints, which gives two linear equations in $a$ and $b$ ; then keep only the case consistent with its sign assumption.

3 · Execute4 carry out the plan

1Remove the absolute value

#13 Convert to Algebra 6.NS.C.7

|ax+2|\le b

means

ax+2

lies between

-b

and

b

. Subtract

2

throughout.

-b\le ax+2\le b \;\Rightarrow\; -b-2\le ax\le b-2

An absolute value at most

b

means the distance from

0

is at most

b

.

2Case $a>0$

#7 Identify Subproblems 8.EE.C.8

Dividing by a positive

a

keeps the signs:

\frac{-b-2}{a}\le x\le\frac{b-2}{a}

. Matching endpoints gives

-b-2=-3a

and

b-2=a

. Adding gives

-4=-2a

, so

a=2

, and then

b=2+a=4

.

-b-2=-3a,\; b-2=a \;\Rightarrow\; a=2,\; b=2+1\cdot 2=4

Here

a=2>0

and

b=4\ge 0

, so the case is consistent.

3Case $a<0$

#3 Eliminate Possibilities 8.EE.C.8

Dividing by a negative

a

flips both signs:

\frac{b-2}{a}\le x\le\frac{-b-2}{a}

. Matching endpoints gives

b-2=-3a

and

-b-2=a

. Adding the two equations gives

-4=-2a

, so

a=2

, which contradicts

a<0

.

b-2=-3a,\; -b-2=a \;\Rightarrow\; -4=-2a \;\Rightarrow\; a=2

The only candidate breaks its own assumption, so this case is eliminated.

4Check

#13 Convert to Algebra 7.EE.B.4

Substitute

a=2

,

b=4

back into the original inequality.

|2x+2|\le 4 \;\Rightarrow\; -4\le 2x+2\le 4 \;\Rightarrow\; -3\le x\le 1

The solution set is exactly the given interval.

Answer:

a=2

,

b=4

4 · Reviewdoes it hold up?

With $a=2$ , $b=4$ : $x=-3$ gives $|-4|=4\le 4$ and $x=1$ gives $|4|=4\le 4$ , both on the boundary, while $x=2$ gives $|6|=6>4$ . So the endpoints are right.

Another way: Think of distance: $|ax+2|=|a|\cdot|x+\frac{2}{a}|$ , so the solutions form an interval centered at $-\frac{2}{a}$ with half-width $\frac{b}{|a|}$ . The interval $-3\le x\le 1$ has center $-1$ and half-width $2$ , so $-\frac{2}{a}=-1$ gives $a=2$ and $\frac{b}{2}=2$ gives $b=4$ .

Standardsmin grade 8

6.NS.C.7 Order and absolute value of rational numbers — Reading $|ax+2|\le b$ as a distance condition and rewriting it as a double inequality.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving and checking the linear inequalities in $x$.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the pair of linear equations for $a$ and $b$ obtained from the endpoints.

💡Takeaway. When you divide by a letter, ask first whether it is positive or negative.

Variant 4 easy

The values of $x$ that satisfy the inequality $|ax+1|\le b$ are exactly $-3\le x\le 1$ . Find the integers $a$ and $b$ . (Assume $a\neq 0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Undo the absolute value to get a double inequality for $ax$ , divide by $a$ , and make the resulting interval equal $-3\le x\le 1$ . Because dividing by $a$ may flip the inequalities, split into the cases $a>0$ and $a<0$ .

Givens

The solutions of $|ax+1|\le b$ are exactly $-3\le x\le 1$ .
$a$ and $b$ are integers and $a\neq 0$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Dividing by a negative $a$ reverses both inequality signs.
Since the inequality has solutions, $b\ge 0$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Split on the sign of $a$ (two subproblems). In each case match the two endpoints, which gives two linear equations in $a$ and $b$ ; then keep only the case consistent with its sign assumption.

3 · Execute4 carry out the plan

1Remove the absolute value

#13 Convert to Algebra 6.NS.C.7

|ax+1|\le b

means

ax+1

lies between

-b

and

b

. Subtract

1

throughout.

-b\le ax+1\le b \;\Rightarrow\; -b-1\le ax\le b-1

An absolute value at most

b

means the distance from

0

is at most

b

.

2Case $a>0$

#7 Identify Subproblems 8.EE.C.8

Dividing by a positive

a

keeps the signs:

\frac{-b-1}{a}\le x\le\frac{b-1}{a}

. Matching endpoints gives

-b-1=-3a

and

b-1=a

. Adding gives

-2=-2a

, so

a=1

, and then

b=1+a=2

.

-b-1=-3a,\; b-1=a \;\Rightarrow\; a=1,\; b=1+1\cdot 1=2

Here

a=1>0

and

b=2\ge 0

, so the case is consistent.

3Case $a<0$

#3 Eliminate Possibilities 8.EE.C.8

Dividing by a negative

a

flips both signs:

\frac{b-1}{a}\le x\le\frac{-b-1}{a}

. Matching endpoints gives

b-1=-3a

and

-b-1=a

. Adding the two equations gives

-2=-2a

, so

a=1

, which contradicts

a<0

.

b-1=-3a,\; -b-1=a \;\Rightarrow\; -2=-2a \;\Rightarrow\; a=1

The only candidate breaks its own assumption, so this case is eliminated.

4Check

#13 Convert to Algebra 7.EE.B.4

Substitute

a=1

,

b=2

back into the original inequality.

|x+1|\le 2 \;\Rightarrow\; -2\le x+1\le 2 \;\Rightarrow\; -3\le x\le 1

The solution set is exactly the given interval.

Answer:

a=1

,

b=2

4 · Reviewdoes it hold up?

With $a=1$ , $b=2$ : $x=-3$ gives $|-2|=2\le 2$ and $x=1$ gives $|2|=2\le 2$ , both on the boundary, while $x=2$ gives $|3|=3>2$ . So the endpoints are right.

Another way: Think of distance: $|ax+1|=|a|\cdot|x+\frac{1}{a}|$ , so the solutions form an interval centered at $-\frac{1}{a}$ with half-width $\frac{b}{|a|}$ . The interval $-3\le x\le 1$ has center $-1$ and half-width $2$ , so $-\frac{1}{a}=-1$ gives $a=1$ and $\frac{b}{1}=2$ gives $b=2$ .

Standardsmin grade 8

6.NS.C.7 Order and absolute value of rational numbers — Reading $|ax+1|\le b$ as a distance condition and rewriting it as a double inequality.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving and checking the linear inequalities in $x$.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the pair of linear equations for $a$ and $b$ obtained from the endpoints.

💡Takeaway. When you divide by a letter, ask first whether it is positive or negative.

Variant 5 medium

The values of $x$ that satisfy the inequality $|ax+1|\le b$ are exactly $-2\le x\le 4$ . Find the integers $a$ and $b$ . (Assume $a\neq 0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Undo the absolute value to get a double inequality for $ax$ , divide by $a$ , and make the resulting interval equal $-2\le x\le 4$ . Because dividing by $a$ may flip the inequalities, split into the cases $a>0$ and $a<0$ .

Givens

The solutions of $|ax+1|\le b$ are exactly $-2\le x\le 4$ .
$a$ and $b$ are integers and $a\neq 0$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Dividing by a negative $a$ reverses both inequality signs.
Since the inequality has solutions, $b\ge 0$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Split on the sign of $a$ (two subproblems). In each case match the two endpoints, which gives two linear equations in $a$ and $b$ ; then keep only the case consistent with its sign assumption.

3 · Execute4 carry out the plan

1Remove the absolute value

#13 Convert to Algebra 6.NS.C.7

|ax+1|\le b

means

ax+1

lies between

-b

and

b

. Subtract

1

throughout.

-b\le ax+1\le b \;\Rightarrow\; -b-1\le ax\le b-1

An absolute value at most

b

means the distance from

0

is at most

b

.

2Case $a>0$

#3 Eliminate Possibilities 8.EE.C.8

Dividing by a positive

a

keeps the signs:

\frac{-b-1}{a}\le x\le\frac{b-1}{a}

. Matching endpoints gives

-b-1=-2a

and

b-1=4a

. Adding the two equations gives

-2=2a

, so

a=-1

, which contradicts

a>0

.

-b-1=-2a,\; b-1=4a \;\Rightarrow\; -2=2a \;\Rightarrow\; a=-1

The only candidate breaks its own assumption, so this case is eliminated.

3Case $a<0$

#7 Identify Subproblems 8.EE.C.8

Dividing by a negative

a

flips both signs:

\frac{b-1}{a}\le x\le\frac{-b-1}{a}

. Matching endpoints gives

b-1=-2a

and

-b-1=4a

. Adding gives

-2=2a

, so

a=-1

, and then

b=1-2a=3

.

b-1=-2a,\; -b-1=4a \;\Rightarrow\; a=-1,\; b=1-2(-1)=3

Here

a=-1<0

and

b=3\ge 0

, so the case is consistent.

4Check

#13 Convert to Algebra 7.EE.B.4

Substitute

a=-1

,

b=3

back into the original inequality.

|-x+1|\le 3 \;\Rightarrow\; -3\le -x+1\le 3 \;\Rightarrow\; -2\le x\le 4

The solution set is exactly the given interval.

Answer:

a=-1

,

b=3

4 · Reviewdoes it hold up?

With $a=-1$ , $b=3$ : $x=-2$ gives $|3|=3\le 3$ and $x=4$ gives $|-3|=3\le 3$ , both on the boundary, while $x=5$ gives $|-4|=4>3$ . So the endpoints are right.

Another way: Think of distance: $|ax+1|=|a|\cdot|x+\frac{1}{a}|$ , so the solutions form an interval centered at $-\frac{1}{a}$ with half-width $\frac{b}{|a|}$ . The interval $-2\le x\le 4$ has center $1$ and half-width $3$ , so $-\frac{1}{a}=1$ gives $a=-1$ and $\frac{b}{1}=3$ gives $b=3$ .

Standardsmin grade 8

6.NS.C.7 Order and absolute value of rational numbers — Reading $|ax+1|\le b$ as a distance condition and rewriting it as a double inequality.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving and checking the linear inequalities in $x$.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the pair of linear equations for $a$ and $b$ obtained from the endpoints.

💡Takeaway. When you divide by a letter, ask first whether it is positive or negative.

Variant 6 medium

The values of $x$ that satisfy the inequality $|ax+3|\le b$ are exactly $-4\le x\le 1$ . Find the integers $a$ and $b$ . (Assume $a\neq 0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Undo the absolute value to get a double inequality for $ax$ , divide by $a$ , and make the resulting interval equal $-4\le x\le 1$ . Because dividing by $a$ may flip the inequalities, split into the cases $a>0$ and $a<0$ .

Givens

The solutions of $|ax+3|\le b$ are exactly $-4\le x\le 1$ .
$a$ and $b$ are integers and $a\neq 0$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Dividing by a negative $a$ reverses both inequality signs.
Since the inequality has solutions, $b\ge 0$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Split on the sign of $a$ (two subproblems). In each case match the two endpoints, which gives two linear equations in $a$ and $b$ ; then keep only the case consistent with its sign assumption.

3 · Execute4 carry out the plan

1Remove the absolute value

#13 Convert to Algebra 6.NS.C.7

|ax+3|\le b

means

ax+3

lies between

-b

and

b

. Subtract

3

throughout.

-b\le ax+3\le b \;\Rightarrow\; -b-3\le ax\le b-3

An absolute value at most

b

means the distance from

0

is at most

b

.

2Case $a>0$

#7 Identify Subproblems 8.EE.C.8

Dividing by a positive

a

keeps the signs:

\frac{-b-3}{a}\le x\le\frac{b-3}{a}

. Matching endpoints gives

-b-3=-4a

and

b-3=a

. Adding gives

-6=-3a

, so

a=2

, and then

b=3+a=5

.

-b-3=-4a,\; b-3=a \;\Rightarrow\; a=2,\; b=3+1\cdot 2=5

Here

a=2>0

and

b=5\ge 0

, so the case is consistent.

3Case $a<0$

#3 Eliminate Possibilities 8.EE.C.8

Dividing by a negative

a

flips both signs:

\frac{b-3}{a}\le x\le\frac{-b-3}{a}

. Matching endpoints gives

b-3=-4a

and

-b-3=a

. Adding the two equations gives

-6=-3a

, so

a=2

, which contradicts

a<0

.

b-3=-4a,\; -b-3=a \;\Rightarrow\; -6=-3a \;\Rightarrow\; a=2

The only candidate breaks its own assumption, so this case is eliminated.

4Check

#13 Convert to Algebra 7.EE.B.4

Substitute

a=2

,

b=5

back into the original inequality.

|2x+3|\le 5 \;\Rightarrow\; -5\le 2x+3\le 5 \;\Rightarrow\; -4\le x\le 1

The solution set is exactly the given interval.

Answer:

a=2

,

b=5

4 · Reviewdoes it hold up?

With $a=2$ , $b=5$ : $x=-4$ gives $|-5|=5\le 5$ and $x=1$ gives $|5|=5\le 5$ , both on the boundary, while $x=2$ gives $|7|=7>5$ . So the endpoints are right.

Another way: Think of distance: $|ax+3|=|a|\cdot|x+\frac{3}{a}|$ , so the solutions form an interval centered at $-\frac{3}{a}$ with half-width $\frac{b}{|a|}$ . The interval $-4\le x\le 1$ has center $-1\frac{1}{2}$ and half-width $2\frac{1}{2}$ , so $-\frac{3}{a}=-1\frac{1}{2}$ gives $a=2$ and $\frac{b}{2}=2\frac{1}{2}$ gives $b=5$ .

Standardsmin grade 8

6.NS.C.7 Order and absolute value of rational numbers — Reading $|ax+3|\le b$ as a distance condition and rewriting it as a double inequality.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving and checking the linear inequalities in $x$.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the pair of linear equations for $a$ and $b$ obtained from the endpoints.

💡Takeaway. When you divide by a letter, ask first whether it is positive or negative.

Variant 7 medium

The values of $x$ that satisfy the inequality $|ax-4|\le b$ are exactly $1\le x\le 3$ . Find the integers $a$ and $b$ . (Assume $a\neq 0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Undo the absolute value to get a double inequality for $ax$ , divide by $a$ , and make the resulting interval equal $1\le x\le 3$ . Because dividing by $a$ may flip the inequalities, split into the cases $a>0$ and $a<0$ .

Givens

The solutions of $|ax-4|\le b$ are exactly $1\le x\le 3$ .
$a$ and $b$ are integers and $a\neq 0$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Dividing by a negative $a$ reverses both inequality signs.
Since the inequality has solutions, $b\ge 0$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Split on the sign of $a$ (two subproblems). In each case match the two endpoints, which gives two linear equations in $a$ and $b$ ; then keep only the case consistent with its sign assumption.

3 · Execute4 carry out the plan

1Remove the absolute value

#13 Convert to Algebra 6.NS.C.7

|ax-4|\le b

means

ax-4

lies between

-b

and

b

. Add

4

throughout.

-b\le ax-4\le b \;\Rightarrow\; -b+4\le ax\le b+4

An absolute value at most

b

means the distance from

0

is at most

b

.

2Case $a>0$

#7 Identify Subproblems 8.EE.C.8

Dividing by a positive

a

keeps the signs:

\frac{-b+4}{a}\le x\le\frac{b+4}{a}

. Matching endpoints gives

-b+4=a

and

b+4=3a

. Adding gives

8=4a

, so

a=2

, and then

b=-4+3a=2

.

-b+4=a,\; b+4=3a \;\Rightarrow\; a=2,\; b=-4+3\cdot 2=2

Here

a=2>0

and

b=2\ge 0

, so the case is consistent.

3Case $a<0$

#3 Eliminate Possibilities 8.EE.C.8

Dividing by a negative

a

flips both signs:

\frac{b+4}{a}\le x\le\frac{-b+4}{a}

. Matching endpoints gives

b+4=a

and

-b+4=3a

. Adding the two equations gives

8=4a

, so

a=2

, which contradicts

a<0

.

b+4=a,\; -b+4=3a \;\Rightarrow\; 8=4a \;\Rightarrow\; a=2

The only candidate breaks its own assumption, so this case is eliminated.

4Check

#13 Convert to Algebra 7.EE.B.4

Substitute

a=2

,

b=2

back into the original inequality.

|2x-4|\le 2 \;\Rightarrow\; -2\le 2x-4\le 2 \;\Rightarrow\; 1\le x\le 3

The solution set is exactly the given interval.

Answer:

a=2

,

b=2

4 · Reviewdoes it hold up?

With $a=2$ , $b=2$ : $x=1$ gives $|-2|=2\le 2$ and $x=3$ gives $|2|=2\le 2$ , both on the boundary, while $x=4$ gives $|4|=4>2$ . So the endpoints are right.

Another way: Think of distance: $|ax-4|=|a|\cdot|x+\frac{-4}{a}|$ , so the solutions form an interval centered at $-\frac{-4}{a}$ with half-width $\frac{b}{|a|}$ . The interval $1\le x\le 3$ has center $2$ and half-width $1$ , so $-\frac{-4}{a}=2$ gives $a=2$ and $\frac{b}{2}=1$ gives $b=2$ .

Standardsmin grade 8

6.NS.C.7 Order and absolute value of rational numbers — Reading $|ax-4|\le b$ as a distance condition and rewriting it as a double inequality.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving and checking the linear inequalities in $x$.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the pair of linear equations for $a$ and $b$ obtained from the endpoints.

💡Takeaway. When you divide by a letter, ask first whether it is positive or negative.

Variant 8 hard

The values of $x$ that satisfy the inequality $|ax+3|\le b$ are exactly $-2\le x\le 5$ . Find the integers $a$ and $b$ . (Assume $a\neq 0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Undo the absolute value to get a double inequality for $ax$ , divide by $a$ , and make the resulting interval equal $-2\le x\le 5$ . Because dividing by $a$ may flip the inequalities, split into the cases $a>0$ and $a<0$ .

Givens

The solutions of $|ax+3|\le b$ are exactly $-2\le x\le 5$ .
$a$ and $b$ are integers and $a\neq 0$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Dividing by a negative $a$ reverses both inequality signs.
Since the inequality has solutions, $b\ge 0$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Split on the sign of $a$ (two subproblems). In each case match the two endpoints, which gives two linear equations in $a$ and $b$ ; then keep only the case consistent with its sign assumption.

3 · Execute4 carry out the plan

1Remove the absolute value

#13 Convert to Algebra 6.NS.C.7

|ax+3|\le b

means

ax+3

lies between

-b

and

b

. Subtract

3

throughout.

-b\le ax+3\le b \;\Rightarrow\; -b-3\le ax\le b-3

An absolute value at most

b

means the distance from

0

is at most

b

.

2Case $a>0$

#3 Eliminate Possibilities 8.EE.C.8

Dividing by a positive

a

keeps the signs:

\frac{-b-3}{a}\le x\le\frac{b-3}{a}

. Matching endpoints gives

-b-3=-2a

and

b-3=5a

. Adding the two equations gives

-6=3a

, so

a=-2

, which contradicts

a>0

.

-b-3=-2a,\; b-3=5a \;\Rightarrow\; -6=3a \;\Rightarrow\; a=-2

The only candidate breaks its own assumption, so this case is eliminated.

3Case $a<0$

#7 Identify Subproblems 8.EE.C.8

Dividing by a negative

a

flips both signs:

\frac{b-3}{a}\le x\le\frac{-b-3}{a}

. Matching endpoints gives

b-3=-2a

and

-b-3=5a

. Adding gives

-6=3a

, so

a=-2

, and then

b=3-2a=7

.

b-3=-2a,\; -b-3=5a \;\Rightarrow\; a=-2,\; b=3-2(-2)=7

Here

a=-2<0

and

b=7\ge 0

, so the case is consistent.

4Check

#13 Convert to Algebra 7.EE.B.4

Substitute

a=-2

,

b=7

back into the original inequality.

|-2x+3|\le 7 \;\Rightarrow\; -7\le -2x+3\le 7 \;\Rightarrow\; -2\le x\le 5

The solution set is exactly the given interval.

Answer:

a=-2

,

b=7

4 · Reviewdoes it hold up?

With $a=-2$ , $b=7$ : $x=-2$ gives $|7|=7\le 7$ and $x=5$ gives $|-7|=7\le 7$ , both on the boundary, while $x=6$ gives $|-9|=9>7$ . So the endpoints are right.

Another way: Think of distance: $|ax+3|=|a|\cdot|x+\frac{3}{a}|$ , so the solutions form an interval centered at $-\frac{3}{a}$ with half-width $\frac{b}{|a|}$ . The interval $-2\le x\le 5$ has center $1\frac{1}{2}$ and half-width $3\frac{1}{2}$ , so $-\frac{3}{a}=1\frac{1}{2}$ gives $a=-2$ and $\frac{b}{2}=3\frac{1}{2}$ gives $b=7$ .

Standardsmin grade 8

6.NS.C.7 Order and absolute value of rational numbers — Reading $|ax+3|\le b$ as a distance condition and rewriting it as a double inequality.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving and checking the linear inequalities in $x$.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the pair of linear equations for $a$ and $b$ obtained from the endpoints.

💡Takeaway. When you divide by a letter, ask first whether it is positive or negative.

Variant 9 hard

The values of $x$ that satisfy the inequality $|ax-3|\le b$ are exactly $-2\le x\le 5$ . Find the integers $a$ and $b$ . (Assume $a\neq 0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Undo the absolute value to get a double inequality for $ax$ , divide by $a$ , and make the resulting interval equal $-2\le x\le 5$ . Because dividing by $a$ may flip the inequalities, split into the cases $a>0$ and $a<0$ .

Givens

The solutions of $|ax-3|\le b$ are exactly $-2\le x\le 5$ .
$a$ and $b$ are integers and $a\neq 0$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Dividing by a negative $a$ reverses both inequality signs.
Since the inequality has solutions, $b\ge 0$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Split on the sign of $a$ (two subproblems). In each case match the two endpoints, which gives two linear equations in $a$ and $b$ ; then keep only the case consistent with its sign assumption.

3 · Execute4 carry out the plan

1Remove the absolute value

#13 Convert to Algebra 6.NS.C.7

|ax-3|\le b

means

ax-3

lies between

-b

and

b

. Add

3

throughout.

-b\le ax-3\le b \;\Rightarrow\; -b+3\le ax\le b+3

An absolute value at most

b

means the distance from

0

is at most

b

.

2Case $a>0$

#7 Identify Subproblems 8.EE.C.8

Dividing by a positive

a

keeps the signs:

\frac{-b+3}{a}\le x\le\frac{b+3}{a}

. Matching endpoints gives

-b+3=-2a

and

b+3=5a

. Adding gives

6=3a

, so

a=2

, and then

b=-3+5a=7

.

-b+3=-2a,\; b+3=5a \;\Rightarrow\; a=2,\; b=-3+5\cdot 2=7

Here

a=2>0

and

b=7\ge 0

, so the case is consistent.

3Case $a<0$

#3 Eliminate Possibilities 8.EE.C.8

Dividing by a negative

a

flips both signs:

\frac{b+3}{a}\le x\le\frac{-b+3}{a}

. Matching endpoints gives

b+3=-2a

and

-b+3=5a

. Adding the two equations gives

6=3a

, so

a=2

, which contradicts

a<0

.

b+3=-2a,\; -b+3=5a \;\Rightarrow\; 6=3a \;\Rightarrow\; a=2

The only candidate breaks its own assumption, so this case is eliminated.

4Check

#13 Convert to Algebra 7.EE.B.4

Substitute

a=2

,

b=7

back into the original inequality.

|2x-3|\le 7 \;\Rightarrow\; -7\le 2x-3\le 7 \;\Rightarrow\; -2\le x\le 5

The solution set is exactly the given interval.

Answer:

a=2

,

b=7

4 · Reviewdoes it hold up?

With $a=2$ , $b=7$ : $x=-2$ gives $|-7|=7\le 7$ and $x=5$ gives $|7|=7\le 7$ , both on the boundary, while $x=6$ gives $|9|=9>7$ . So the endpoints are right.

Another way: Think of distance: $|ax-3|=|a|\cdot|x+\frac{-3}{a}|$ , so the solutions form an interval centered at $-\frac{-3}{a}$ with half-width $\frac{b}{|a|}$ . The interval $-2\le x\le 5$ has center $1\frac{1}{2}$ and half-width $3\frac{1}{2}$ , so $-\frac{-3}{a}=1\frac{1}{2}$ gives $a=2$ and $\frac{b}{2}=3\frac{1}{2}$ gives $b=7$ .

Standardsmin grade 8

6.NS.C.7 Order and absolute value of rational numbers — Reading $|ax-3|\le b$ as a distance condition and rewriting it as a double inequality.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving and checking the linear inequalities in $x$.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the pair of linear equations for $a$ and $b$ obtained from the endpoints.

💡Takeaway. When you divide by a letter, ask first whether it is positive or negative.

Variant 10 hard

The values of $x$ that satisfy the inequality $|ax-2|\le b$ are exactly $-3\le x\le 5$ . Find the integers $a$ and $b$ . (Assume $a\neq 0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Undo the absolute value to get a double inequality for $ax$ , divide by $a$ , and make the resulting interval equal $-3\le x\le 5$ . Because dividing by $a$ may flip the inequalities, split into the cases $a>0$ and $a<0$ .

Givens

The solutions of $|ax-2|\le b$ are exactly $-3\le x\le 5$ .
$a$ and $b$ are integers and $a\neq 0$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Dividing by a negative $a$ reverses both inequality signs.
Since the inequality has solutions, $b\ge 0$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Split on the sign of $a$ (two subproblems). In each case match the two endpoints, which gives two linear equations in $a$ and $b$ ; then keep only the case consistent with its sign assumption.

3 · Execute4 carry out the plan

1Remove the absolute value

#13 Convert to Algebra 6.NS.C.7

|ax-2|\le b

means

ax-2

lies between

-b

and

b

. Add

2

throughout.

-b\le ax-2\le b \;\Rightarrow\; -b+2\le ax\le b+2

An absolute value at most

b

means the distance from

0

is at most

b

.

2Case $a>0$

#7 Identify Subproblems 8.EE.C.8

Dividing by a positive

a

keeps the signs:

\frac{-b+2}{a}\le x\le\frac{b+2}{a}

. Matching endpoints gives

-b+2=-3a

and

b+2=5a

. Adding gives

4=2a

, so

a=2

, and then

b=-2+5a=8

.

-b+2=-3a,\; b+2=5a \;\Rightarrow\; a=2,\; b=-2+5\cdot 2=8

Here

a=2>0

and

b=8\ge 0

, so the case is consistent.

3Case $a<0$

#3 Eliminate Possibilities 8.EE.C.8

Dividing by a negative

a

flips both signs:

\frac{b+2}{a}\le x\le\frac{-b+2}{a}

. Matching endpoints gives

b+2=-3a

and

-b+2=5a

. Adding the two equations gives

4=2a

, so

a=2

, which contradicts

a<0

.

b+2=-3a,\; -b+2=5a \;\Rightarrow\; 4=2a \;\Rightarrow\; a=2

The only candidate breaks its own assumption, so this case is eliminated.

4Check

#13 Convert to Algebra 7.EE.B.4

Substitute

a=2

,

b=8

back into the original inequality.

|2x-2|\le 8 \;\Rightarrow\; -8\le 2x-2\le 8 \;\Rightarrow\; -3\le x\le 5

The solution set is exactly the given interval.

Answer:

a=2

,

b=8

4 · Reviewdoes it hold up?

With $a=2$ , $b=8$ : $x=-3$ gives $|-8|=8\le 8$ and $x=5$ gives $|8|=8\le 8$ , both on the boundary, while $x=6$ gives $|10|=10>8$ . So the endpoints are right.

Another way: Think of distance: $|ax-2|=|a|\cdot|x+\frac{-2}{a}|$ , so the solutions form an interval centered at $-\frac{-2}{a}$ with half-width $\frac{b}{|a|}$ . The interval $-3\le x\le 5$ has center $1$ and half-width $4$ , so $-\frac{-2}{a}=1$ gives $a=2$ and $\frac{b}{2}=4$ gives $b=8$ .

Standardsmin grade 8

6.NS.C.7 Order and absolute value of rational numbers — Reading $|ax-2|\le b$ as a distance condition and rewriting it as a double inequality.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving and checking the linear inequalities in $x$.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the pair of linear equations for $a$ and $b$ obtained from the endpoints.

💡Takeaway. When you divide by a letter, ask first whether it is positive or negative.

Variant 11 medium

The values of $x$ that satisfy the inequality $|ax+2|\le b$ are exactly $-1\le x\le 5$ . Find the integers $a$ and $b$ . (Assume $a\neq 0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Undo the absolute value to get a double inequality for $ax$ , divide by $a$ , and make the resulting interval equal $-1\le x\le 5$ . Because dividing by $a$ may flip the inequalities, split into the cases $a>0$ and $a<0$ .

Givens

The solutions of $|ax+2|\le b$ are exactly $-1\le x\le 5$ .
$a$ and $b$ are integers and $a\neq 0$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Dividing by a negative $a$ reverses both inequality signs.
Since the inequality has solutions, $b\ge 0$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Split on the sign of $a$ (two subproblems). In each case match the two endpoints, which gives two linear equations in $a$ and $b$ ; then keep only the case consistent with its sign assumption.

3 · Execute4 carry out the plan

1Remove the absolute value

#13 Convert to Algebra 6.NS.C.7

|ax+2|\le b

means

ax+2

lies between

-b

and

b

. Subtract

2

throughout.

-b\le ax+2\le b \;\Rightarrow\; -b-2\le ax\le b-2

An absolute value at most

b

means the distance from

0

is at most

b

.

2Case $a>0$

#3 Eliminate Possibilities 8.EE.C.8

Dividing by a positive

a

keeps the signs:

\frac{-b-2}{a}\le x\le\frac{b-2}{a}

. Matching endpoints gives

-b-2=-a

and

b-2=5a

. Adding the two equations gives

-4=4a

, so

a=-1

, which contradicts

a>0

.

-b-2=-a,\; b-2=5a \;\Rightarrow\; -4=4a \;\Rightarrow\; a=-1

The only candidate breaks its own assumption, so this case is eliminated.

3Case $a<0$

#7 Identify Subproblems 8.EE.C.8

Dividing by a negative

a

flips both signs:

\frac{b-2}{a}\le x\le\frac{-b-2}{a}

. Matching endpoints gives

b-2=-a

and

-b-2=5a

. Adding gives

-4=4a

, so

a=-1

, and then

b=2-a=3

.

b-2=-a,\; -b-2=5a \;\Rightarrow\; a=-1,\; b=2-1(-1)=3

Here

a=-1<0

and

b=3\ge 0

, so the case is consistent.

4Check

#13 Convert to Algebra 7.EE.B.4

Substitute

a=-1

,

b=3

back into the original inequality.

|-x+2|\le 3 \;\Rightarrow\; -3\le -x+2\le 3 \;\Rightarrow\; -1\le x\le 5

The solution set is exactly the given interval.

Answer:

a=-1

,

b=3

4 · Reviewdoes it hold up?

With $a=-1$ , $b=3$ : $x=-1$ gives $|3|=3\le 3$ and $x=5$ gives $|-3|=3\le 3$ , both on the boundary, while $x=6$ gives $|-4|=4>3$ . So the endpoints are right.

Another way: Think of distance: $|ax+2|=|a|\cdot|x+\frac{2}{a}|$ , so the solutions form an interval centered at $-\frac{2}{a}$ with half-width $\frac{b}{|a|}$ . The interval $-1\le x\le 5$ has center $2$ and half-width $3$ , so $-\frac{2}{a}=2$ gives $a=-1$ and $\frac{b}{1}=3$ gives $b=3$ .

Standardsmin grade 8

6.NS.C.7 Order and absolute value of rational numbers — Reading $|ax+2|\le b$ as a distance condition and rewriting it as a double inequality.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving and checking the linear inequalities in $x$.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the pair of linear equations for $a$ and $b$ obtained from the endpoints.

💡Takeaway. When you divide by a letter, ask first whether it is positive or negative.

Variant 12 hard

The values of $x$ that satisfy the inequality $|ax+6|\le b$ are exactly $-5\le x\le 1$ . Find the integers $a$ and $b$ . (Assume $a\neq 0$ .)

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Undo the absolute value to get a double inequality for $ax$ , divide by $a$ , and make the resulting interval equal $-5\le x\le 1$ . Because dividing by $a$ may flip the inequalities, split into the cases $a>0$ and $a<0$ .

Givens

The solutions of $|ax+6|\le b$ are exactly $-5\le x\le 1$ .
$a$ and $b$ are integers and $a\neq 0$ .

Unknowns

The values of $a$ and $b$ .

Constraints

Dividing by a negative $a$ reverses both inequality signs.
Since the inequality has solutions, $b\ge 0$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#3 Eliminate Possibilities

Split on the sign of $a$ (two subproblems). In each case match the two endpoints, which gives two linear equations in $a$ and $b$ ; then keep only the case consistent with its sign assumption.

3 · Execute4 carry out the plan

1Remove the absolute value

#13 Convert to Algebra 6.NS.C.7

|ax+6|\le b

means

ax+6

lies between

-b

and

b

. Subtract

6

throughout.

-b\le ax+6\le b \;\Rightarrow\; -b-6\le ax\le b-6

An absolute value at most

b

means the distance from

0

is at most

b

.

2Case $a>0$

#7 Identify Subproblems 8.EE.C.8

Dividing by a positive

a

keeps the signs:

\frac{-b-6}{a}\le x\le\frac{b-6}{a}

. Matching endpoints gives

-b-6=-5a

and

b-6=a

. Adding gives

-12=-4a

, so

a=3

, and then

b=6+a=9

.

-b-6=-5a,\; b-6=a \;\Rightarrow\; a=3,\; b=6+1\cdot 3=9

Here

a=3>0

and

b=9\ge 0

, so the case is consistent.

3Case $a<0$

#3 Eliminate Possibilities 8.EE.C.8

Dividing by a negative

a

flips both signs:

\frac{b-6}{a}\le x\le\frac{-b-6}{a}

. Matching endpoints gives

b-6=-5a

and

-b-6=a

. Adding the two equations gives

-12=-4a

, so

a=3

, which contradicts

a<0

.

b-6=-5a,\; -b-6=a \;\Rightarrow\; -12=-4a \;\Rightarrow\; a=3

The only candidate breaks its own assumption, so this case is eliminated.

4Check

#13 Convert to Algebra 7.EE.B.4

Substitute

a=3

,

b=9

back into the original inequality.

|3x+6|\le 9 \;\Rightarrow\; -9\le 3x+6\le 9 \;\Rightarrow\; -5\le x\le 1

The solution set is exactly the given interval.

Answer:

a=3

,

b=9

4 · Reviewdoes it hold up?

With $a=3$ , $b=9$ : $x=-5$ gives $|-9|=9\le 9$ and $x=1$ gives $|9|=9\le 9$ , both on the boundary, while $x=2$ gives $|12|=12>9$ . So the endpoints are right.

Another way: Think of distance: $|ax+6|=|a|\cdot|x+\frac{6}{a}|$ , so the solutions form an interval centered at $-\frac{6}{a}$ with half-width $\frac{b}{|a|}$ . The interval $-5\le x\le 1$ has center $-2$ and half-width $3$ , so $-\frac{6}{a}=-2$ gives $a=3$ and $\frac{b}{3}=3$ gives $b=9$ .

Standardsmin grade 8

6.NS.C.7 Order and absolute value of rational numbers — Reading $|ax+6|\le b$ as a distance condition and rewriting it as a double inequality.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving and checking the linear inequalities in $x$.
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the pair of linear equations for $a$ and $b$ obtained from the endpoints.

💡Takeaway. When you divide by a letter, ask first whether it is positive or negative.

Match interval endpoints, splitting on the sign · 12 practice problems

Generated variants — 12

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Where next?