Practice · ac=1 makes a and c same-sign reciprocals · Sensim Math

Variant 1 medium

Three rational numbers $a$ , $b$ , $c$ satisfy $bc=1$ , $a\times b\times c<0$ and $a+b+c>0$ . If $|b|<1$ , how are $a$ , $b$ and $c$ ordered? Write the order using inequality signs.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

$b$ and $c$ multiply to $1$ , the product $abc$ is negative, the sum $a + b + c$ is positive, and $b$ is less than $1$ in size. Put $a$ , $b$ , $c$ in order.

Givens

$bc = 1$
$a \times b \times c < 0$
$a + b + c > 0$
$|b| < 1$

Unknowns

The order of $a$ , $b$ , $c$ from least to greatest.

Constraints

Since $bc = 1$ , neither $b$ nor $c$ is $0$ , and $c$ is the reciprocal of $b$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

Work out signs before sizes. Because $bc = 1$ , the product $abc$ is just $a$ , which settles the sign of $a$ . The sum then forces $b$ and $c$ to be positive, and the reciprocal relation turns $|b| < 1$ into $c > 1$ .

3 · Execute5 carry out the plan

1b and c share a sign

#7 Identify Subproblems 7.NS.A.2

Their product

1

is positive, so they are both positive or both negative.

bc = 1 > 0

Only like signs give a positive product.

2The sign of a

#13 Convert to Algebra 7.NS.A.2

Group the product as

a \times (bc)

; since

bc = 1

, the whole product equals

a

.

a \times b \times c = a \times (bc) = a \times 1 = a < 0

abc < 0

means simply

a < 0

, so

a

is negative.

3The signs of b and c

#7 Identify Subproblems 6.NS.C.7

From

a + b + c > 0

we get

b + c > -a

, and

-a

is positive. So

b + c > 0

, which is impossible if

b

and

c

are both negative.

b + c > -a > 0 \Rightarrow b > 0,\ c > 0

So

b > 0

and

c > 0

.

4Compare b and c

#13 Convert to Algebra 7.NS.A.2

c

is the reciprocal of

b

. With

0 < b < 1

, dividing

1

by a number less than

1

gives more than

1

.

c = \frac{1}{b},\ 0 < b < 1 \Rightarrow c > 1

For example

b = \frac{1}{2}

gives

c = 2

.

5Put them in order

#7 Identify Subproblems 6.NS.C.7

a

is negative,

b

is between

0

and

1

, and

c

is greater than

1

.

a < 0 < b < 1 < c

So

a < b < c

.

Answer:

a<b<c

4 · Reviewdoes it hold up?

Try $b = \frac{1}{2}$ , $c = 2$ , $a = -1$ : $bc = 1$ , $abc = -1 < 0$ , $a + b + c = 1\frac{1}{2} > 0$ , $|b| < 1$ . The order is $-1 < \frac{1}{2} < 2$ , that is $a < b < c$ .

Another way: Check the two sign patterns allowed by $bc = 1$ . If $b, c < 0$ , then $abc$ still has the sign of $a$ , so $a < 0$ too, and three negatives cannot have a positive sum. So $b, c > 0$ and $a < 0$ , which leads to the same order.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Sign rules for products and the meaning of reciprocals ($bc = 1$).
6.NS.C.7 Order and absolute value of rational numbers — Ordering the rational numbers once their signs and sizes are known.

💡Takeaway. When two numbers multiply to

1

, they share a sign and one is big exactly when the other is small.

Variant 2 medium

Three rational numbers $a$ , $b$ , $c$ satisfy $ac=1$ , $a\times b\times c<0$ and $a+b+c>0$ . If $|c|>1$ , how are $a$ , $b$ and $c$ ordered? Write the order using inequality signs.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

$a$ and $c$ multiply to $1$ , the product $abc$ is negative, the sum $a + b + c$ is positive, and $c$ is more than $1$ in size. Put $a$ , $b$ , $c$ in order.

Givens

$ac = 1$
$a \times b \times c < 0$
$a + b + c > 0$
$|c| > 1$

Unknowns

The order of $a$ , $b$ , $c$ from least to greatest.

Constraints

Since $ac = 1$ , neither $a$ nor $c$ is $0$ , and $a$ is the reciprocal of $c$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

Work out signs before sizes. Because $ac = 1$ , the product $abc$ is just $b$ , which settles the sign of $b$ . The sum then forces $a$ and $c$ to be positive, and the reciprocal relation turns $|c| > 1$ into $0 < a < 1$ .

3 · Execute5 carry out the plan

1a and c share a sign

#7 Identify Subproblems 7.NS.A.2

Their product

1

is positive, so they are both positive or both negative.

ac = 1 > 0

Only like signs give a positive product.

2The sign of b

#13 Convert to Algebra 7.NS.A.2

Group the product as

b \times (ac)

; since

ac = 1

, the whole product equals

b

.

a \times b \times c = b \times (ac) = b \times 1 = b < 0

abc < 0

means simply

b < 0

, so

b

is negative.

3The signs of a and c

#7 Identify Subproblems 6.NS.C.7

From

a + b + c > 0

we get

a + c > -b

, and

-b

is positive. So

a + c > 0

, which is impossible if

a

and

c

are both negative.

a + c > -b > 0 \Rightarrow a > 0,\ c > 0

So

a > 0

and

c > 0

.

4Compare c and a

#13 Convert to Algebra 7.NS.A.2

a

is the reciprocal of

c

. With

c > 1

, dividing

1

by a number greater than

1

gives less than

1

, but still more than

0

.

a = \frac{1}{c},\ c > 1 \Rightarrow 0 < a < 1

For example

c = 2

gives

a = \frac{1}{2}

.

5Put them in order

#7 Identify Subproblems 6.NS.C.7

b

is negative,

a

is between

0

and

1

, and

c

is greater than

1

.

b < 0 < a < 1 < c

So

b < a < c

.

Answer:

b<a<c

4 · Reviewdoes it hold up?

Try $c = 2$ , $a = \frac{1}{2}$ , $b = -1$ : $ac = 1$ , $abc = -1 < 0$ , $a + b + c = 1\frac{1}{2} > 0$ , $|c| > 1$ . The order is $-1 < \frac{1}{2} < 2$ , that is $b < a < c$ .

Another way: Check the two sign patterns allowed by $ac = 1$ . If $a, c < 0$ , then $abc$ still has the sign of $b$ , so $b < 0$ too, and three negatives cannot have a positive sum. So $a, c > 0$ and $b < 0$ , which leads to the same order.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Sign rules for products and the meaning of reciprocals ($ac = 1$).
6.NS.C.7 Order and absolute value of rational numbers — Ordering the rational numbers once their signs and sizes are known.

💡Takeaway. When two numbers multiply to

1

, they share a sign and one is big exactly when the other is small.

Variant 3 medium

Three rational numbers $a$ , $b$ , $c$ satisfy $ab=1$ , $a\times b\times c>0$ and $a+b+c<0$ . If $|b|<1$ , how are $a$ , $b$ and $c$ ordered? Write the order using inequality signs.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

$a$ and $b$ multiply to $1$ , the product $abc$ is positive, the sum $a + b + c$ is negative, and $b$ is less than $1$ in size. Put $a$ , $b$ , $c$ in order.

Givens

$ab = 1$
$a \times b \times c > 0$
$a + b + c < 0$
$|b| < 1$

Unknowns

The order of $a$ , $b$ , $c$ from least to greatest.

Constraints

Since $ab = 1$ , neither $a$ nor $b$ is $0$ , and $a$ is the reciprocal of $b$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

Work out signs before sizes. Because $ab = 1$ , the product $abc$ is just $c$ , which settles the sign of $c$ . The sum then forces $a$ and $b$ to be negative, and the reciprocal relation turns $|b| < 1$ into $a < -1$ .

3 · Execute5 carry out the plan

1a and b share a sign

#7 Identify Subproblems 7.NS.A.2

Their product

1

is positive, so they are both positive or both negative.

ab = 1 > 0

Only like signs give a positive product.

2The sign of c

#13 Convert to Algebra 7.NS.A.2

Group the product as

c \times (ab)

; since

ab = 1

, the whole product equals

c

.

a \times b \times c = c \times (ab) = c \times 1 = c > 0

abc > 0

means simply

c > 0

, so

c

is positive.

3The signs of a and b

#7 Identify Subproblems 6.NS.C.7

From

a + b + c < 0

we get

a + b < -c

, and

-c

is negative. So

a + b < 0

, which is impossible if

a

and

b

are both positive.

a + b < -c < 0 \Rightarrow a < 0,\ b < 0

So

a < 0

and

b < 0

.

4Compare b and a

#13 Convert to Algebra 7.NS.A.2

a

is the reciprocal of

b

. With

-1 < b < 0

, the size

|a| = \frac{1}{|b|}

is more than

1

, and

a

is negative too.

a = \frac{1}{b},\ -1 < b < 0 \Rightarrow a < -1

For example

b = -\frac{1}{2}

gives

a = -2

.

5Put them in order

#7 Identify Subproblems 6.NS.C.7

a

is less than

-1

,

b

is between

-1

and

0

, and

c

is positive.

a < -1 < b < 0 < c

So

a < b < c

.

Answer:

a<b<c

4 · Reviewdoes it hold up?

Try $b = -\frac{1}{2}$ , $a = -2$ , $c = 1$ : $ab = 1$ , $abc = 1 > 0$ , $a + b + c = -1\frac{1}{2} < 0$ , $|b| < 1$ . The order is $-2 < -\frac{1}{2} < 1$ , that is $a < b < c$ .

Another way: Check the two sign patterns allowed by $ab = 1$ . If $a, b > 0$ , then $abc$ still has the sign of $c$ , so $c > 0$ too, and three positives cannot have a negative sum. So $a, b < 0$ and $c > 0$ , which leads to the same order.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Sign rules for products and the meaning of reciprocals ($ab = 1$).
6.NS.C.7 Order and absolute value of rational numbers — Ordering the rational numbers once their signs and sizes are known.

💡Takeaway. When two numbers multiply to

1

, they share a sign and one is big exactly when the other is small.

Variant 4 medium

Three rational numbers $a$ , $b$ , $c$ satisfy $ab=1$ , $a\times b\times c<0$ and $a+b+c>0$ . If $|a|<1$ , how are $a$ , $b$ and $c$ ordered? Write the order using inequality signs.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

$a$ and $b$ multiply to $1$ , the product $abc$ is negative, the sum $a + b + c$ is positive, and $a$ is less than $1$ in size. Put $a$ , $b$ , $c$ in order.

Givens

$ab = 1$
$a \times b \times c < 0$
$a + b + c > 0$
$|a| < 1$

Unknowns

The order of $a$ , $b$ , $c$ from least to greatest.

Constraints

Since $ab = 1$ , neither $a$ nor $b$ is $0$ , and $b$ is the reciprocal of $a$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

Work out signs before sizes. Because $ab = 1$ , the product $abc$ is just $c$ , which settles the sign of $c$ . The sum then forces $a$ and $b$ to be positive, and the reciprocal relation turns $|a| < 1$ into $b > 1$ .

3 · Execute5 carry out the plan

1a and b share a sign

#7 Identify Subproblems 7.NS.A.2

Their product

1

is positive, so they are both positive or both negative.

ab = 1 > 0

Only like signs give a positive product.

2The sign of c

#13 Convert to Algebra 7.NS.A.2

Group the product as

c \times (ab)

; since

ab = 1

, the whole product equals

c

.

a \times b \times c = c \times (ab) = c \times 1 = c < 0

abc < 0

means simply

c < 0

, so

c

is negative.

3The signs of a and b

#7 Identify Subproblems 6.NS.C.7

From

a + b + c > 0

we get

a + b > -c

, and

-c

is positive. So

a + b > 0

, which is impossible if

a

and

b

are both negative.

a + b > -c > 0 \Rightarrow a > 0,\ b > 0

So

a > 0

and

b > 0

.

4Compare a and b

#13 Convert to Algebra 7.NS.A.2

b

is the reciprocal of

a

. With

0 < a < 1

, dividing

1

by a number less than

1

gives more than

1

.

b = \frac{1}{a},\ 0 < a < 1 \Rightarrow b > 1

For example

a = \frac{1}{2}

gives

b = 2

.

5Put them in order

#7 Identify Subproblems 6.NS.C.7

c

is negative,

a

is between

0

and

1

, and

b

is greater than

1

.

c < 0 < a < 1 < b

So

c < a < b

.

Answer:

c<a<b

4 · Reviewdoes it hold up?

Try $a = \frac{1}{2}$ , $b = 2$ , $c = -1$ : $ab = 1$ , $abc = -1 < 0$ , $a + b + c = 1\frac{1}{2} > 0$ , $|a| < 1$ . The order is $-1 < \frac{1}{2} < 2$ , that is $c < a < b$ .

Another way: Check the two sign patterns allowed by $ab = 1$ . If $a, b < 0$ , then $abc$ still has the sign of $c$ , so $c < 0$ too, and three negatives cannot have a positive sum. So $a, b > 0$ and $c < 0$ , which leads to the same order.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Sign rules for products and the meaning of reciprocals ($ab = 1$).
6.NS.C.7 Order and absolute value of rational numbers — Ordering the rational numbers once their signs and sizes are known.

💡Takeaway. When two numbers multiply to

1

, they share a sign and one is big exactly when the other is small.

Variant 5 medium

Three rational numbers $a$ , $b$ , $c$ satisfy $bc=1$ , $a\times b\times c>0$ and $a+b+c<0$ . If $|b|>1$ , how are $a$ , $b$ and $c$ ordered? Write the order using inequality signs.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

$b$ and $c$ multiply to $1$ , the product $abc$ is positive, the sum $a + b + c$ is negative, and $b$ is more than $1$ in size. Put $a$ , $b$ , $c$ in order.

Givens

$bc = 1$
$a \times b \times c > 0$
$a + b + c < 0$
$|b| > 1$

Unknowns

The order of $a$ , $b$ , $c$ from least to greatest.

Constraints

Since $bc = 1$ , neither $b$ nor $c$ is $0$ , and $c$ is the reciprocal of $b$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

Work out signs before sizes. Because $bc = 1$ , the product $abc$ is just $a$ , which settles the sign of $a$ . The sum then forces $b$ and $c$ to be negative, and the reciprocal relation turns $|b| > 1$ into $-1 < c < 0$ .

3 · Execute5 carry out the plan

1b and c share a sign

#7 Identify Subproblems 7.NS.A.2

Their product

1

is positive, so they are both positive or both negative.

bc = 1 > 0

Only like signs give a positive product.

2The sign of a

#13 Convert to Algebra 7.NS.A.2

Group the product as

a \times (bc)

; since

bc = 1

, the whole product equals

a

.

a \times b \times c = a \times (bc) = a \times 1 = a > 0

abc > 0

means simply

a > 0

, so

a

is positive.

3The signs of b and c

#7 Identify Subproblems 6.NS.C.7

From

a + b + c < 0

we get

b + c < -a

, and

-a

is negative. So

b + c < 0

, which is impossible if

b

and

c

are both positive.

b + c < -a < 0 \Rightarrow b < 0,\ c < 0

So

b < 0

and

c < 0

.

4Compare b and c

#13 Convert to Algebra 7.NS.A.2

c

is the reciprocal of

b

. With

b < -1

, the size

|c| = \frac{1}{|b|}

is less than

1

, and

c

is negative too.

c = \frac{1}{b},\ b < -1 \Rightarrow -1 < c < 0

For example

b = -2

gives

c = -\frac{1}{2}

.

5Put them in order

#7 Identify Subproblems 6.NS.C.7

b

is less than

-1

,

c

is between

-1

and

0

, and

a

is positive.

b < -1 < c < 0 < a

So

b < c < a

.

Answer:

b<c<a

4 · Reviewdoes it hold up?

Try $b = -2$ , $c = -\frac{1}{2}$ , $a = 1$ : $bc = 1$ , $abc = 1 > 0$ , $a + b + c = -1\frac{1}{2} < 0$ , $|b| > 1$ . The order is $-2 < -\frac{1}{2} < 1$ , that is $b < c < a$ .

Another way: Check the two sign patterns allowed by $bc = 1$ . If $b, c > 0$ , then $abc$ still has the sign of $a$ , so $a > 0$ too, and three positives cannot have a negative sum. So $b, c < 0$ and $a > 0$ , which leads to the same order.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Sign rules for products and the meaning of reciprocals ($bc = 1$).
6.NS.C.7 Order and absolute value of rational numbers — Ordering the rational numbers once their signs and sizes are known.

💡Takeaway. When two numbers multiply to

1

, they share a sign and one is big exactly when the other is small.

Variant 6 medium

Three rational numbers $a$ , $b$ , $c$ satisfy $bc=1$ , $a\times b\times c>0$ and $a+b+c<0$ . If $|c|<1$ , how are $a$ , $b$ and $c$ ordered? Write the order using inequality signs.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

$b$ and $c$ multiply to $1$ , the product $abc$ is positive, the sum $a + b + c$ is negative, and $c$ is less than $1$ in size. Put $a$ , $b$ , $c$ in order.

Givens

$bc = 1$
$a \times b \times c > 0$
$a + b + c < 0$
$|c| < 1$

Unknowns

The order of $a$ , $b$ , $c$ from least to greatest.

Constraints

Since $bc = 1$ , neither $b$ nor $c$ is $0$ , and $b$ is the reciprocal of $c$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

Work out signs before sizes. Because $bc = 1$ , the product $abc$ is just $a$ , which settles the sign of $a$ . The sum then forces $b$ and $c$ to be negative, and the reciprocal relation turns $|c| < 1$ into $b < -1$ .

3 · Execute5 carry out the plan

1b and c share a sign

#7 Identify Subproblems 7.NS.A.2

Their product

1

is positive, so they are both positive or both negative.

bc = 1 > 0

Only like signs give a positive product.

2The sign of a

#13 Convert to Algebra 7.NS.A.2

Group the product as

a \times (bc)

; since

bc = 1

, the whole product equals

a

.

a \times b \times c = a \times (bc) = a \times 1 = a > 0

abc > 0

means simply

a > 0

, so

a

is positive.

3The signs of b and c

#7 Identify Subproblems 6.NS.C.7

From

a + b + c < 0

we get

b + c < -a

, and

-a

is negative. So

b + c < 0

, which is impossible if

b

and

c

are both positive.

b + c < -a < 0 \Rightarrow b < 0,\ c < 0

So

b < 0

and

c < 0

.

4Compare c and b

#13 Convert to Algebra 7.NS.A.2

b

is the reciprocal of

c

. With

-1 < c < 0

, the size

|b| = \frac{1}{|c|}

is more than

1

, and

b

is negative too.

b = \frac{1}{c},\ -1 < c < 0 \Rightarrow b < -1

For example

c = -\frac{1}{2}

gives

b = -2

.

5Put them in order

#7 Identify Subproblems 6.NS.C.7

b

is less than

-1

,

c

is between

-1

and

0

, and

a

is positive.

b < -1 < c < 0 < a

So

b < c < a

.

Answer:

b<c<a

4 · Reviewdoes it hold up?

Try $c = -\frac{1}{2}$ , $b = -2$ , $a = 1$ : $bc = 1$ , $abc = 1 > 0$ , $a + b + c = -1\frac{1}{2} < 0$ , $|c| < 1$ . The order is $-2 < -\frac{1}{2} < 1$ , that is $b < c < a$ .

Another way: Check the two sign patterns allowed by $bc = 1$ . If $b, c > 0$ , then $abc$ still has the sign of $a$ , so $a > 0$ too, and three positives cannot have a negative sum. So $b, c < 0$ and $a > 0$ , which leads to the same order.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Sign rules for products and the meaning of reciprocals ($bc = 1$).
6.NS.C.7 Order and absolute value of rational numbers — Ordering the rational numbers once their signs and sizes are known.

💡Takeaway. When two numbers multiply to

1

, they share a sign and one is big exactly when the other is small.

Variant 7 medium

Three rational numbers $a$ , $b$ , $c$ satisfy $bc=1$ , $a\times b\times c<0$ and $a+b+c>0$ . If $|c|>1$ , how are $a$ , $b$ and $c$ ordered? Write the order using inequality signs.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

$b$ and $c$ multiply to $1$ , the product $abc$ is negative, the sum $a + b + c$ is positive, and $c$ is more than $1$ in size. Put $a$ , $b$ , $c$ in order.

Givens

$bc = 1$
$a \times b \times c < 0$
$a + b + c > 0$
$|c| > 1$

Unknowns

The order of $a$ , $b$ , $c$ from least to greatest.

Constraints

Since $bc = 1$ , neither $b$ nor $c$ is $0$ , and $b$ is the reciprocal of $c$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

Work out signs before sizes. Because $bc = 1$ , the product $abc$ is just $a$ , which settles the sign of $a$ . The sum then forces $b$ and $c$ to be positive, and the reciprocal relation turns $|c| > 1$ into $0 < b < 1$ .

3 · Execute5 carry out the plan

1b and c share a sign

#7 Identify Subproblems 7.NS.A.2

Their product

1

is positive, so they are both positive or both negative.

bc = 1 > 0

Only like signs give a positive product.

2The sign of a

#13 Convert to Algebra 7.NS.A.2

Group the product as

a \times (bc)

; since

bc = 1

, the whole product equals

a

.

a \times b \times c = a \times (bc) = a \times 1 = a < 0

abc < 0

means simply

a < 0

, so

a

is negative.

3The signs of b and c

#7 Identify Subproblems 6.NS.C.7

From

a + b + c > 0

we get

b + c > -a

, and

-a

is positive. So

b + c > 0

, which is impossible if

b

and

c

are both negative.

b + c > -a > 0 \Rightarrow b > 0,\ c > 0

So

b > 0

and

c > 0

.

4Compare c and b

#13 Convert to Algebra 7.NS.A.2

b

is the reciprocal of

c

. With

c > 1

, dividing

1

by a number greater than

1

gives less than

1

, but still more than

0

.

b = \frac{1}{c},\ c > 1 \Rightarrow 0 < b < 1

For example

c = 2

gives

b = \frac{1}{2}

.

5Put them in order

#7 Identify Subproblems 6.NS.C.7

a

is negative,

b

is between

0

and

1

, and

c

is greater than

1

.

a < 0 < b < 1 < c

So

a < b < c

.

Answer:

a<b<c

4 · Reviewdoes it hold up?

Try $c = 2$ , $b = \frac{1}{2}$ , $a = -1$ : $bc = 1$ , $abc = -1 < 0$ , $a + b + c = 1\frac{1}{2} > 0$ , $|c| > 1$ . The order is $-1 < \frac{1}{2} < 2$ , that is $a < b < c$ .

Another way: Check the two sign patterns allowed by $bc = 1$ . If $b, c < 0$ , then $abc$ still has the sign of $a$ , so $a < 0$ too, and three negatives cannot have a positive sum. So $b, c > 0$ and $a < 0$ , which leads to the same order.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Sign rules for products and the meaning of reciprocals ($bc = 1$).
6.NS.C.7 Order and absolute value of rational numbers — Ordering the rational numbers once their signs and sizes are known.

💡Takeaway. When two numbers multiply to

1

, they share a sign and one is big exactly when the other is small.

Variant 8 medium

Three rational numbers $a$ , $b$ , $c$ satisfy $ac=1$ , $a\times b\times c<0$ and $a+b+c>0$ . If $|a|<1$ , how are $a$ , $b$ and $c$ ordered? Write the order using inequality signs.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

$a$ and $c$ multiply to $1$ , the product $abc$ is negative, the sum $a + b + c$ is positive, and $a$ is less than $1$ in size. Put $a$ , $b$ , $c$ in order.

Givens

$ac = 1$
$a \times b \times c < 0$
$a + b + c > 0$
$|a| < 1$

Unknowns

The order of $a$ , $b$ , $c$ from least to greatest.

Constraints

Since $ac = 1$ , neither $a$ nor $c$ is $0$ , and $c$ is the reciprocal of $a$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

Work out signs before sizes. Because $ac = 1$ , the product $abc$ is just $b$ , which settles the sign of $b$ . The sum then forces $a$ and $c$ to be positive, and the reciprocal relation turns $|a| < 1$ into $c > 1$ .

3 · Execute5 carry out the plan

1a and c share a sign

#7 Identify Subproblems 7.NS.A.2

Their product

1

is positive, so they are both positive or both negative.

ac = 1 > 0

Only like signs give a positive product.

2The sign of b

#13 Convert to Algebra 7.NS.A.2

Group the product as

b \times (ac)

; since

ac = 1

, the whole product equals

b

.

a \times b \times c = b \times (ac) = b \times 1 = b < 0

abc < 0

means simply

b < 0

, so

b

is negative.

3The signs of a and c

#7 Identify Subproblems 6.NS.C.7

From

a + b + c > 0

we get

a + c > -b

, and

-b

is positive. So

a + c > 0

, which is impossible if

a

and

c

are both negative.

a + c > -b > 0 \Rightarrow a > 0,\ c > 0

So

a > 0

and

c > 0

.

4Compare a and c

#13 Convert to Algebra 7.NS.A.2

c

is the reciprocal of

a

. With

0 < a < 1

, dividing

1

by a number less than

1

gives more than

1

.

c = \frac{1}{a},\ 0 < a < 1 \Rightarrow c > 1

For example

a = \frac{1}{2}

gives

c = 2

.

5Put them in order

#7 Identify Subproblems 6.NS.C.7

b

is negative,

a

is between

0

and

1

, and

c

is greater than

1

.

b < 0 < a < 1 < c

So

b < a < c

.

Answer:

b<a<c

4 · Reviewdoes it hold up?

Try $a = \frac{1}{2}$ , $c = 2$ , $b = -1$ : $ac = 1$ , $abc = -1 < 0$ , $a + b + c = 1\frac{1}{2} > 0$ , $|a| < 1$ . The order is $-1 < \frac{1}{2} < 2$ , that is $b < a < c$ .

Another way: Check the two sign patterns allowed by $ac = 1$ . If $a, c < 0$ , then $abc$ still has the sign of $b$ , so $b < 0$ too, and three negatives cannot have a positive sum. So $a, c > 0$ and $b < 0$ , which leads to the same order.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Sign rules for products and the meaning of reciprocals ($ac = 1$).
6.NS.C.7 Order and absolute value of rational numbers — Ordering the rational numbers once their signs and sizes are known.

💡Takeaway. When two numbers multiply to

1

, they share a sign and one is big exactly when the other is small.

Variant 9 medium

Three rational numbers $a$ , $b$ , $c$ satisfy $ac=1$ , $a\times b\times c>0$ and $a+b+c<0$ . If $|c|>1$ , how are $a$ , $b$ and $c$ ordered? Write the order using inequality signs.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

$a$ and $c$ multiply to $1$ , the product $abc$ is positive, the sum $a + b + c$ is negative, and $c$ is more than $1$ in size. Put $a$ , $b$ , $c$ in order.

Givens

$ac = 1$
$a \times b \times c > 0$
$a + b + c < 0$
$|c| > 1$

Unknowns

The order of $a$ , $b$ , $c$ from least to greatest.

Constraints

Since $ac = 1$ , neither $a$ nor $c$ is $0$ , and $a$ is the reciprocal of $c$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

Work out signs before sizes. Because $ac = 1$ , the product $abc$ is just $b$ , which settles the sign of $b$ . The sum then forces $a$ and $c$ to be negative, and the reciprocal relation turns $|c| > 1$ into $-1 < a < 0$ .

3 · Execute5 carry out the plan

1a and c share a sign

#7 Identify Subproblems 7.NS.A.2

Their product

1

is positive, so they are both positive or both negative.

ac = 1 > 0

Only like signs give a positive product.

2The sign of b

#13 Convert to Algebra 7.NS.A.2

Group the product as

b \times (ac)

; since

ac = 1

, the whole product equals

b

.

a \times b \times c = b \times (ac) = b \times 1 = b > 0

abc > 0

means simply

b > 0

, so

b

is positive.

3The signs of a and c

#7 Identify Subproblems 6.NS.C.7

From

a + b + c < 0

we get

a + c < -b

, and

-b

is negative. So

a + c < 0

, which is impossible if

a

and

c

are both positive.

a + c < -b < 0 \Rightarrow a < 0,\ c < 0

So

a < 0

and

c < 0

.

4Compare c and a

#13 Convert to Algebra 7.NS.A.2

a

is the reciprocal of

c

. With

c < -1

, the size

|a| = \frac{1}{|c|}

is less than

1

, and

a

is negative too.

a = \frac{1}{c},\ c < -1 \Rightarrow -1 < a < 0

For example

c = -2

gives

a = -\frac{1}{2}

.

5Put them in order

#7 Identify Subproblems 6.NS.C.7

c

is less than

-1

,

a

is between

-1

and

0

, and

b

is positive.

c < -1 < a < 0 < b

So

c < a < b

.

Answer:

c<a<b

4 · Reviewdoes it hold up?

Try $c = -2$ , $a = -\frac{1}{2}$ , $b = 1$ : $ac = 1$ , $abc = 1 > 0$ , $a + b + c = -1\frac{1}{2} < 0$ , $|c| > 1$ . The order is $-2 < -\frac{1}{2} < 1$ , that is $c < a < b$ .

Another way: Check the two sign patterns allowed by $ac = 1$ . If $a, c > 0$ , then $abc$ still has the sign of $b$ , so $b > 0$ too, and three positives cannot have a negative sum. So $a, c < 0$ and $b > 0$ , which leads to the same order.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Sign rules for products and the meaning of reciprocals ($ac = 1$).
6.NS.C.7 Order and absolute value of rational numbers — Ordering the rational numbers once their signs and sizes are known.

💡Takeaway. When two numbers multiply to

1

, they share a sign and one is big exactly when the other is small.

Variant 10 medium

Three rational numbers $a$ , $b$ , $c$ satisfy $ac=1$ , $a\times b\times c>0$ and $a+b+c<0$ . If $|a|<1$ , how are $a$ , $b$ and $c$ ordered? Write the order using inequality signs.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

$a$ and $c$ multiply to $1$ , the product $abc$ is positive, the sum $a + b + c$ is negative, and $a$ is less than $1$ in size. Put $a$ , $b$ , $c$ in order.

Givens

$ac = 1$
$a \times b \times c > 0$
$a + b + c < 0$
$|a| < 1$

Unknowns

The order of $a$ , $b$ , $c$ from least to greatest.

Constraints

Since $ac = 1$ , neither $a$ nor $c$ is $0$ , and $c$ is the reciprocal of $a$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

Work out signs before sizes. Because $ac = 1$ , the product $abc$ is just $b$ , which settles the sign of $b$ . The sum then forces $a$ and $c$ to be negative, and the reciprocal relation turns $|a| < 1$ into $c < -1$ .

3 · Execute5 carry out the plan

1a and c share a sign

#7 Identify Subproblems 7.NS.A.2

Their product

1

is positive, so they are both positive or both negative.

ac = 1 > 0

Only like signs give a positive product.

2The sign of b

#13 Convert to Algebra 7.NS.A.2

Group the product as

b \times (ac)

; since

ac = 1

, the whole product equals

b

.

a \times b \times c = b \times (ac) = b \times 1 = b > 0

abc > 0

means simply

b > 0

, so

b

is positive.

3The signs of a and c

#7 Identify Subproblems 6.NS.C.7

From

a + b + c < 0

we get

a + c < -b

, and

-b

is negative. So

a + c < 0

, which is impossible if

a

and

c

are both positive.

a + c < -b < 0 \Rightarrow a < 0,\ c < 0

So

a < 0

and

c < 0

.

4Compare a and c

#13 Convert to Algebra 7.NS.A.2

c

is the reciprocal of

a

. With

-1 < a < 0

, the size

|c| = \frac{1}{|a|}

is more than

1

, and

c

is negative too.

c = \frac{1}{a},\ -1 < a < 0 \Rightarrow c < -1

For example

a = -\frac{1}{2}

gives

c = -2

.

5Put them in order

#7 Identify Subproblems 6.NS.C.7

c

is less than

-1

,

a

is between

-1

and

0

, and

b

is positive.

c < -1 < a < 0 < b

So

c < a < b

.

Answer:

c<a<b

4 · Reviewdoes it hold up?

Try $a = -\frac{1}{2}$ , $c = -2$ , $b = 1$ : $ac = 1$ , $abc = 1 > 0$ , $a + b + c = -1\frac{1}{2} < 0$ , $|a| < 1$ . The order is $-2 < -\frac{1}{2} < 1$ , that is $c < a < b$ .

Another way: Check the two sign patterns allowed by $ac = 1$ . If $a, c > 0$ , then $abc$ still has the sign of $b$ , so $b > 0$ too, and three positives cannot have a negative sum. So $a, c < 0$ and $b > 0$ , which leads to the same order.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Sign rules for products and the meaning of reciprocals ($ac = 1$).
6.NS.C.7 Order and absolute value of rational numbers — Ordering the rational numbers once their signs and sizes are known.

💡Takeaway. When two numbers multiply to

1

, they share a sign and one is big exactly when the other is small.

Variant 11 medium

Three rational numbers $a$ , $b$ , $c$ satisfy $ab=1$ , $a\times b\times c<0$ and $a+b+c>0$ . If $|b|>1$ , how are $a$ , $b$ and $c$ ordered? Write the order using inequality signs.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

$a$ and $b$ multiply to $1$ , the product $abc$ is negative, the sum $a + b + c$ is positive, and $b$ is more than $1$ in size. Put $a$ , $b$ , $c$ in order.

Givens

$ab = 1$
$a \times b \times c < 0$
$a + b + c > 0$
$|b| > 1$

Unknowns

The order of $a$ , $b$ , $c$ from least to greatest.

Constraints

Since $ab = 1$ , neither $a$ nor $b$ is $0$ , and $a$ is the reciprocal of $b$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

Work out signs before sizes. Because $ab = 1$ , the product $abc$ is just $c$ , which settles the sign of $c$ . The sum then forces $a$ and $b$ to be positive, and the reciprocal relation turns $|b| > 1$ into $0 < a < 1$ .

3 · Execute5 carry out the plan

1a and b share a sign

#7 Identify Subproblems 7.NS.A.2

Their product

1

is positive, so they are both positive or both negative.

ab = 1 > 0

Only like signs give a positive product.

2The sign of c

#13 Convert to Algebra 7.NS.A.2

Group the product as

c \times (ab)

; since

ab = 1

, the whole product equals

c

.

a \times b \times c = c \times (ab) = c \times 1 = c < 0

abc < 0

means simply

c < 0

, so

c

is negative.

3The signs of a and b

#7 Identify Subproblems 6.NS.C.7

From

a + b + c > 0

we get

a + b > -c

, and

-c

is positive. So

a + b > 0

, which is impossible if

a

and

b

are both negative.

a + b > -c > 0 \Rightarrow a > 0,\ b > 0

So

a > 0

and

b > 0

.

4Compare b and a

#13 Convert to Algebra 7.NS.A.2

a

is the reciprocal of

b

. With

b > 1

, dividing

1

by a number greater than

1

gives less than

1

, but still more than

0

.

a = \frac{1}{b},\ b > 1 \Rightarrow 0 < a < 1

For example

b = 2

gives

a = \frac{1}{2}

.

5Put them in order

#7 Identify Subproblems 6.NS.C.7

c

is negative,

a

is between

0

and

1

, and

b

is greater than

1

.

c < 0 < a < 1 < b

So

c < a < b

.

Answer:

c<a<b

4 · Reviewdoes it hold up?

Try $b = 2$ , $a = \frac{1}{2}$ , $c = -1$ : $ab = 1$ , $abc = -1 < 0$ , $a + b + c = 1\frac{1}{2} > 0$ , $|b| > 1$ . The order is $-1 < \frac{1}{2} < 2$ , that is $c < a < b$ .

Another way: Check the two sign patterns allowed by $ab = 1$ . If $a, b < 0$ , then $abc$ still has the sign of $c$ , so $c < 0$ too, and three negatives cannot have a positive sum. So $a, b > 0$ and $c < 0$ , which leads to the same order.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Sign rules for products and the meaning of reciprocals ($ab = 1$).
6.NS.C.7 Order and absolute value of rational numbers — Ordering the rational numbers once their signs and sizes are known.

💡Takeaway. When two numbers multiply to

1

, they share a sign and one is big exactly when the other is small.

Variant 12 medium

Three rational numbers $a$ , $b$ , $c$ satisfy $ab=1$ , $a\times b\times c>0$ and $a+b+c<0$ . If $|a|>1$ , how are $a$ , $b$ and $c$ ordered? Write the order using inequality signs.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

$a$ and $b$ multiply to $1$ , the product $abc$ is positive, the sum $a + b + c$ is negative, and $a$ is more than $1$ in size. Put $a$ , $b$ , $c$ in order.

Givens

$ab = 1$
$a \times b \times c > 0$
$a + b + c < 0$
$|a| > 1$

Unknowns

The order of $a$ , $b$ , $c$ from least to greatest.

Constraints

Since $ab = 1$ , neither $a$ nor $b$ is $0$ , and $b$ is the reciprocal of $a$ .

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

Work out signs before sizes. Because $ab = 1$ , the product $abc$ is just $c$ , which settles the sign of $c$ . The sum then forces $a$ and $b$ to be negative, and the reciprocal relation turns $|a| > 1$ into $-1 < b < 0$ .

3 · Execute5 carry out the plan

1a and b share a sign

#7 Identify Subproblems 7.NS.A.2

Their product

1

is positive, so they are both positive or both negative.

ab = 1 > 0

Only like signs give a positive product.

2The sign of c

#13 Convert to Algebra 7.NS.A.2

Group the product as

c \times (ab)

; since

ab = 1

, the whole product equals

c

.

a \times b \times c = c \times (ab) = c \times 1 = c > 0

abc > 0

means simply

c > 0

, so

c

is positive.

3The signs of a and b

#7 Identify Subproblems 6.NS.C.7

From

a + b + c < 0

we get

a + b < -c

, and

-c

is negative. So

a + b < 0

, which is impossible if

a

and

b

are both positive.

a + b < -c < 0 \Rightarrow a < 0,\ b < 0

So

a < 0

and

b < 0

.

4Compare a and b

#13 Convert to Algebra 7.NS.A.2

b

is the reciprocal of

a

. With

a < -1

, the size

|b| = \frac{1}{|a|}

is less than

1

, and

b

is negative too.

b = \frac{1}{a},\ a < -1 \Rightarrow -1 < b < 0

For example

a = -2

gives

b = -\frac{1}{2}

.

5Put them in order

#7 Identify Subproblems 6.NS.C.7

a

is less than

-1

,

b

is between

-1

and

0

, and

c

is positive.

a < -1 < b < 0 < c

So

a < b < c

.

Answer:

a<b<c

4 · Reviewdoes it hold up?

Try $a = -2$ , $b = -\frac{1}{2}$ , $c = 1$ : $ab = 1$ , $abc = 1 > 0$ , $a + b + c = -1\frac{1}{2} < 0$ , $|a| > 1$ . The order is $-2 < -\frac{1}{2} < 1$ , that is $a < b < c$ .

Another way: Check the two sign patterns allowed by $ab = 1$ . If $a, b > 0$ , then $abc$ still has the sign of $c$ , so $c > 0$ too, and three positives cannot have a negative sum. So $a, b < 0$ and $c > 0$ , which leads to the same order.

Standardsmin grade 7

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Sign rules for products and the meaning of reciprocals ($ab = 1$).
6.NS.C.7 Order and absolute value of rational numbers — Ordering the rational numbers once their signs and sizes are known.

💡Takeaway. When two numbers multiply to

1

, they share a sign and one is big exactly when the other is small.

ac=1 makes a and c same-sign reciprocals · 12 practice problems

Generated variants — 12

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Where next?