Practice · No solution forces a zero coefficient · Sensim Math

Variant 1 easy

The inequality $ax+1<-3x+b$ in $x$ has no solution. Find the greatest possible value of $a+b$ . ( $a$ and $b$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Gather the $x$ -terms of $ax+1<-3x+b$ on one side. Decide what the coefficients must be so that no value of $x$ works, then make $a+b$ as large as possible.

Givens

$ax+1<-3x+b$ has no solution.
$a$ and $b$ are constants.

Unknowns

The greatest possible value of $a+b$ .

Constraints

Every real number $x$ must make the inequality false.

2 · Planchoose the strategy

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

Write the inequality as (number) $\cdot x<$ (number). A nonzero coefficient always leaves a whole ray of solutions, so rule that out; then the inequality has no $x$ left in it and is simply true or false.

3 · Execute4 carry out the plan

1Collect the $x$ -terms

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

Add

3x

and subtract

1

on both sides.

ax+1<-3x+b \;\Rightarrow\; (a+3)x<b-1

Now the coefficient of

x

is

a+3

and the constant is

b-1

.

2A nonzero coefficient always gives solutions

#3 Eliminate Possibilities 7.EE.B.4

If

a+3\neq 0

, dividing by

a+3

gives

x<\frac{b-1}{a+3}

(when

a+3>0

) or

x>\frac{b-1}{a+3}

(when

a+3<0

). Either way there are solutions, so this case is impossible.

a+3=0

Dividing by a positive or a negative number only changes the direction of the ray; it never makes the ray disappear.

3Make the leftover statement false

#7 Identify Subproblems 8.EE.C.7

With

a=-3

the inequality reads

0\cdot x<b-1

, that is

0<b-1

, for every

x

. It has no solution exactly when this statement is false.

0<b-1 \text{ is false} \;\Leftrightarrow\; b-1\le 0 \;\Leftrightarrow\; b\le 1

Like a linear equation with no solution, the

x

vanishes and only a false number statement is left.

4Maximize $a+b$

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

Since

a=-3

and

b\le 1

, the sum is at most

-3+1

.

a+b=-3+b\le -3+1=-2

The largest allowed

b

is

1

, giving

a+b=-2

.

Answer:

-2

4 · Reviewdoes it hold up?

Try $a=-3$ , $b=1$ : the inequality becomes $-3x+1<-3x+1$ , i.e. $1<1$ , false for every $x$ , so there is indeed no solution. With $b=1.5$ it would read $1<1.5$ , true for every $x$ , so $b$ cannot exceed $1$ .

Another way: Picture the two lines $y=ax+1$ and $y=-3x+b$ . The left side is never below the right only if the lines are parallel ( $a=-3$ ) and the first line lies on or above the second ( $1\ge b$ ); that gives the same maximum $-3+1=-2$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Rearranging the inequality and dividing by its coefficient.
8.EE.C.7 Solve linear equations in one variable — Recognizing the case where $x$ cancels and only a true or false statement remains.

💡Takeaway. "No solution" is a strong clue: the

x

has to cancel out completely.

Variant 2 easy

The inequality $ax-3<2x+b$ in $x$ has no solution. Find the greatest possible value of $a+b$ . ( $a$ and $b$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Gather the $x$ -terms of $ax-3<2x+b$ on one side. Decide what the coefficients must be so that no value of $x$ works, then make $a+b$ as large as possible.

Givens

$ax-3<2x+b$ has no solution.
$a$ and $b$ are constants.

Unknowns

The greatest possible value of $a+b$ .

Constraints

Every real number $x$ must make the inequality false.

2 · Planchoose the strategy

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

Write the inequality as (number) $\cdot x<$ (number). A nonzero coefficient always leaves a whole ray of solutions, so rule that out; then the inequality has no $x$ left in it and is simply true or false.

3 · Execute4 carry out the plan

1Collect the $x$ -terms

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

Subtract

2x

and add

3

on both sides.

ax-3<2x+b \;\Rightarrow\; (a-2)x<b+3

Now the coefficient of

x

is

a-2

and the constant is

b+3

.

2A nonzero coefficient always gives solutions

#3 Eliminate Possibilities 7.EE.B.4

If

a-2\neq 0

, dividing by

a-2

gives

x<\frac{b+3}{a-2}

(when

a-2>0

) or

x>\frac{b+3}{a-2}

(when

a-2<0

). Either way there are solutions, so this case is impossible.

a-2=0

Dividing by a positive or a negative number only changes the direction of the ray; it never makes the ray disappear.

3Make the leftover statement false

#7 Identify Subproblems 8.EE.C.7

With

a=2

the inequality reads

0\cdot x<b+3

, that is

0<b+3

, for every

x

. It has no solution exactly when this statement is false.

0<b+3 \text{ is false} \;\Leftrightarrow\; b+3\le 0 \;\Leftrightarrow\; b\le -3

Like a linear equation with no solution, the

x

vanishes and only a false number statement is left.

4Maximize $a+b$

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

Since

a=2

and

b\le -3

, the sum is at most

2-3

.

a+b=2+b\le 2-3=-1

The largest allowed

b

is

-3

, giving

a+b=-1

.

Answer:

-1

4 · Reviewdoes it hold up?

Try $a=2$ , $b=-3$ : the inequality becomes $2x-3<2x-3$ , i.e. $-3<-3$ , false for every $x$ , so there is indeed no solution. With $b=-2.5$ it would read $-3<-2.5$ , true for every $x$ , so $b$ cannot exceed $-3$ .

Another way: Picture the two lines $y=ax-3$ and $y=2x+b$ . The left side is never below the right only if the lines are parallel ( $a=2$ ) and the first line lies on or above the second ( $-3\ge b$ ); that gives the same maximum $2-3=-1$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Rearranging the inequality and dividing by its coefficient.
8.EE.C.7 Solve linear equations in one variable — Recognizing the case where $x$ cancels and only a true or false statement remains.

💡Takeaway. "No solution" is a strong clue: the

x

has to cancel out completely.

Variant 3 easy

The inequality $ax+1<3x+b$ in $x$ has no solution. Find the greatest possible value of $a+b$ . ( $a$ and $b$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Gather the $x$ -terms of $ax+1<3x+b$ on one side. Decide what the coefficients must be so that no value of $x$ works, then make $a+b$ as large as possible.

Givens

$ax+1<3x+b$ has no solution.
$a$ and $b$ are constants.

Unknowns

The greatest possible value of $a+b$ .

Constraints

Every real number $x$ must make the inequality false.

2 · Planchoose the strategy

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

Write the inequality as (number) $\cdot x<$ (number). A nonzero coefficient always leaves a whole ray of solutions, so rule that out; then the inequality has no $x$ left in it and is simply true or false.

3 · Execute4 carry out the plan

1Collect the $x$ -terms

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

Subtract

3x

and

1

from both sides.

ax+1<3x+b \;\Rightarrow\; (a-3)x<b-1

Now the coefficient of

x

is

a-3

and the constant is

b-1

.

2A nonzero coefficient always gives solutions

#3 Eliminate Possibilities 7.EE.B.4

If

a-3\neq 0

, dividing by

a-3

gives

x<\frac{b-1}{a-3}

(when

a-3>0

) or

x>\frac{b-1}{a-3}

(when

a-3<0

). Either way there are solutions, so this case is impossible.

a-3=0

Dividing by a positive or a negative number only changes the direction of the ray; it never makes the ray disappear.

3Make the leftover statement false

#7 Identify Subproblems 8.EE.C.7

With

a=3

the inequality reads

0\cdot x<b-1

, that is

0<b-1

, for every

x

. It has no solution exactly when this statement is false.

0<b-1 \text{ is false} \;\Leftrightarrow\; b-1\le 0 \;\Leftrightarrow\; b\le 1

Like a linear equation with no solution, the

x

vanishes and only a false number statement is left.

4Maximize $a+b$

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

Since

a=3

and

b\le 1

, the sum is at most

3+1

.

a+b=3+b\le 3+1=4

The largest allowed

b

is

1

, giving

a+b=4

.

Answer:

4

4 · Reviewdoes it hold up?

Try $a=3$ , $b=1$ : the inequality becomes $3x+1<3x+1$ , i.e. $1<1$ , false for every $x$ , so there is indeed no solution. With $b=1.5$ it would read $1<1.5$ , true for every $x$ , so $b$ cannot exceed $1$ .

Another way: Picture the two lines $y=ax+1$ and $y=3x+b$ . The left side is never below the right only if the lines are parallel ( $a=3$ ) and the first line lies on or above the second ( $1\ge b$ ); that gives the same maximum $3+1=4$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Rearranging the inequality and dividing by its coefficient.
8.EE.C.7 Solve linear equations in one variable — Recognizing the case where $x$ cancels and only a true or false statement remains.

💡Takeaway. "No solution" is a strong clue: the

x

has to cancel out completely.

Variant 4 easy

The inequality $ax+3<-2x+b$ in $x$ has no solution. Find the greatest possible value of $a+b$ . ( $a$ and $b$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Gather the $x$ -terms of $ax+3<-2x+b$ on one side. Decide what the coefficients must be so that no value of $x$ works, then make $a+b$ as large as possible.

Givens

$ax+3<-2x+b$ has no solution.
$a$ and $b$ are constants.

Unknowns

The greatest possible value of $a+b$ .

Constraints

Every real number $x$ must make the inequality false.

2 · Planchoose the strategy

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

Write the inequality as (number) $\cdot x<$ (number). A nonzero coefficient always leaves a whole ray of solutions, so rule that out; then the inequality has no $x$ left in it and is simply true or false.

3 · Execute4 carry out the plan

1Collect the $x$ -terms

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

Add

2x

and subtract

3

on both sides.

ax+3<-2x+b \;\Rightarrow\; (a+2)x<b-3

Now the coefficient of

x

is

a+2

and the constant is

b-3

.

2A nonzero coefficient always gives solutions

#3 Eliminate Possibilities 7.EE.B.4

If

a+2\neq 0

, dividing by

a+2

gives

x<\frac{b-3}{a+2}

(when

a+2>0

) or

x>\frac{b-3}{a+2}

(when

a+2<0

). Either way there are solutions, so this case is impossible.

a+2=0

Dividing by a positive or a negative number only changes the direction of the ray; it never makes the ray disappear.

3Make the leftover statement false

#7 Identify Subproblems 8.EE.C.7

With

a=-2

the inequality reads

0\cdot x<b-3

, that is

0<b-3

, for every

x

. It has no solution exactly when this statement is false.

0<b-3 \text{ is false} \;\Leftrightarrow\; b-3\le 0 \;\Leftrightarrow\; b\le 3

Like a linear equation with no solution, the

x

vanishes and only a false number statement is left.

4Maximize $a+b$

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

Since

a=-2

and

b\le 3

, the sum is at most

-2+3

.

a+b=-2+b\le -2+3=1

The largest allowed

b

is

3

, giving

a+b=1

.

Answer:

1

4 · Reviewdoes it hold up?

Try $a=-2$ , $b=3$ : the inequality becomes $-2x+3<-2x+3$ , i.e. $3<3$ , false for every $x$ , so there is indeed no solution. With $b=3.5$ it would read $3<3.5$ , true for every $x$ , so $b$ cannot exceed $3$ .

Another way: Picture the two lines $y=ax+3$ and $y=-2x+b$ . The left side is never below the right only if the lines are parallel ( $a=-2$ ) and the first line lies on or above the second ( $3\ge b$ ); that gives the same maximum $-2+3=1$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Rearranging the inequality and dividing by its coefficient.
8.EE.C.7 Solve linear equations in one variable — Recognizing the case where $x$ cancels and only a true or false statement remains.

💡Takeaway. "No solution" is a strong clue: the

x

has to cancel out completely.

Variant 5 medium

The inequality $ax-1<4x+b$ in $x$ has no solution. Find the greatest possible value of $a+b$ . ( $a$ and $b$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Gather the $x$ -terms of $ax-1<4x+b$ on one side. Decide what the coefficients must be so that no value of $x$ works, then make $a+b$ as large as possible.

Givens

$ax-1<4x+b$ has no solution.
$a$ and $b$ are constants.

Unknowns

The greatest possible value of $a+b$ .

Constraints

Every real number $x$ must make the inequality false.

2 · Planchoose the strategy

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

Write the inequality as (number) $\cdot x<$ (number). A nonzero coefficient always leaves a whole ray of solutions, so rule that out; then the inequality has no $x$ left in it and is simply true or false.

3 · Execute4 carry out the plan

1Collect the $x$ -terms

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

Subtract

4x

and add

1

on both sides.

ax-1<4x+b \;\Rightarrow\; (a-4)x<b+1

Now the coefficient of

x

is

a-4

and the constant is

b+1

.

2A nonzero coefficient always gives solutions

#3 Eliminate Possibilities 7.EE.B.4

If

a-4\neq 0

, dividing by

a-4

gives

x<\frac{b+1}{a-4}

(when

a-4>0

) or

x>\frac{b+1}{a-4}

(when

a-4<0

). Either way there are solutions, so this case is impossible.

a-4=0

Dividing by a positive or a negative number only changes the direction of the ray; it never makes the ray disappear.

3Make the leftover statement false

#7 Identify Subproblems 8.EE.C.7

With

a=4

the inequality reads

0\cdot x<b+1

, that is

0<b+1

, for every

x

. It has no solution exactly when this statement is false.

0<b+1 \text{ is false} \;\Leftrightarrow\; b+1\le 0 \;\Leftrightarrow\; b\le -1

Like a linear equation with no solution, the

x

vanishes and only a false number statement is left.

4Maximize $a+b$

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

Since

a=4

and

b\le -1

, the sum is at most

4-1

.

a+b=4+b\le 4-1=3

The largest allowed

b

is

-1

, giving

a+b=3

.

Answer:

3

4 · Reviewdoes it hold up?

Try $a=4$ , $b=-1$ : the inequality becomes $4x-1<4x-1$ , i.e. $-1<-1$ , false for every $x$ , so there is indeed no solution. With $b=-0.5$ it would read $-1<-0.5$ , true for every $x$ , so $b$ cannot exceed $-1$ .

Another way: Picture the two lines $y=ax-1$ and $y=4x+b$ . The left side is never below the right only if the lines are parallel ( $a=4$ ) and the first line lies on or above the second ( $-1\ge b$ ); that gives the same maximum $4-1=3$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Rearranging the inequality and dividing by its coefficient.
8.EE.C.7 Solve linear equations in one variable — Recognizing the case where $x$ cancels and only a true or false statement remains.

💡Takeaway. "No solution" is a strong clue: the

x

has to cancel out completely.

Variant 6 medium

The inequality $ax+5<2x+b$ in $x$ has no solution. Find the greatest possible value of $a+b$ . ( $a$ and $b$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Gather the $x$ -terms of $ax+5<2x+b$ on one side. Decide what the coefficients must be so that no value of $x$ works, then make $a+b$ as large as possible.

Givens

$ax+5<2x+b$ has no solution.
$a$ and $b$ are constants.

Unknowns

The greatest possible value of $a+b$ .

Constraints

Every real number $x$ must make the inequality false.

2 · Planchoose the strategy

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

Write the inequality as (number) $\cdot x<$ (number). A nonzero coefficient always leaves a whole ray of solutions, so rule that out; then the inequality has no $x$ left in it and is simply true or false.

3 · Execute4 carry out the plan

1Collect the $x$ -terms

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

Subtract

2x

and

5

from both sides.

ax+5<2x+b \;\Rightarrow\; (a-2)x<b-5

Now the coefficient of

x

is

a-2

and the constant is

b-5

.

2A nonzero coefficient always gives solutions

#3 Eliminate Possibilities 7.EE.B.4

If

a-2\neq 0

, dividing by

a-2

gives

x<\frac{b-5}{a-2}

(when

a-2>0

) or

x>\frac{b-5}{a-2}

(when

a-2<0

). Either way there are solutions, so this case is impossible.

a-2=0

Dividing by a positive or a negative number only changes the direction of the ray; it never makes the ray disappear.

3Make the leftover statement false

#7 Identify Subproblems 8.EE.C.7

With

a=2

the inequality reads

0\cdot x<b-5

, that is

0<b-5

, for every

x

. It has no solution exactly when this statement is false.

0<b-5 \text{ is false} \;\Leftrightarrow\; b-5\le 0 \;\Leftrightarrow\; b\le 5

Like a linear equation with no solution, the

x

vanishes and only a false number statement is left.

4Maximize $a+b$

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

Since

a=2

and

b\le 5

, the sum is at most

2+5

.

a+b=2+b\le 2+5=7

The largest allowed

b

is

5

, giving

a+b=7

.

Answer:

7

4 · Reviewdoes it hold up?

Try $a=2$ , $b=5$ : the inequality becomes $2x+5<2x+5$ , i.e. $5<5$ , false for every $x$ , so there is indeed no solution. With $b=5.5$ it would read $5<5.5$ , true for every $x$ , so $b$ cannot exceed $5$ .

Another way: Picture the two lines $y=ax+5$ and $y=2x+b$ . The left side is never below the right only if the lines are parallel ( $a=2$ ) and the first line lies on or above the second ( $5\ge b$ ); that gives the same maximum $2+5=7$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Rearranging the inequality and dividing by its coefficient.
8.EE.C.7 Solve linear equations in one variable — Recognizing the case where $x$ cancels and only a true or false statement remains.

💡Takeaway. "No solution" is a strong clue: the

x

has to cancel out completely.

Variant 7 medium

The inequality $ax+2<5x+b$ in $x$ has no solution. Find the greatest possible value of $a+b$ . ( $a$ and $b$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Gather the $x$ -terms of $ax+2<5x+b$ on one side. Decide what the coefficients must be so that no value of $x$ works, then make $a+b$ as large as possible.

Givens

$ax+2<5x+b$ has no solution.
$a$ and $b$ are constants.

Unknowns

The greatest possible value of $a+b$ .

Constraints

Every real number $x$ must make the inequality false.

2 · Planchoose the strategy

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

Write the inequality as (number) $\cdot x<$ (number). A nonzero coefficient always leaves a whole ray of solutions, so rule that out; then the inequality has no $x$ left in it and is simply true or false.

3 · Execute4 carry out the plan

1Collect the $x$ -terms

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

Subtract

5x

and

2

from both sides.

ax+2<5x+b \;\Rightarrow\; (a-5)x<b-2

Now the coefficient of

x

is

a-5

and the constant is

b-2

.

2A nonzero coefficient always gives solutions

#3 Eliminate Possibilities 7.EE.B.4

If

a-5\neq 0

, dividing by

a-5

gives

x<\frac{b-2}{a-5}

(when

a-5>0

) or

x>\frac{b-2}{a-5}

(when

a-5<0

). Either way there are solutions, so this case is impossible.

a-5=0

Dividing by a positive or a negative number only changes the direction of the ray; it never makes the ray disappear.

3Make the leftover statement false

#7 Identify Subproblems 8.EE.C.7

With

a=5

the inequality reads

0\cdot x<b-2

, that is

0<b-2

, for every

x

. It has no solution exactly when this statement is false.

0<b-2 \text{ is false} \;\Leftrightarrow\; b-2\le 0 \;\Leftrightarrow\; b\le 2

Like a linear equation with no solution, the

x

vanishes and only a false number statement is left.

4Maximize $a+b$

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

Since

a=5

and

b\le 2

, the sum is at most

5+2

.

a+b=5+b\le 5+2=7

The largest allowed

b

is

2

, giving

a+b=7

.

Answer:

7

4 · Reviewdoes it hold up?

Try $a=5$ , $b=2$ : the inequality becomes $5x+2<5x+2$ , i.e. $2<2$ , false for every $x$ , so there is indeed no solution. With $b=2.5$ it would read $2<2.5$ , true for every $x$ , so $b$ cannot exceed $2$ .

Another way: Picture the two lines $y=ax+2$ and $y=5x+b$ . The left side is never below the right only if the lines are parallel ( $a=5$ ) and the first line lies on or above the second ( $2\ge b$ ); that gives the same maximum $5+2=7$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Rearranging the inequality and dividing by its coefficient.
8.EE.C.7 Solve linear equations in one variable — Recognizing the case where $x$ cancels and only a true or false statement remains.

💡Takeaway. "No solution" is a strong clue: the

x

has to cancel out completely.

Variant 8 medium

The inequality $ax+2<-5x+b$ in $x$ has no solution. Find the greatest possible value of $a+b$ . ( $a$ and $b$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Gather the $x$ -terms of $ax+2<-5x+b$ on one side. Decide what the coefficients must be so that no value of $x$ works, then make $a+b$ as large as possible.

Givens

$ax+2<-5x+b$ has no solution.
$a$ and $b$ are constants.

Unknowns

The greatest possible value of $a+b$ .

Constraints

Every real number $x$ must make the inequality false.

2 · Planchoose the strategy

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

Write the inequality as (number) $\cdot x<$ (number). A nonzero coefficient always leaves a whole ray of solutions, so rule that out; then the inequality has no $x$ left in it and is simply true or false.

3 · Execute4 carry out the plan

1Collect the $x$ -terms

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

Add

5x

and subtract

2

on both sides.

ax+2<-5x+b \;\Rightarrow\; (a+5)x<b-2

Now the coefficient of

x

is

a+5

and the constant is

b-2

.

2A nonzero coefficient always gives solutions

#3 Eliminate Possibilities 7.EE.B.4

If

a+5\neq 0

, dividing by

a+5

gives

x<\frac{b-2}{a+5}

(when

a+5>0

) or

x>\frac{b-2}{a+5}

(when

a+5<0

). Either way there are solutions, so this case is impossible.

a+5=0

Dividing by a positive or a negative number only changes the direction of the ray; it never makes the ray disappear.

3Make the leftover statement false

#7 Identify Subproblems 8.EE.C.7

With

a=-5

the inequality reads

0\cdot x<b-2

, that is

0<b-2

, for every

x

. It has no solution exactly when this statement is false.

0<b-2 \text{ is false} \;\Leftrightarrow\; b-2\le 0 \;\Leftrightarrow\; b\le 2

Like a linear equation with no solution, the

x

vanishes and only a false number statement is left.

4Maximize $a+b$

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

Since

a=-5

and

b\le 2

, the sum is at most

-5+2

.

a+b=-5+b\le -5+2=-3

The largest allowed

b

is

2

, giving

a+b=-3

.

Answer:

-3

4 · Reviewdoes it hold up?

Try $a=-5$ , $b=2$ : the inequality becomes $-5x+2<-5x+2$ , i.e. $2<2$ , false for every $x$ , so there is indeed no solution. With $b=2.5$ it would read $2<2.5$ , true for every $x$ , so $b$ cannot exceed $2$ .

Another way: Picture the two lines $y=ax+2$ and $y=-5x+b$ . The left side is never below the right only if the lines are parallel ( $a=-5$ ) and the first line lies on or above the second ( $2\ge b$ ); that gives the same maximum $-5+2=-3$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Rearranging the inequality and dividing by its coefficient.
8.EE.C.7 Solve linear equations in one variable — Recognizing the case where $x$ cancels and only a true or false statement remains.

💡Takeaway. "No solution" is a strong clue: the

x

has to cancel out completely.

Variant 9 hard

The inequality $ax+6<4x+b$ in $x$ has no solution. Find the greatest possible value of $a+b$ . ( $a$ and $b$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Gather the $x$ -terms of $ax+6<4x+b$ on one side. Decide what the coefficients must be so that no value of $x$ works, then make $a+b$ as large as possible.

Givens

$ax+6<4x+b$ has no solution.
$a$ and $b$ are constants.

Unknowns

The greatest possible value of $a+b$ .

Constraints

Every real number $x$ must make the inequality false.

2 · Planchoose the strategy

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

Write the inequality as (number) $\cdot x<$ (number). A nonzero coefficient always leaves a whole ray of solutions, so rule that out; then the inequality has no $x$ left in it and is simply true or false.

3 · Execute4 carry out the plan

1Collect the $x$ -terms

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

Subtract

4x

and

6

from both sides.

ax+6<4x+b \;\Rightarrow\; (a-4)x<b-6

Now the coefficient of

x

is

a-4

and the constant is

b-6

.

2A nonzero coefficient always gives solutions

#3 Eliminate Possibilities 7.EE.B.4

If

a-4\neq 0

, dividing by

a-4

gives

x<\frac{b-6}{a-4}

(when

a-4>0

) or

x>\frac{b-6}{a-4}

(when

a-4<0

). Either way there are solutions, so this case is impossible.

a-4=0

Dividing by a positive or a negative number only changes the direction of the ray; it never makes the ray disappear.

3Make the leftover statement false

#7 Identify Subproblems 8.EE.C.7

With

a=4

the inequality reads

0\cdot x<b-6

, that is

0<b-6

, for every

x

. It has no solution exactly when this statement is false.

0<b-6 \text{ is false} \;\Leftrightarrow\; b-6\le 0 \;\Leftrightarrow\; b\le 6

Like a linear equation with no solution, the

x

vanishes and only a false number statement is left.

4Maximize $a+b$

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

Since

a=4

and

b\le 6

, the sum is at most

4+6

.

a+b=4+b\le 4+6=10

The largest allowed

b

is

6

, giving

a+b=10

.

Answer:

10

4 · Reviewdoes it hold up?

Try $a=4$ , $b=6$ : the inequality becomes $4x+6<4x+6$ , i.e. $6<6$ , false for every $x$ , so there is indeed no solution. With $b=6.5$ it would read $6<6.5$ , true for every $x$ , so $b$ cannot exceed $6$ .

Another way: Picture the two lines $y=ax+6$ and $y=4x+b$ . The left side is never below the right only if the lines are parallel ( $a=4$ ) and the first line lies on or above the second ( $6\ge b$ ); that gives the same maximum $4+6=10$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Rearranging the inequality and dividing by its coefficient.
8.EE.C.7 Solve linear equations in one variable — Recognizing the case where $x$ cancels and only a true or false statement remains.

💡Takeaway. "No solution" is a strong clue: the

x

has to cancel out completely.

Variant 10 hard

The inequality $ax-4<6x+b$ in $x$ has no solution. Find the greatest possible value of $a+b$ . ( $a$ and $b$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Gather the $x$ -terms of $ax-4<6x+b$ on one side. Decide what the coefficients must be so that no value of $x$ works, then make $a+b$ as large as possible.

Givens

$ax-4<6x+b$ has no solution.
$a$ and $b$ are constants.

Unknowns

The greatest possible value of $a+b$ .

Constraints

Every real number $x$ must make the inequality false.

2 · Planchoose the strategy

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

Write the inequality as (number) $\cdot x<$ (number). A nonzero coefficient always leaves a whole ray of solutions, so rule that out; then the inequality has no $x$ left in it and is simply true or false.

3 · Execute4 carry out the plan

1Collect the $x$ -terms

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

Subtract

6x

and add

4

on both sides.

ax-4<6x+b \;\Rightarrow\; (a-6)x<b+4

Now the coefficient of

x

is

a-6

and the constant is

b+4

.

2A nonzero coefficient always gives solutions

#3 Eliminate Possibilities 7.EE.B.4

If

a-6\neq 0

, dividing by

a-6

gives

x<\frac{b+4}{a-6}

(when

a-6>0

) or

x>\frac{b+4}{a-6}

(when

a-6<0

). Either way there are solutions, so this case is impossible.

a-6=0

Dividing by a positive or a negative number only changes the direction of the ray; it never makes the ray disappear.

3Make the leftover statement false

#7 Identify Subproblems 8.EE.C.7

With

a=6

the inequality reads

0\cdot x<b+4

, that is

0<b+4

, for every

x

. It has no solution exactly when this statement is false.

0<b+4 \text{ is false} \;\Leftrightarrow\; b+4\le 0 \;\Leftrightarrow\; b\le -4

Like a linear equation with no solution, the

x

vanishes and only a false number statement is left.

4Maximize $a+b$

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

Since

a=6

and

b\le -4

, the sum is at most

6-4

.

a+b=6+b\le 6-4=2

The largest allowed

b

is

-4

, giving

a+b=2

.

Answer:

2

4 · Reviewdoes it hold up?

Try $a=6$ , $b=-4$ : the inequality becomes $6x-4<6x-4$ , i.e. $-4<-4$ , false for every $x$ , so there is indeed no solution. With $b=-3.5$ it would read $-4<-3.5$ , true for every $x$ , so $b$ cannot exceed $-4$ .

Another way: Picture the two lines $y=ax-4$ and $y=6x+b$ . The left side is never below the right only if the lines are parallel ( $a=6$ ) and the first line lies on or above the second ( $-4\ge b$ ); that gives the same maximum $6-4=2$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Rearranging the inequality and dividing by its coefficient.
8.EE.C.7 Solve linear equations in one variable — Recognizing the case where $x$ cancels and only a true or false statement remains.

💡Takeaway. "No solution" is a strong clue: the

x

has to cancel out completely.

Variant 11 hard

The inequality $ax+4<7x+b$ in $x$ has no solution. Find the greatest possible value of $a+b$ . ( $a$ and $b$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Gather the $x$ -terms of $ax+4<7x+b$ on one side. Decide what the coefficients must be so that no value of $x$ works, then make $a+b$ as large as possible.

Givens

$ax+4<7x+b$ has no solution.
$a$ and $b$ are constants.

Unknowns

The greatest possible value of $a+b$ .

Constraints

Every real number $x$ must make the inequality false.

2 · Planchoose the strategy

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

Write the inequality as (number) $\cdot x<$ (number). A nonzero coefficient always leaves a whole ray of solutions, so rule that out; then the inequality has no $x$ left in it and is simply true or false.

3 · Execute4 carry out the plan

1Collect the $x$ -terms

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

Subtract

7x

and

4

from both sides.

ax+4<7x+b \;\Rightarrow\; (a-7)x<b-4

Now the coefficient of

x

is

a-7

and the constant is

b-4

.

2A nonzero coefficient always gives solutions

#3 Eliminate Possibilities 7.EE.B.4

If

a-7\neq 0

, dividing by

a-7

gives

x<\frac{b-4}{a-7}

(when

a-7>0

) or

x>\frac{b-4}{a-7}

(when

a-7<0

). Either way there are solutions, so this case is impossible.

a-7=0

Dividing by a positive or a negative number only changes the direction of the ray; it never makes the ray disappear.

3Make the leftover statement false

#7 Identify Subproblems 8.EE.C.7

With

a=7

the inequality reads

0\cdot x<b-4

, that is

0<b-4

, for every

x

. It has no solution exactly when this statement is false.

0<b-4 \text{ is false} \;\Leftrightarrow\; b-4\le 0 \;\Leftrightarrow\; b\le 4

Like a linear equation with no solution, the

x

vanishes and only a false number statement is left.

4Maximize $a+b$

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

Since

a=7

and

b\le 4

, the sum is at most

7+4

.

a+b=7+b\le 7+4=11

The largest allowed

b

is

4

, giving

a+b=11

.

Answer:

11

4 · Reviewdoes it hold up?

Try $a=7$ , $b=4$ : the inequality becomes $7x+4<7x+4$ , i.e. $4<4$ , false for every $x$ , so there is indeed no solution. With $b=4.5$ it would read $4<4.5$ , true for every $x$ , so $b$ cannot exceed $4$ .

Another way: Picture the two lines $y=ax+4$ and $y=7x+b$ . The left side is never below the right only if the lines are parallel ( $a=7$ ) and the first line lies on or above the second ( $4\ge b$ ); that gives the same maximum $7+4=11$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Rearranging the inequality and dividing by its coefficient.
8.EE.C.7 Solve linear equations in one variable — Recognizing the case where $x$ cancels and only a true or false statement remains.

💡Takeaway. "No solution" is a strong clue: the

x

has to cancel out completely.

Variant 12 hard

The inequality $ax-2<9x+b$ in $x$ has no solution. Find the greatest possible value of $a+b$ . ( $a$ and $b$ are constants.)

Your answer

Show solution

1 · Understandwhat's really being asked

Gather the $x$ -terms of $ax-2<9x+b$ on one side. Decide what the coefficients must be so that no value of $x$ works, then make $a+b$ as large as possible.

Givens

$ax-2<9x+b$ has no solution.
$a$ and $b$ are constants.

Unknowns

The greatest possible value of $a+b$ .

Constraints

Every real number $x$ must make the inequality false.

2 · Planchoose the strategy

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

Write the inequality as (number) $\cdot x<$ (number). A nonzero coefficient always leaves a whole ray of solutions, so rule that out; then the inequality has no $x$ left in it and is simply true or false.

3 · Execute4 carry out the plan

1Collect the $x$ -terms

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

Subtract

9x

and add

2

on both sides.

ax-2<9x+b \;\Rightarrow\; (a-9)x<b+2

Now the coefficient of

x

is

a-9

and the constant is

b+2

.

2A nonzero coefficient always gives solutions

#3 Eliminate Possibilities 7.EE.B.4

If

a-9\neq 0

, dividing by

a-9

gives

x<\frac{b+2}{a-9}

(when

a-9>0

) or

x>\frac{b+2}{a-9}

(when

a-9<0

). Either way there are solutions, so this case is impossible.

a-9=0

Dividing by a positive or a negative number only changes the direction of the ray; it never makes the ray disappear.

3Make the leftover statement false

#7 Identify Subproblems 8.EE.C.7

With

a=9

the inequality reads

0\cdot x<b+2

, that is

0<b+2

, for every

x

. It has no solution exactly when this statement is false.

0<b+2 \text{ is false} \;\Leftrightarrow\; b+2\le 0 \;\Leftrightarrow\; b\le -2

Like a linear equation with no solution, the

x

vanishes and only a false number statement is left.

4Maximize $a+b$

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

Since

a=9

and

b\le -2

, the sum is at most

9-2

.

a+b=9+b\le 9-2=7

The largest allowed

b

is

-2

, giving

a+b=7

.

Answer:

7

4 · Reviewdoes it hold up?

Try $a=9$ , $b=-2$ : the inequality becomes $9x-2<9x-2$ , i.e. $-2<-2$ , false for every $x$ , so there is indeed no solution. With $b=-1.5$ it would read $-2<-1.5$ , true for every $x$ , so $b$ cannot exceed $-2$ .

Another way: Picture the two lines $y=ax-2$ and $y=9x+b$ . The left side is never below the right only if the lines are parallel ( $a=9$ ) and the first line lies on or above the second ( $-2\ge b$ ); that gives the same maximum $9-2=7$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Rearranging the inequality and dividing by its coefficient.
8.EE.C.7 Solve linear equations in one variable — Recognizing the case where $x$ cancels and only a true or false statement remains.

💡Takeaway. "No solution" is a strong clue: the

x

has to cancel out completely.

No solution forces a zero coefficient · 12 practice problems

Generated variants — 12

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