Practice · Subtract team rates to isolate one worker · Sensim Math

Variant 1 easy

A job takes $3$ days when A, B, and C work together, $5$ days when A and C work together, and $6$ days when B and C work together. When A and B work on this job together, what fraction of the whole job can they do in one day?

Your answer

Show solution

1 · Understandwhat's really being asked

Three team-ups of A, B, C finish a job in 3, 5, and 6 days. Find what fraction of the job A and B together finish in one day.

Givens

A, B, and C together: $3$ days.
A and C together: $5$ days.
B and C together: $6$ days.

Unknowns

The fraction of the job A and B do together in one day.

Constraints

Each person works at a steady rate, and rates of people working together add.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Call the whole job $1$ and let each person's daily share be an unknown. Each team-up then says what a sum of daily shares equals. Taking a two-person team away from the three-person team leaves one person's share.

3 · Execute4 carry out the plan

1Turn days into daily shares

#13 Convert to Algebra 8.EE.C.8

Let A, B, C do

x

,

y

,

z

of the job per day. A team that needs

3

days does

\frac{1}{3}

of the job each day.

x + y + z = \frac{1}{3},\quad x + z = \frac{1}{5},\quad y + z = \frac{1}{6}

Finishing in

n

days means doing

\frac{1}{n}

of the job per day.

2Find A's share

#7 Identify Subproblems 7.NS.A.3

Take the B-and-C team away from the whole team; what is left is A.

x = \frac{1}{3} - \frac{1}{6} = \frac{2}{6} - \frac{1}{6} = \frac{1}{6}

Whole team minus a sub-team leaves the missing member.

3Find B's share

#7 Identify Subproblems 7.NS.A.3

Take the A-and-C team away from the whole team; what is left is B.

y = \frac{1}{3} - \frac{1}{5} = \frac{5}{15} - \frac{3}{15} = \frac{2}{15}

Same subtraction, different sub-team.

4Add A and B

#13 Convert to Algebra 7.NS.A.3

Together A and B do the sum of their daily shares.

\frac{1}{6} + \frac{2}{15} = \frac{5}{30} + \frac{4}{30} = \frac{9}{30} = \frac{3}{10}

Rates of people working side by side add.

Answer:

\frac{3}{10}

4 · Reviewdoes it hold up?

C's share is $\frac{1}{5} - \frac{1}{6} = \frac{1}{30}$ . Then $\frac{5}{30} + \frac{4}{30} + \frac{1}{30} = \frac{10}{30} = \frac{1}{3}$ , matching the $3$ -day team. A and B alone would need $\frac{10}{3} = 3\frac{1}{3}$ days, more than the $3$ days with C's help, as expected.

Another way: Add the two pair equations and compare with twice the whole-team equation: $2(x + y + z) - (x + z) - (y + z) = x + y$ , so $x + y = \frac{2}{3} - \frac{1}{5} - \frac{1}{6} = \frac{20}{30} - \frac{6}{30} - \frac{5}{30} = \frac{9}{30} = \frac{3}{10}$ without finding anyone separately.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and combining the three team equations.
7.NS.A.3 Solve real-world problems involving the four operations with rational numbers — Subtracting and adding the fractional daily shares.

💡Takeaway. Call the whole job

1

; then every team is just a sum of daily shares, and subtracting teams picks out individuals.

Variant 2 easy

A job takes $4$ days when A, B, and C work together, $6$ days when A and C work together, and $9$ days when B and C work together. When A and B work on this job together, what fraction of the whole job can they do in one day?

Your answer

Show solution

1 · Understandwhat's really being asked

Three team-ups of A, B, C finish a job in 4, 6, and 9 days. Find what fraction of the job A and B together finish in one day.

Givens

A, B, and C together: $4$ days.
A and C together: $6$ days.
B and C together: $9$ days.

Unknowns

The fraction of the job A and B do together in one day.

Constraints

Each person works at a steady rate, and rates of people working together add.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Call the whole job $1$ and let each person's daily share be an unknown. Each team-up then says what a sum of daily shares equals. Taking a two-person team away from the three-person team leaves one person's share.

3 · Execute4 carry out the plan

1Turn days into daily shares

#13 Convert to Algebra 8.EE.C.8

Let A, B, C do

x

,

y

,

z

of the job per day. A team that needs

4

days does

\frac{1}{4}

of the job each day.

x + y + z = \frac{1}{4},\quad x + z = \frac{1}{6},\quad y + z = \frac{1}{9}

Finishing in

n

days means doing

\frac{1}{n}

of the job per day.

2Find A's share

#7 Identify Subproblems 7.NS.A.3

Take the B-and-C team away from the whole team; what is left is A.

x = \frac{1}{4} - \frac{1}{9} = \frac{9}{36} - \frac{4}{36} = \frac{5}{36}

Whole team minus a sub-team leaves the missing member.

3Find B's share

#7 Identify Subproblems 7.NS.A.3

Take the A-and-C team away from the whole team; what is left is B.

y = \frac{1}{4} - \frac{1}{6} = \frac{3}{12} - \frac{2}{12} = \frac{1}{12}

Same subtraction, different sub-team.

4Add A and B

#13 Convert to Algebra 7.NS.A.3

Together A and B do the sum of their daily shares.

\frac{5}{36} + \frac{1}{12} = \frac{5}{36} + \frac{3}{36} = \frac{8}{36} = \frac{2}{9}

Rates of people working side by side add.

Answer:

\frac{2}{9}

4 · Reviewdoes it hold up?

C's share is $\frac{1}{6} - \frac{5}{36} = \frac{1}{36}$ . Then $\frac{5}{36} + \frac{3}{36} + \frac{1}{36} = \frac{9}{36} = \frac{1}{4}$ , matching the $4$ -day team. A and B alone would need $\frac{9}{2} = 4\frac{1}{2}$ days, more than the $4$ days with C's help, as expected.

Another way: Add the two pair equations and compare with twice the whole-team equation: $2(x + y + z) - (x + z) - (y + z) = x + y$ , so $x + y = \frac{2}{4} - \frac{1}{6} - \frac{1}{9} = \frac{18}{36} - \frac{6}{36} - \frac{4}{36} = \frac{8}{36} = \frac{2}{9}$ without finding anyone separately.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and combining the three team equations.
7.NS.A.3 Solve real-world problems involving the four operations with rational numbers — Subtracting and adding the fractional daily shares.

💡Takeaway. Call the whole job

1

; then every team is just a sum of daily shares, and subtracting teams picks out individuals.

Variant 3 easy

A job takes $5$ days when A, B, and C work together, $6$ days when A and C work together, and $10$ days when B and C work together. When A and B work on this job together, what fraction of the whole job can they do in one day?

Your answer

Show solution

1 · Understandwhat's really being asked

Three team-ups of A, B, C finish a job in 5, 6, and 10 days. Find what fraction of the job A and B together finish in one day.

Givens

A, B, and C together: $5$ days.
A and C together: $6$ days.
B and C together: $10$ days.

Unknowns

The fraction of the job A and B do together in one day.

Constraints

Each person works at a steady rate, and rates of people working together add.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Call the whole job $1$ and let each person's daily share be an unknown. Each team-up then says what a sum of daily shares equals. Taking a two-person team away from the three-person team leaves one person's share.

3 · Execute4 carry out the plan

1Turn days into daily shares

#13 Convert to Algebra 8.EE.C.8

Let A, B, C do

x

,

y

,

z

of the job per day. A team that needs

5

days does

\frac{1}{5}

of the job each day.

x + y + z = \frac{1}{5},\quad x + z = \frac{1}{6},\quad y + z = \frac{1}{10}

Finishing in

n

days means doing

\frac{1}{n}

of the job per day.

2Find A's share

#7 Identify Subproblems 7.NS.A.3

Take the B-and-C team away from the whole team; what is left is A.

x = \frac{1}{5} - \frac{1}{10} = \frac{2}{10} - \frac{1}{10} = \frac{1}{10}

Whole team minus a sub-team leaves the missing member.

3Find B's share

#7 Identify Subproblems 7.NS.A.3

Take the A-and-C team away from the whole team; what is left is B.

y = \frac{1}{5} - \frac{1}{6} = \frac{6}{30} - \frac{5}{30} = \frac{1}{30}

Same subtraction, different sub-team.

4Add A and B

#13 Convert to Algebra 7.NS.A.3

Together A and B do the sum of their daily shares.

\frac{1}{10} + \frac{1}{30} = \frac{3}{30} + \frac{1}{30} = \frac{4}{30} = \frac{2}{15}

Rates of people working side by side add.

Answer:

\frac{2}{15}

4 · Reviewdoes it hold up?

C's share is $\frac{1}{6} - \frac{1}{10} = \frac{1}{15}$ . Then $\frac{3}{30} + \frac{1}{30} + \frac{2}{30} = \frac{6}{30} = \frac{1}{5}$ , matching the $5$ -day team. A and B alone would need $\frac{15}{2} = 7\frac{1}{2}$ days, more than the $5$ days with C's help, as expected.

Another way: Add the two pair equations and compare with twice the whole-team equation: $2(x + y + z) - (x + z) - (y + z) = x + y$ , so $x + y = \frac{2}{5} - \frac{1}{6} - \frac{1}{10} = \frac{12}{30} - \frac{5}{30} - \frac{3}{30} = \frac{4}{30} = \frac{2}{15}$ without finding anyone separately.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and combining the three team equations.
7.NS.A.3 Solve real-world problems involving the four operations with rational numbers — Subtracting and adding the fractional daily shares.

💡Takeaway. Call the whole job

1

; then every team is just a sum of daily shares, and subtracting teams picks out individuals.

Variant 4 easy

A job takes $5$ days when A, B, and C work together, $7$ days when A and C work together, and $10$ days when B and C work together. When A and B work on this job together, what fraction of the whole job can they do in one day?

Your answer

Show solution

1 · Understandwhat's really being asked

Three team-ups of A, B, C finish a job in 5, 7, and 10 days. Find what fraction of the job A and B together finish in one day.

Givens

A, B, and C together: $5$ days.
A and C together: $7$ days.
B and C together: $10$ days.

Unknowns

The fraction of the job A and B do together in one day.

Constraints

Each person works at a steady rate, and rates of people working together add.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Call the whole job $1$ and let each person's daily share be an unknown. Each team-up then says what a sum of daily shares equals. Taking a two-person team away from the three-person team leaves one person's share.

3 · Execute4 carry out the plan

1Turn days into daily shares

#13 Convert to Algebra 8.EE.C.8

Let A, B, C do

x

,

y

,

z

of the job per day. A team that needs

5

days does

\frac{1}{5}

of the job each day.

x + y + z = \frac{1}{5},\quad x + z = \frac{1}{7},\quad y + z = \frac{1}{10}

Finishing in

n

days means doing

\frac{1}{n}

of the job per day.

2Find A's share

#7 Identify Subproblems 7.NS.A.3

Take the B-and-C team away from the whole team; what is left is A.

x = \frac{1}{5} - \frac{1}{10} = \frac{2}{10} - \frac{1}{10} = \frac{1}{10}

Whole team minus a sub-team leaves the missing member.

3Find B's share

#7 Identify Subproblems 7.NS.A.3

Take the A-and-C team away from the whole team; what is left is B.

y = \frac{1}{5} - \frac{1}{7} = \frac{7}{35} - \frac{5}{35} = \frac{2}{35}

Same subtraction, different sub-team.

4Add A and B

#13 Convert to Algebra 7.NS.A.3

Together A and B do the sum of their daily shares.

\frac{1}{10} + \frac{2}{35} = \frac{7}{70} + \frac{4}{70} = \frac{11}{70}

Rates of people working side by side add.

Answer:

\frac{11}{70}

4 · Reviewdoes it hold up?

C's share is $\frac{1}{7} - \frac{1}{10} = \frac{3}{70}$ . Then $\frac{7}{70} + \frac{4}{70} + \frac{3}{70} = \frac{14}{70} = \frac{1}{5}$ , matching the $5$ -day team. A and B alone would need $\frac{70}{11} = 6\frac{4}{11}$ days, more than the $5$ days with C's help, as expected.

Another way: Add the two pair equations and compare with twice the whole-team equation: $2(x + y + z) - (x + z) - (y + z) = x + y$ , so $x + y = \frac{2}{5} - \frac{1}{7} - \frac{1}{10} = \frac{28}{70} - \frac{10}{70} - \frac{7}{70} = \frac{11}{70}$ without finding anyone separately.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and combining the three team equations.
7.NS.A.3 Solve real-world problems involving the four operations with rational numbers — Subtracting and adding the fractional daily shares.

💡Takeaway. Call the whole job

1

; then every team is just a sum of daily shares, and subtracting teams picks out individuals.

Variant 5 medium

A job takes $8$ days when A, B, and C work together, $9$ days when A and C work together, and $12$ days when B and C work together. When A and B work on this job together, what fraction of the whole job can they do in one day?

Your answer

Show solution

1 · Understandwhat's really being asked

Three team-ups of A, B, C finish a job in 8, 9, and 12 days. Find what fraction of the job A and B together finish in one day.

Givens

A, B, and C together: $8$ days.
A and C together: $9$ days.
B and C together: $12$ days.

Unknowns

The fraction of the job A and B do together in one day.

Constraints

Each person works at a steady rate, and rates of people working together add.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Call the whole job $1$ and let each person's daily share be an unknown. Each team-up then says what a sum of daily shares equals. Taking a two-person team away from the three-person team leaves one person's share.

3 · Execute4 carry out the plan

1Turn days into daily shares

#13 Convert to Algebra 8.EE.C.8

Let A, B, C do

x

,

y

,

z

of the job per day. A team that needs

8

days does

\frac{1}{8}

of the job each day.

x + y + z = \frac{1}{8},\quad x + z = \frac{1}{9},\quad y + z = \frac{1}{12}

Finishing in

n

days means doing

\frac{1}{n}

of the job per day.

2Find A's share

#7 Identify Subproblems 7.NS.A.3

Take the B-and-C team away from the whole team; what is left is A.

x = \frac{1}{8} - \frac{1}{12} = \frac{3}{24} - \frac{2}{24} = \frac{1}{24}

Whole team minus a sub-team leaves the missing member.

3Find B's share

#7 Identify Subproblems 7.NS.A.3

Take the A-and-C team away from the whole team; what is left is B.

y = \frac{1}{8} - \frac{1}{9} = \frac{9}{72} - \frac{8}{72} = \frac{1}{72}

Same subtraction, different sub-team.

4Add A and B

#13 Convert to Algebra 7.NS.A.3

Together A and B do the sum of their daily shares.

\frac{1}{24} + \frac{1}{72} = \frac{3}{72} + \frac{1}{72} = \frac{4}{72} = \frac{1}{18}

Rates of people working side by side add.

Answer:

\frac{1}{18}

4 · Reviewdoes it hold up?

C's share is $\frac{1}{9} - \frac{1}{24} = \frac{5}{72}$ . Then $\frac{3}{72} + \frac{1}{72} + \frac{5}{72} = \frac{9}{72} = \frac{1}{8}$ , matching the $8$ -day team. A and B alone would need $18$ days, more than the $8$ days with C's help, as expected.

Another way: Add the two pair equations and compare with twice the whole-team equation: $2(x + y + z) - (x + z) - (y + z) = x + y$ , so $x + y = \frac{2}{8} - \frac{1}{9} - \frac{1}{12} = \frac{18}{72} - \frac{8}{72} - \frac{6}{72} = \frac{4}{72} = \frac{1}{18}$ without finding anyone separately.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and combining the three team equations.
7.NS.A.3 Solve real-world problems involving the four operations with rational numbers — Subtracting and adding the fractional daily shares.

💡Takeaway. Call the whole job

1

; then every team is just a sum of daily shares, and subtracting teams picks out individuals.

Variant 6 medium

A job takes $6$ days when A, B, and C work together, $9$ days when A and C work together, and $12$ days when B and C work together. When A and B work on this job together, what fraction of the whole job can they do in one day?

Your answer

Show solution

1 · Understandwhat's really being asked

Three team-ups of A, B, C finish a job in 6, 9, and 12 days. Find what fraction of the job A and B together finish in one day.

Givens

A, B, and C together: $6$ days.
A and C together: $9$ days.
B and C together: $12$ days.

Unknowns

The fraction of the job A and B do together in one day.

Constraints

Each person works at a steady rate, and rates of people working together add.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Call the whole job $1$ and let each person's daily share be an unknown. Each team-up then says what a sum of daily shares equals. Taking a two-person team away from the three-person team leaves one person's share.

3 · Execute4 carry out the plan

1Turn days into daily shares

#13 Convert to Algebra 8.EE.C.8

Let A, B, C do

x

,

y

,

z

of the job per day. A team that needs

6

days does

\frac{1}{6}

of the job each day.

x + y + z = \frac{1}{6},\quad x + z = \frac{1}{9},\quad y + z = \frac{1}{12}

Finishing in

n

days means doing

\frac{1}{n}

of the job per day.

2Find A's share

#7 Identify Subproblems 7.NS.A.3

Take the B-and-C team away from the whole team; what is left is A.

x = \frac{1}{6} - \frac{1}{12} = \frac{2}{12} - \frac{1}{12} = \frac{1}{12}

Whole team minus a sub-team leaves the missing member.

3Find B's share

#7 Identify Subproblems 7.NS.A.3

Take the A-and-C team away from the whole team; what is left is B.

y = \frac{1}{6} - \frac{1}{9} = \frac{3}{18} - \frac{2}{18} = \frac{1}{18}

Same subtraction, different sub-team.

4Add A and B

#13 Convert to Algebra 7.NS.A.3

Together A and B do the sum of their daily shares.

\frac{1}{12} + \frac{1}{18} = \frac{3}{36} + \frac{2}{36} = \frac{5}{36}

Rates of people working side by side add.

Answer:

\frac{5}{36}

4 · Reviewdoes it hold up?

C's share is $\frac{1}{9} - \frac{1}{12} = \frac{1}{36}$ . Then $\frac{3}{36} + \frac{2}{36} + \frac{1}{36} = \frac{6}{36} = \frac{1}{6}$ , matching the $6$ -day team. A and B alone would need $\frac{36}{5} = 7\frac{1}{5}$ days, more than the $6$ days with C's help, as expected.

Another way: Add the two pair equations and compare with twice the whole-team equation: $2(x + y + z) - (x + z) - (y + z) = x + y$ , so $x + y = \frac{2}{6} - \frac{1}{9} - \frac{1}{12} = \frac{12}{36} - \frac{4}{36} - \frac{3}{36} = \frac{5}{36}$ without finding anyone separately.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and combining the three team equations.
7.NS.A.3 Solve real-world problems involving the four operations with rational numbers — Subtracting and adding the fractional daily shares.

💡Takeaway. Call the whole job

1

; then every team is just a sum of daily shares, and subtracting teams picks out individuals.

Variant 7 medium

A job takes $10$ days when A, B, and C work together, $12$ days when A and C work together, and $15$ days when B and C work together. When A and B work on this job together, what fraction of the whole job can they do in one day?

Your answer

Show solution

1 · Understandwhat's really being asked

Three team-ups of A, B, C finish a job in 10, 12, and 15 days. Find what fraction of the job A and B together finish in one day.

Givens

A, B, and C together: $10$ days.
A and C together: $12$ days.
B and C together: $15$ days.

Unknowns

The fraction of the job A and B do together in one day.

Constraints

Each person works at a steady rate, and rates of people working together add.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Call the whole job $1$ and let each person's daily share be an unknown. Each team-up then says what a sum of daily shares equals. Taking a two-person team away from the three-person team leaves one person's share.

3 · Execute4 carry out the plan

1Turn days into daily shares

#13 Convert to Algebra 8.EE.C.8

Let A, B, C do

x

,

y

,

z

of the job per day. A team that needs

10

days does

\frac{1}{10}

of the job each day.

x + y + z = \frac{1}{10},\quad x + z = \frac{1}{12},\quad y + z = \frac{1}{15}

Finishing in

n

days means doing

\frac{1}{n}

of the job per day.

2Find A's share

#7 Identify Subproblems 7.NS.A.3

Take the B-and-C team away from the whole team; what is left is A.

x = \frac{1}{10} - \frac{1}{15} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30}

Whole team minus a sub-team leaves the missing member.

3Find B's share

#7 Identify Subproblems 7.NS.A.3

Take the A-and-C team away from the whole team; what is left is B.

y = \frac{1}{10} - \frac{1}{12} = \frac{6}{60} - \frac{5}{60} = \frac{1}{60}

Same subtraction, different sub-team.

4Add A and B

#13 Convert to Algebra 7.NS.A.3

Together A and B do the sum of their daily shares.

\frac{1}{30} + \frac{1}{60} = \frac{2}{60} + \frac{1}{60} = \frac{3}{60} = \frac{1}{20}

Rates of people working side by side add.

Answer:

\frac{1}{20}

4 · Reviewdoes it hold up?

C's share is $\frac{1}{12} - \frac{1}{30} = \frac{1}{20}$ . Then $\frac{2}{60} + \frac{1}{60} + \frac{3}{60} = \frac{6}{60} = \frac{1}{10}$ , matching the $10$ -day team. A and B alone would need $20$ days, more than the $10$ days with C's help, as expected.

Another way: Add the two pair equations and compare with twice the whole-team equation: $2(x + y + z) - (x + z) - (y + z) = x + y$ , so $x + y = \frac{2}{10} - \frac{1}{12} - \frac{1}{15} = \frac{12}{60} - \frac{5}{60} - \frac{4}{60} = \frac{3}{60} = \frac{1}{20}$ without finding anyone separately.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and combining the three team equations.
7.NS.A.3 Solve real-world problems involving the four operations with rational numbers — Subtracting and adding the fractional daily shares.

💡Takeaway. Call the whole job

1

; then every team is just a sum of daily shares, and subtracting teams picks out individuals.

Variant 8 medium

A job takes $6$ days when A, B, and C work together, $16$ days when A and C work together, and $8$ days when B and C work together. When A and B work on this job together, what fraction of the whole job can they do in one day?

Your answer

Show solution

1 · Understandwhat's really being asked

Three team-ups of A, B, C finish a job in 6, 16, and 8 days. Find what fraction of the job A and B together finish in one day.

Givens

A, B, and C together: $6$ days.
A and C together: $16$ days.
B and C together: $8$ days.

Unknowns

The fraction of the job A and B do together in one day.

Constraints

Each person works at a steady rate, and rates of people working together add.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Call the whole job $1$ and let each person's daily share be an unknown. Each team-up then says what a sum of daily shares equals. Taking a two-person team away from the three-person team leaves one person's share.

3 · Execute4 carry out the plan

1Turn days into daily shares

#13 Convert to Algebra 8.EE.C.8

Let A, B, C do

x

,

y

,

z

of the job per day. A team that needs

6

days does

\frac{1}{6}

of the job each day.

x + y + z = \frac{1}{6},\quad x + z = \frac{1}{16},\quad y + z = \frac{1}{8}

Finishing in

n

days means doing

\frac{1}{n}

of the job per day.

2Find A's share

#7 Identify Subproblems 7.NS.A.3

Take the B-and-C team away from the whole team; what is left is A.

x = \frac{1}{6} - \frac{1}{8} = \frac{4}{24} - \frac{3}{24} = \frac{1}{24}

Whole team minus a sub-team leaves the missing member.

3Find B's share

#7 Identify Subproblems 7.NS.A.3

Take the A-and-C team away from the whole team; what is left is B.

y = \frac{1}{6} - \frac{1}{16} = \frac{8}{48} - \frac{3}{48} = \frac{5}{48}

Same subtraction, different sub-team.

4Add A and B

#13 Convert to Algebra 7.NS.A.3

Together A and B do the sum of their daily shares.

\frac{1}{24} + \frac{5}{48} = \frac{2}{48} + \frac{5}{48} = \frac{7}{48}

Rates of people working side by side add.

Answer:

\frac{7}{48}

4 · Reviewdoes it hold up?

C's share is $\frac{1}{16} - \frac{1}{24} = \frac{1}{48}$ . Then $\frac{2}{48} + \frac{5}{48} + \frac{1}{48} = \frac{8}{48} = \frac{1}{6}$ , matching the $6$ -day team. A and B alone would need $\frac{48}{7} = 6\frac{6}{7}$ days, more than the $6$ days with C's help, as expected.

Another way: Add the two pair equations and compare with twice the whole-team equation: $2(x + y + z) - (x + z) - (y + z) = x + y$ , so $x + y = \frac{2}{6} - \frac{1}{16} - \frac{1}{8} = \frac{16}{48} - \frac{3}{48} - \frac{6}{48} = \frac{7}{48}$ without finding anyone separately.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and combining the three team equations.
7.NS.A.3 Solve real-world problems involving the four operations with rational numbers — Subtracting and adding the fractional daily shares.

💡Takeaway. Call the whole job

1

; then every team is just a sum of daily shares, and subtracting teams picks out individuals.

Variant 9 hard

A job takes $8$ days when A, B, and C work together, $12$ days when A and C work together, and $18$ days when B and C work together. When A and B work on this job together, what fraction of the whole job can they do in one day?

Your answer

Show solution

1 · Understandwhat's really being asked

Three team-ups of A, B, C finish a job in 8, 12, and 18 days. Find what fraction of the job A and B together finish in one day.

Givens

A, B, and C together: $8$ days.
A and C together: $12$ days.
B and C together: $18$ days.

Unknowns

The fraction of the job A and B do together in one day.

Constraints

Each person works at a steady rate, and rates of people working together add.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Call the whole job $1$ and let each person's daily share be an unknown. Each team-up then says what a sum of daily shares equals. Taking a two-person team away from the three-person team leaves one person's share.

3 · Execute4 carry out the plan

1Turn days into daily shares

#13 Convert to Algebra 8.EE.C.8

Let A, B, C do

x

,

y

,

z

of the job per day. A team that needs

8

days does

\frac{1}{8}

of the job each day.

x + y + z = \frac{1}{8},\quad x + z = \frac{1}{12},\quad y + z = \frac{1}{18}

Finishing in

n

days means doing

\frac{1}{n}

of the job per day.

2Find A's share

#7 Identify Subproblems 7.NS.A.3

Take the B-and-C team away from the whole team; what is left is A.

x = \frac{1}{8} - \frac{1}{18} = \frac{9}{72} - \frac{4}{72} = \frac{5}{72}

Whole team minus a sub-team leaves the missing member.

3Find B's share

#7 Identify Subproblems 7.NS.A.3

Take the A-and-C team away from the whole team; what is left is B.

y = \frac{1}{8} - \frac{1}{12} = \frac{3}{24} - \frac{2}{24} = \frac{1}{24}

Same subtraction, different sub-team.

4Add A and B

#13 Convert to Algebra 7.NS.A.3

Together A and B do the sum of their daily shares.

\frac{5}{72} + \frac{1}{24} = \frac{5}{72} + \frac{3}{72} = \frac{8}{72} = \frac{1}{9}

Rates of people working side by side add.

Answer:

\frac{1}{9}

4 · Reviewdoes it hold up?

C's share is $\frac{1}{12} - \frac{5}{72} = \frac{1}{72}$ . Then $\frac{5}{72} + \frac{3}{72} + \frac{1}{72} = \frac{9}{72} = \frac{1}{8}$ , matching the $8$ -day team. A and B alone would need $9$ days, more than the $8$ days with C's help, as expected.

Another way: Add the two pair equations and compare with twice the whole-team equation: $2(x + y + z) - (x + z) - (y + z) = x + y$ , so $x + y = \frac{2}{8} - \frac{1}{12} - \frac{1}{18} = \frac{18}{72} - \frac{6}{72} - \frac{4}{72} = \frac{8}{72} = \frac{1}{9}$ without finding anyone separately.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and combining the three team equations.
7.NS.A.3 Solve real-world problems involving the four operations with rational numbers — Subtracting and adding the fractional daily shares.

💡Takeaway. Call the whole job

1

; then every team is just a sum of daily shares, and subtracting teams picks out individuals.

Variant 10 hard

A job takes $10$ days when A, B, and C work together, $15$ days when A and C work together, and $20$ days when B and C work together. When A and B work on this job together, what fraction of the whole job can they do in one day?

Your answer

Show solution

1 · Understandwhat's really being asked

Three team-ups of A, B, C finish a job in 10, 15, and 20 days. Find what fraction of the job A and B together finish in one day.

Givens

A, B, and C together: $10$ days.
A and C together: $15$ days.
B and C together: $20$ days.

Unknowns

The fraction of the job A and B do together in one day.

Constraints

Each person works at a steady rate, and rates of people working together add.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Call the whole job $1$ and let each person's daily share be an unknown. Each team-up then says what a sum of daily shares equals. Taking a two-person team away from the three-person team leaves one person's share.

3 · Execute4 carry out the plan

1Turn days into daily shares

#13 Convert to Algebra 8.EE.C.8

Let A, B, C do

x

,

y

,

z

of the job per day. A team that needs

10

days does

\frac{1}{10}

of the job each day.

x + y + z = \frac{1}{10},\quad x + z = \frac{1}{15},\quad y + z = \frac{1}{20}

Finishing in

n

days means doing

\frac{1}{n}

of the job per day.

2Find A's share

#7 Identify Subproblems 7.NS.A.3

Take the B-and-C team away from the whole team; what is left is A.

x = \frac{1}{10} - \frac{1}{20} = \frac{2}{20} - \frac{1}{20} = \frac{1}{20}

Whole team minus a sub-team leaves the missing member.

3Find B's share

#7 Identify Subproblems 7.NS.A.3

Take the A-and-C team away from the whole team; what is left is B.

y = \frac{1}{10} - \frac{1}{15} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30}

Same subtraction, different sub-team.

4Add A and B

#13 Convert to Algebra 7.NS.A.3

Together A and B do the sum of their daily shares.

\frac{1}{20} + \frac{1}{30} = \frac{3}{60} + \frac{2}{60} = \frac{5}{60} = \frac{1}{12}

Rates of people working side by side add.

Answer:

\frac{1}{12}

4 · Reviewdoes it hold up?

C's share is $\frac{1}{15} - \frac{1}{20} = \frac{1}{60}$ . Then $\frac{3}{60} + \frac{2}{60} + \frac{1}{60} = \frac{6}{60} = \frac{1}{10}$ , matching the $10$ -day team. A and B alone would need $12$ days, more than the $10$ days with C's help, as expected.

Another way: Add the two pair equations and compare with twice the whole-team equation: $2(x + y + z) - (x + z) - (y + z) = x + y$ , so $x + y = \frac{2}{10} - \frac{1}{15} - \frac{1}{20} = \frac{12}{60} - \frac{4}{60} - \frac{3}{60} = \frac{5}{60} = \frac{1}{12}$ without finding anyone separately.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and combining the three team equations.
7.NS.A.3 Solve real-world problems involving the four operations with rational numbers — Subtracting and adding the fractional daily shares.

💡Takeaway. Call the whole job

1

; then every team is just a sum of daily shares, and subtracting teams picks out individuals.

Variant 11 hard

A job takes $12$ days when A, B, and C work together, $20$ days when A and C work together, and $15$ days when B and C work together. When A and B work on this job together, what fraction of the whole job can they do in one day?

Your answer

Show solution

1 · Understandwhat's really being asked

Three team-ups of A, B, C finish a job in 12, 20, and 15 days. Find what fraction of the job A and B together finish in one day.

Givens

A, B, and C together: $12$ days.
A and C together: $20$ days.
B and C together: $15$ days.

Unknowns

The fraction of the job A and B do together in one day.

Constraints

Each person works at a steady rate, and rates of people working together add.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Call the whole job $1$ and let each person's daily share be an unknown. Each team-up then says what a sum of daily shares equals. Taking a two-person team away from the three-person team leaves one person's share.

3 · Execute4 carry out the plan

1Turn days into daily shares

#13 Convert to Algebra 8.EE.C.8

Let A, B, C do

x

,

y

,

z

of the job per day. A team that needs

12

days does

\frac{1}{12}

of the job each day.

x + y + z = \frac{1}{12},\quad x + z = \frac{1}{20},\quad y + z = \frac{1}{15}

Finishing in

n

days means doing

\frac{1}{n}

of the job per day.

2Find A's share

#7 Identify Subproblems 7.NS.A.3

Take the B-and-C team away from the whole team; what is left is A.

x = \frac{1}{12} - \frac{1}{15} = \frac{5}{60} - \frac{4}{60} = \frac{1}{60}

Whole team minus a sub-team leaves the missing member.

3Find B's share

#7 Identify Subproblems 7.NS.A.3

Take the A-and-C team away from the whole team; what is left is B.

y = \frac{1}{12} - \frac{1}{20} = \frac{5}{60} - \frac{3}{60} = \frac{2}{60} = \frac{1}{30}

Same subtraction, different sub-team.

4Add A and B

#13 Convert to Algebra 7.NS.A.3

Together A and B do the sum of their daily shares.

\frac{1}{60} + \frac{1}{30} = \frac{1}{60} + \frac{2}{60} = \frac{3}{60} = \frac{1}{20}

Rates of people working side by side add.

Answer:

\frac{1}{20}

4 · Reviewdoes it hold up?

C's share is $\frac{1}{20} - \frac{1}{60} = \frac{1}{30}$ . Then $\frac{1}{60} + \frac{2}{60} + \frac{2}{60} = \frac{5}{60} = \frac{1}{12}$ , matching the $12$ -day team. A and B alone would need $20$ days, more than the $12$ days with C's help, as expected.

Another way: Add the two pair equations and compare with twice the whole-team equation: $2(x + y + z) - (x + z) - (y + z) = x + y$ , so $x + y = \frac{2}{12} - \frac{1}{20} - \frac{1}{15} = \frac{10}{60} - \frac{3}{60} - \frac{4}{60} = \frac{3}{60} = \frac{1}{20}$ without finding anyone separately.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and combining the three team equations.
7.NS.A.3 Solve real-world problems involving the four operations with rational numbers — Subtracting and adding the fractional daily shares.

💡Takeaway. Call the whole job

1

; then every team is just a sum of daily shares, and subtracting teams picks out individuals.

Variant 12 hard

A job takes $12$ days when A, B, and C work together, $24$ days when A and C work together, and $18$ days when B and C work together. When A and B work on this job together, what fraction of the whole job can they do in one day?

Your answer

Show solution

1 · Understandwhat's really being asked

Three team-ups of A, B, C finish a job in 12, 24, and 18 days. Find what fraction of the job A and B together finish in one day.

Givens

A, B, and C together: $12$ days.
A and C together: $24$ days.
B and C together: $18$ days.

Unknowns

The fraction of the job A and B do together in one day.

Constraints

Each person works at a steady rate, and rates of people working together add.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#15 Organize Information in More Ways

Call the whole job $1$ and let each person's daily share be an unknown. Each team-up then says what a sum of daily shares equals. Taking a two-person team away from the three-person team leaves one person's share.

3 · Execute4 carry out the plan

1Turn days into daily shares

#13 Convert to Algebra 8.EE.C.8

Let A, B, C do

x

,

y

,

z

of the job per day. A team that needs

12

days does

\frac{1}{12}

of the job each day.

x + y + z = \frac{1}{12},\quad x + z = \frac{1}{24},\quad y + z = \frac{1}{18}

Finishing in

n

days means doing

\frac{1}{n}

of the job per day.

2Find A's share

#7 Identify Subproblems 7.NS.A.3

Take the B-and-C team away from the whole team; what is left is A.

x = \frac{1}{12} - \frac{1}{18} = \frac{3}{36} - \frac{2}{36} = \frac{1}{36}

Whole team minus a sub-team leaves the missing member.

3Find B's share

#7 Identify Subproblems 7.NS.A.3

Take the A-and-C team away from the whole team; what is left is B.

y = \frac{1}{12} - \frac{1}{24} = \frac{2}{24} - \frac{1}{24} = \frac{1}{24}

Same subtraction, different sub-team.

4Add A and B

#13 Convert to Algebra 7.NS.A.3

Together A and B do the sum of their daily shares.

\frac{1}{36} + \frac{1}{24} = \frac{2}{72} + \frac{3}{72} = \frac{5}{72}

Rates of people working side by side add.

Answer:

\frac{5}{72}

4 · Reviewdoes it hold up?

C's share is $\frac{1}{24} - \frac{1}{36} = \frac{1}{72}$ . Then $\frac{2}{72} + \frac{3}{72} + \frac{1}{72} = \frac{6}{72} = \frac{1}{12}$ , matching the $12$ -day team. A and B alone would need $\frac{72}{5} = 14\frac{2}{5}$ days, more than the $12$ days with C's help, as expected.

Another way: Add the two pair equations and compare with twice the whole-team equation: $2(x + y + z) - (x + z) - (y + z) = x + y$ , so $x + y = \frac{2}{12} - \frac{1}{24} - \frac{1}{18} = \frac{12}{72} - \frac{3}{72} - \frac{4}{72} = \frac{5}{72}$ without finding anyone separately.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and combining the three team equations.
7.NS.A.3 Solve real-world problems involving the four operations with rational numbers — Subtracting and adding the fractional daily shares.

💡Takeaway. Call the whole job

1

; then every team is just a sum of daily shares, and subtracting teams picks out individuals.

Subtract team rates to isolate one worker · 12 practice problems

Generated variants — 12

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