Practice · Solve in terms of a, then demand positivity · Sensim Math

Variant 1 easy

For the system $\begin{cases} x+y=3 \\ y+z=7 \\ z+x=a \end{cases}$ , the solution values $x$ , $y$ , $z$ are all positive. How many integers $a$ make this possible?

Your answer

Show solution

1 · Understandwhat's really being asked

Solve the system for $x$ , $y$ , $z$ in terms of $a$ . Then require each of them to be positive, which gives inequalities for $a$ , and count the integers that satisfy all of them.

Givens

$x+y=3$ , $y+z=7$ , $z+x=a$
$x$ , $y$ , $z$ are all positive.
$a$ is an integer.

Unknowns

The number of integers $a$ .

Constraints

All three positivity conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#2 Make a Systematic List

The system is symmetric: each variable appears in exactly two equations. Adding all three gives $x+y+z$ , and subtracting each equation from that total isolates one variable. Then positivity is three linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

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

Each of

x

,

y

,

z

appears twice on the left.

2(x+y+z)=3+7+a=10+a \;\Rightarrow\; x+y+z=5+\frac{a}{2}

The total of all three is the easiest thing to find first.

2Isolate each variable

#7 Identify Subproblems 8.EE.C.8

Subtract each original equation from the total.

x=\left(5+\frac{a}{2}\right)-7=\frac{a}{2}-2,\; y=\left(5+\frac{a}{2}\right)-a=5-\frac{a}{2},\; z=\left(5+\frac{a}{2}\right)-3=2+\frac{a}{2}

Removing the pair that leaves out a variable leaves exactly that variable.

3Require positivity

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

Set each expression greater than

0

and solve.

\frac{a}{2}-2>0 \Rightarrow a>4,\quad 5-\frac{a}{2}>0 \Rightarrow a<10,\quad 2+\frac{a}{2}>0 \Rightarrow a>-4

All three must hold, so the binding conditions are

a>4

and

a<10

:

4<a<10

.

4Count the integers

#2 Make a Systematic List 7.EE.B.4

The integers with

4<a<10

are

5, 6, \dots, 9

.

9-5+1=5

Both ends are excluded, so

4

and

10

are not counted.

Answer:

5

4 · Reviewdoes it hold up?

Check the extremes: $a=5$ gives $x=\frac{1}{2}$ , $y=2\frac{1}{2}$ , $z=4\frac{1}{2}$ , all positive; $a=4$ gives $x=0$ (not positive), and $a=10$ gives $y=0$ (not positive). So $5$ to $9$ is exactly right.

Another way: Reason without solving: $x=3-y$ and $z=7-y$ , so $a=x+z=10-2y$ . Since $x$ and $z$ are positive, $y<3$ , so $0<y<3$ . As $y$ runs over that range, $a=10-2y$ runs over $4<a<10$ , giving the same $5$ integers.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the system of linear equations for $x$, $y$, $z$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning "positive" into linear inequalities for $a$ and solving them.

💡Takeaway. First let the letter

a

ride along as if it were a number, then ask what it must be.

Variant 2 easy

For the system $\begin{cases} x+y=7 \\ y+z=3 \\ z+x=a \end{cases}$ , the solution values $x$ , $y$ , $z$ are all positive. How many integers $a$ make this possible?

Your answer

Show solution

1 · Understandwhat's really being asked

Solve the system for $x$ , $y$ , $z$ in terms of $a$ . Then require each of them to be positive, which gives inequalities for $a$ , and count the integers that satisfy all of them.

Givens

$x+y=7$ , $y+z=3$ , $z+x=a$
$x$ , $y$ , $z$ are all positive.
$a$ is an integer.

Unknowns

The number of integers $a$ .

Constraints

All three positivity conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#2 Make a Systematic List

The system is symmetric: each variable appears in exactly two equations. Adding all three gives $x+y+z$ , and subtracting each equation from that total isolates one variable. Then positivity is three linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

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

Each of

x

,

y

,

z

appears twice on the left.

2(x+y+z)=7+3+a=10+a \;\Rightarrow\; x+y+z=5+\frac{a}{2}

The total of all three is the easiest thing to find first.

2Isolate each variable

#7 Identify Subproblems 8.EE.C.8

Subtract each original equation from the total.

x=\left(5+\frac{a}{2}\right)-3=2+\frac{a}{2},\; y=\left(5+\frac{a}{2}\right)-a=5-\frac{a}{2},\; z=\left(5+\frac{a}{2}\right)-7=\frac{a}{2}-2

Removing the pair that leaves out a variable leaves exactly that variable.

3Require positivity

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

Set each expression greater than

0

and solve.

2+\frac{a}{2}>0 \Rightarrow a>-4,\quad 5-\frac{a}{2}>0 \Rightarrow a<10,\quad \frac{a}{2}-2>0 \Rightarrow a>4

All three must hold, so the binding conditions are

a>4

and

a<10

:

4<a<10

.

4Count the integers

#2 Make a Systematic List 7.EE.B.4

The integers with

4<a<10

are

5, 6, \dots, 9

.

9-5+1=5

Both ends are excluded, so

4

and

10

are not counted.

Answer:

5

4 · Reviewdoes it hold up?

Check the extremes: $a=5$ gives $x=4\frac{1}{2}$ , $y=2\frac{1}{2}$ , $z=\frac{1}{2}$ , all positive; $a=4$ gives $z=0$ (not positive), and $a=10$ gives $y=0$ (not positive). So $5$ to $9$ is exactly right.

Another way: Reason without solving: $x=7-y$ and $z=3-y$ , so $a=x+z=10-2y$ . Since $x$ and $z$ are positive, $y<3$ , so $0<y<3$ . As $y$ runs over that range, $a=10-2y$ runs over $4<a<10$ , giving the same $5$ integers.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the system of linear equations for $x$, $y$, $z$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning "positive" into linear inequalities for $a$ and solving them.

💡Takeaway. First let the letter

a

ride along as if it were a number, then ask what it must be.

Variant 3 easy

For the system $\begin{cases} x+y=4 \\ y+z=8 \\ z+x=a \end{cases}$ , the solution values $x$ , $y$ , $z$ are all positive. How many integers $a$ make this possible?

Your answer

Show solution

1 · Understandwhat's really being asked

Solve the system for $x$ , $y$ , $z$ in terms of $a$ . Then require each of them to be positive, which gives inequalities for $a$ , and count the integers that satisfy all of them.

Givens

$x+y=4$ , $y+z=8$ , $z+x=a$
$x$ , $y$ , $z$ are all positive.
$a$ is an integer.

Unknowns

The number of integers $a$ .

Constraints

All three positivity conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#2 Make a Systematic List

The system is symmetric: each variable appears in exactly two equations. Adding all three gives $x+y+z$ , and subtracting each equation from that total isolates one variable. Then positivity is three linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

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

Each of

x

,

y

,

z

appears twice on the left.

2(x+y+z)=4+8+a=12+a \;\Rightarrow\; x+y+z=6+\frac{a}{2}

The total of all three is the easiest thing to find first.

2Isolate each variable

#7 Identify Subproblems 8.EE.C.8

Subtract each original equation from the total.

x=\left(6+\frac{a}{2}\right)-8=\frac{a}{2}-2,\; y=\left(6+\frac{a}{2}\right)-a=6-\frac{a}{2},\; z=\left(6+\frac{a}{2}\right)-4=2+\frac{a}{2}

Removing the pair that leaves out a variable leaves exactly that variable.

3Require positivity

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

Set each expression greater than

0

and solve.

\frac{a}{2}-2>0 \Rightarrow a>4,\quad 6-\frac{a}{2}>0 \Rightarrow a<12,\quad 2+\frac{a}{2}>0 \Rightarrow a>-4

All three must hold, so the binding conditions are

a>4

and

a<12

:

4<a<12

.

4Count the integers

#2 Make a Systematic List 7.EE.B.4

The integers with

4<a<12

are

5, 6, \dots, 11

.

11-5+1=7

Both ends are excluded, so

4

and

12

are not counted.

Answer:

7

4 · Reviewdoes it hold up?

Check the extremes: $a=5$ gives $x=\frac{1}{2}$ , $y=3\frac{1}{2}$ , $z=4\frac{1}{2}$ , all positive; $a=4$ gives $x=0$ (not positive), and $a=12$ gives $y=0$ (not positive). So $5$ to $11$ is exactly right.

Another way: Reason without solving: $x=4-y$ and $z=8-y$ , so $a=x+z=12-2y$ . Since $x$ and $z$ are positive, $y<4$ , so $0<y<4$ . As $y$ runs over that range, $a=12-2y$ runs over $4<a<12$ , giving the same $7$ integers.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the system of linear equations for $x$, $y$, $z$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning "positive" into linear inequalities for $a$ and solving them.

💡Takeaway. First let the letter

a

ride along as if it were a number, then ask what it must be.

Variant 4 easy

For the system $\begin{cases} x+y=5 \\ y+z=9 \\ z+x=a \end{cases}$ , the solution values $x$ , $y$ , $z$ are all positive. How many integers $a$ make this possible?

Your answer

Show solution

1 · Understandwhat's really being asked

Solve the system for $x$ , $y$ , $z$ in terms of $a$ . Then require each of them to be positive, which gives inequalities for $a$ , and count the integers that satisfy all of them.

Givens

$x+y=5$ , $y+z=9$ , $z+x=a$
$x$ , $y$ , $z$ are all positive.
$a$ is an integer.

Unknowns

The number of integers $a$ .

Constraints

All three positivity conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#2 Make a Systematic List

The system is symmetric: each variable appears in exactly two equations. Adding all three gives $x+y+z$ , and subtracting each equation from that total isolates one variable. Then positivity is three linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

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

Each of

x

,

y

,

z

appears twice on the left.

2(x+y+z)=5+9+a=14+a \;\Rightarrow\; x+y+z=7+\frac{a}{2}

The total of all three is the easiest thing to find first.

2Isolate each variable

#7 Identify Subproblems 8.EE.C.8

Subtract each original equation from the total.

x=\left(7+\frac{a}{2}\right)-9=\frac{a}{2}-2,\; y=\left(7+\frac{a}{2}\right)-a=7-\frac{a}{2},\; z=\left(7+\frac{a}{2}\right)-5=2+\frac{a}{2}

Removing the pair that leaves out a variable leaves exactly that variable.

3Require positivity

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

Set each expression greater than

0

and solve.

\frac{a}{2}-2>0 \Rightarrow a>4,\quad 7-\frac{a}{2}>0 \Rightarrow a<14,\quad 2+\frac{a}{2}>0 \Rightarrow a>-4

All three must hold, so the binding conditions are

a>4

and

a<14

:

4<a<14

.

4Count the integers

#2 Make a Systematic List 7.EE.B.4

The integers with

4<a<14

are

5, 6, \dots, 13

.

13-5+1=9

Both ends are excluded, so

4

and

14

are not counted.

Answer:

9

4 · Reviewdoes it hold up?

Check the extremes: $a=5$ gives $x=\frac{1}{2}$ , $y=4\frac{1}{2}$ , $z=4\frac{1}{2}$ , all positive; $a=4$ gives $x=0$ (not positive), and $a=14$ gives $y=0$ (not positive). So $5$ to $13$ is exactly right.

Another way: Reason without solving: $x=5-y$ and $z=9-y$ , so $a=x+z=14-2y$ . Since $x$ and $z$ are positive, $y<5$ , so $0<y<5$ . As $y$ runs over that range, $a=14-2y$ runs over $4<a<14$ , giving the same $9$ integers.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the system of linear equations for $x$, $y$, $z$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning "positive" into linear inequalities for $a$ and solving them.

💡Takeaway. First let the letter

a

ride along as if it were a number, then ask what it must be.

Variant 5 medium

For the system $\begin{cases} x+y=10 \\ y+z=6 \\ z+x=a \end{cases}$ , the solution values $x$ , $y$ , $z$ are all positive. How many integers $a$ make this possible?

Your answer

Show solution

1 · Understandwhat's really being asked

Solve the system for $x$ , $y$ , $z$ in terms of $a$ . Then require each of them to be positive, which gives inequalities for $a$ , and count the integers that satisfy all of them.

Givens

$x+y=10$ , $y+z=6$ , $z+x=a$
$x$ , $y$ , $z$ are all positive.
$a$ is an integer.

Unknowns

The number of integers $a$ .

Constraints

All three positivity conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#2 Make a Systematic List

The system is symmetric: each variable appears in exactly two equations. Adding all three gives $x+y+z$ , and subtracting each equation from that total isolates one variable. Then positivity is three linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

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

Each of

x

,

y

,

z

appears twice on the left.

2(x+y+z)=10+6+a=16+a \;\Rightarrow\; x+y+z=8+\frac{a}{2}

The total of all three is the easiest thing to find first.

2Isolate each variable

#7 Identify Subproblems 8.EE.C.8

Subtract each original equation from the total.

x=\left(8+\frac{a}{2}\right)-6=2+\frac{a}{2},\; y=\left(8+\frac{a}{2}\right)-a=8-\frac{a}{2},\; z=\left(8+\frac{a}{2}\right)-10=\frac{a}{2}-2

Removing the pair that leaves out a variable leaves exactly that variable.

3Require positivity

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

Set each expression greater than

0

and solve.

2+\frac{a}{2}>0 \Rightarrow a>-4,\quad 8-\frac{a}{2}>0 \Rightarrow a<16,\quad \frac{a}{2}-2>0 \Rightarrow a>4

All three must hold, so the binding conditions are

a>4

and

a<16

:

4<a<16

.

4Count the integers

#2 Make a Systematic List 7.EE.B.4

The integers with

4<a<16

are

5, 6, \dots, 15

.

15-5+1=11

Both ends are excluded, so

4

and

16

are not counted.

Answer:

11

4 · Reviewdoes it hold up?

Check the extremes: $a=5$ gives $x=4\frac{1}{2}$ , $y=5\frac{1}{2}$ , $z=\frac{1}{2}$ , all positive; $a=4$ gives $z=0$ (not positive), and $a=16$ gives $y=0$ (not positive). So $5$ to $15$ is exactly right.

Another way: Reason without solving: $x=10-y$ and $z=6-y$ , so $a=x+z=16-2y$ . Since $x$ and $z$ are positive, $y<6$ , so $0<y<6$ . As $y$ runs over that range, $a=16-2y$ runs over $4<a<16$ , giving the same $11$ integers.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the system of linear equations for $x$, $y$, $z$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning "positive" into linear inequalities for $a$ and solving them.

💡Takeaway. First let the letter

a

ride along as if it were a number, then ask what it must be.

Variant 6 medium

For the system $\begin{cases} x+y=2 \\ y+z=10 \\ z+x=a \end{cases}$ , the solution values $x$ , $y$ , $z$ are all positive. How many integers $a$ make this possible?

Your answer

Show solution

1 · Understandwhat's really being asked

Solve the system for $x$ , $y$ , $z$ in terms of $a$ . Then require each of them to be positive, which gives inequalities for $a$ , and count the integers that satisfy all of them.

Givens

$x+y=2$ , $y+z=10$ , $z+x=a$
$x$ , $y$ , $z$ are all positive.
$a$ is an integer.

Unknowns

The number of integers $a$ .

Constraints

All three positivity conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#2 Make a Systematic List

The system is symmetric: each variable appears in exactly two equations. Adding all three gives $x+y+z$ , and subtracting each equation from that total isolates one variable. Then positivity is three linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

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

Each of

x

,

y

,

z

appears twice on the left.

2(x+y+z)=2+10+a=12+a \;\Rightarrow\; x+y+z=6+\frac{a}{2}

The total of all three is the easiest thing to find first.

2Isolate each variable

#7 Identify Subproblems 8.EE.C.8

Subtract each original equation from the total.

x=\left(6+\frac{a}{2}\right)-10=\frac{a}{2}-4,\; y=\left(6+\frac{a}{2}\right)-a=6-\frac{a}{2},\; z=\left(6+\frac{a}{2}\right)-2=4+\frac{a}{2}

Removing the pair that leaves out a variable leaves exactly that variable.

3Require positivity

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

Set each expression greater than

0

and solve.

\frac{a}{2}-4>0 \Rightarrow a>8,\quad 6-\frac{a}{2}>0 \Rightarrow a<12,\quad 4+\frac{a}{2}>0 \Rightarrow a>-8

All three must hold, so the binding conditions are

a>8

and

a<12

:

8<a<12

.

4Count the integers

#2 Make a Systematic List 7.EE.B.4

The integers with

8<a<12

are

9, 10, \dots, 11

.

11-9+1=3

Both ends are excluded, so

8

and

12

are not counted.

Answer:

3

4 · Reviewdoes it hold up?

Check the extremes: $a=9$ gives $x=\frac{1}{2}$ , $y=1\frac{1}{2}$ , $z=8\frac{1}{2}$ , all positive; $a=8$ gives $x=0$ (not positive), and $a=12$ gives $y=0$ (not positive). So $9$ to $11$ is exactly right.

Another way: Reason without solving: $x=2-y$ and $z=10-y$ , so $a=x+z=12-2y$ . Since $x$ and $z$ are positive, $y<2$ , so $0<y<2$ . As $y$ runs over that range, $a=12-2y$ runs over $8<a<12$ , giving the same $3$ integers.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the system of linear equations for $x$, $y$, $z$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning "positive" into linear inequalities for $a$ and solving them.

💡Takeaway. First let the letter

a

ride along as if it were a number, then ask what it must be.

Variant 7 medium

For the system $\begin{cases} x+y=6 \\ y+z=10 \\ z+x=a \end{cases}$ , the solution values $x$ , $y$ , $z$ are all positive. How many integers $a$ make this possible?

Your answer

Show solution

1 · Understandwhat's really being asked

Solve the system for $x$ , $y$ , $z$ in terms of $a$ . Then require each of them to be positive, which gives inequalities for $a$ , and count the integers that satisfy all of them.

Givens

$x+y=6$ , $y+z=10$ , $z+x=a$
$x$ , $y$ , $z$ are all positive.
$a$ is an integer.

Unknowns

The number of integers $a$ .

Constraints

All three positivity conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#2 Make a Systematic List

The system is symmetric: each variable appears in exactly two equations. Adding all three gives $x+y+z$ , and subtracting each equation from that total isolates one variable. Then positivity is three linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

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

Each of

x

,

y

,

z

appears twice on the left.

2(x+y+z)=6+10+a=16+a \;\Rightarrow\; x+y+z=8+\frac{a}{2}

The total of all three is the easiest thing to find first.

2Isolate each variable

#7 Identify Subproblems 8.EE.C.8

Subtract each original equation from the total.

x=\left(8+\frac{a}{2}\right)-10=\frac{a}{2}-2,\; y=\left(8+\frac{a}{2}\right)-a=8-\frac{a}{2},\; z=\left(8+\frac{a}{2}\right)-6=2+\frac{a}{2}

Removing the pair that leaves out a variable leaves exactly that variable.

3Require positivity

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

Set each expression greater than

0

and solve.

\frac{a}{2}-2>0 \Rightarrow a>4,\quad 8-\frac{a}{2}>0 \Rightarrow a<16,\quad 2+\frac{a}{2}>0 \Rightarrow a>-4

All three must hold, so the binding conditions are

a>4

and

a<16

:

4<a<16

.

4Count the integers

#2 Make a Systematic List 7.EE.B.4

The integers with

4<a<16

are

5, 6, \dots, 15

.

15-5+1=11

Both ends are excluded, so

4

and

16

are not counted.

Answer:

11

4 · Reviewdoes it hold up?

Check the extremes: $a=5$ gives $x=\frac{1}{2}$ , $y=5\frac{1}{2}$ , $z=4\frac{1}{2}$ , all positive; $a=4$ gives $x=0$ (not positive), and $a=16$ gives $y=0$ (not positive). So $5$ to $15$ is exactly right.

Another way: Reason without solving: $x=6-y$ and $z=10-y$ , so $a=x+z=16-2y$ . Since $x$ and $z$ are positive, $y<6$ , so $0<y<6$ . As $y$ runs over that range, $a=16-2y$ runs over $4<a<16$ , giving the same $11$ integers.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the system of linear equations for $x$, $y$, $z$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning "positive" into linear inequalities for $a$ and solving them.

💡Takeaway. First let the letter

a

ride along as if it were a number, then ask what it must be.

Variant 8 hard

For the system $\begin{cases} x+y=11 \\ y+z=7 \\ z+x=a \end{cases}$ , the solution values $x$ , $y$ , $z$ are all positive. How many integers $a$ make this possible?

Your answer

Show solution

1 · Understandwhat's really being asked

Solve the system for $x$ , $y$ , $z$ in terms of $a$ . Then require each of them to be positive, which gives inequalities for $a$ , and count the integers that satisfy all of them.

Givens

$x+y=11$ , $y+z=7$ , $z+x=a$
$x$ , $y$ , $z$ are all positive.
$a$ is an integer.

Unknowns

The number of integers $a$ .

Constraints

All three positivity conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#2 Make a Systematic List

The system is symmetric: each variable appears in exactly two equations. Adding all three gives $x+y+z$ , and subtracting each equation from that total isolates one variable. Then positivity is three linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

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

Each of

x

,

y

,

z

appears twice on the left.

2(x+y+z)=11+7+a=18+a \;\Rightarrow\; x+y+z=9+\frac{a}{2}

The total of all three is the easiest thing to find first.

2Isolate each variable

#7 Identify Subproblems 8.EE.C.8

Subtract each original equation from the total.

x=\left(9+\frac{a}{2}\right)-7=2+\frac{a}{2},\; y=\left(9+\frac{a}{2}\right)-a=9-\frac{a}{2},\; z=\left(9+\frac{a}{2}\right)-11=\frac{a}{2}-2

Removing the pair that leaves out a variable leaves exactly that variable.

3Require positivity

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

Set each expression greater than

0

and solve.

2+\frac{a}{2}>0 \Rightarrow a>-4,\quad 9-\frac{a}{2}>0 \Rightarrow a<18,\quad \frac{a}{2}-2>0 \Rightarrow a>4

All three must hold, so the binding conditions are

a>4

and

a<18

:

4<a<18

.

4Count the integers

#2 Make a Systematic List 7.EE.B.4

The integers with

4<a<18

are

5, 6, \dots, 17

.

17-5+1=13

Both ends are excluded, so

4

and

18

are not counted.

Answer:

13

4 · Reviewdoes it hold up?

Check the extremes: $a=5$ gives $x=4\frac{1}{2}$ , $y=6\frac{1}{2}$ , $z=\frac{1}{2}$ , all positive; $a=4$ gives $z=0$ (not positive), and $a=18$ gives $y=0$ (not positive). So $5$ to $17$ is exactly right.

Another way: Reason without solving: $x=11-y$ and $z=7-y$ , so $a=x+z=18-2y$ . Since $x$ and $z$ are positive, $y<7$ , so $0<y<7$ . As $y$ runs over that range, $a=18-2y$ runs over $4<a<18$ , giving the same $13$ integers.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the system of linear equations for $x$, $y$, $z$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning "positive" into linear inequalities for $a$ and solving them.

💡Takeaway. First let the letter

a

ride along as if it were a number, then ask what it must be.

Variant 9 medium

For the system $\begin{cases} x+y=5 \\ y+z=11 \\ z+x=a \end{cases}$ , the solution values $x$ , $y$ , $z$ are all positive. How many integers $a$ make this possible?

Your answer

Show solution

1 · Understandwhat's really being asked

Solve the system for $x$ , $y$ , $z$ in terms of $a$ . Then require each of them to be positive, which gives inequalities for $a$ , and count the integers that satisfy all of them.

Givens

$x+y=5$ , $y+z=11$ , $z+x=a$
$x$ , $y$ , $z$ are all positive.
$a$ is an integer.

Unknowns

The number of integers $a$ .

Constraints

All three positivity conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#2 Make a Systematic List

The system is symmetric: each variable appears in exactly two equations. Adding all three gives $x+y+z$ , and subtracting each equation from that total isolates one variable. Then positivity is three linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

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

Each of

x

,

y

,

z

appears twice on the left.

2(x+y+z)=5+11+a=16+a \;\Rightarrow\; x+y+z=8+\frac{a}{2}

The total of all three is the easiest thing to find first.

2Isolate each variable

#7 Identify Subproblems 8.EE.C.8

Subtract each original equation from the total.

x=\left(8+\frac{a}{2}\right)-11=\frac{a}{2}-3,\; y=\left(8+\frac{a}{2}\right)-a=8-\frac{a}{2},\; z=\left(8+\frac{a}{2}\right)-5=3+\frac{a}{2}

Removing the pair that leaves out a variable leaves exactly that variable.

3Require positivity

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

Set each expression greater than

0

and solve.

\frac{a}{2}-3>0 \Rightarrow a>6,\quad 8-\frac{a}{2}>0 \Rightarrow a<16,\quad 3+\frac{a}{2}>0 \Rightarrow a>-6

All three must hold, so the binding conditions are

a>6

and

a<16

:

6<a<16

.

4Count the integers

#2 Make a Systematic List 7.EE.B.4

The integers with

6<a<16

are

7, 8, \dots, 15

.

15-7+1=9

Both ends are excluded, so

6

and

16

are not counted.

Answer:

9

4 · Reviewdoes it hold up?

Check the extremes: $a=7$ gives $x=\frac{1}{2}$ , $y=4\frac{1}{2}$ , $z=6\frac{1}{2}$ , all positive; $a=6$ gives $x=0$ (not positive), and $a=16$ gives $y=0$ (not positive). So $7$ to $15$ is exactly right.

Another way: Reason without solving: $x=5-y$ and $z=11-y$ , so $a=x+z=16-2y$ . Since $x$ and $z$ are positive, $y<5$ , so $0<y<5$ . As $y$ runs over that range, $a=16-2y$ runs over $6<a<16$ , giving the same $9$ integers.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the system of linear equations for $x$, $y$, $z$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning "positive" into linear inequalities for $a$ and solving them.

💡Takeaway. First let the letter

a

ride along as if it were a number, then ask what it must be.

Variant 10 hard

For the system $\begin{cases} x+y=8 \\ y+z=12 \\ z+x=a \end{cases}$ , the solution values $x$ , $y$ , $z$ are all positive. How many integers $a$ make this possible?

Your answer

Show solution

1 · Understandwhat's really being asked

Solve the system for $x$ , $y$ , $z$ in terms of $a$ . Then require each of them to be positive, which gives inequalities for $a$ , and count the integers that satisfy all of them.

Givens

$x+y=8$ , $y+z=12$ , $z+x=a$
$x$ , $y$ , $z$ are all positive.
$a$ is an integer.

Unknowns

The number of integers $a$ .

Constraints

All three positivity conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#2 Make a Systematic List

The system is symmetric: each variable appears in exactly two equations. Adding all three gives $x+y+z$ , and subtracting each equation from that total isolates one variable. Then positivity is three linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

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

Each of

x

,

y

,

z

appears twice on the left.

2(x+y+z)=8+12+a=20+a \;\Rightarrow\; x+y+z=10+\frac{a}{2}

The total of all three is the easiest thing to find first.

2Isolate each variable

#7 Identify Subproblems 8.EE.C.8

Subtract each original equation from the total.

x=\left(10+\frac{a}{2}\right)-12=\frac{a}{2}-2,\; y=\left(10+\frac{a}{2}\right)-a=10-\frac{a}{2},\; z=\left(10+\frac{a}{2}\right)-8=2+\frac{a}{2}

Removing the pair that leaves out a variable leaves exactly that variable.

3Require positivity

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

Set each expression greater than

0

and solve.

\frac{a}{2}-2>0 \Rightarrow a>4,\quad 10-\frac{a}{2}>0 \Rightarrow a<20,\quad 2+\frac{a}{2}>0 \Rightarrow a>-4

All three must hold, so the binding conditions are

a>4

and

a<20

:

4<a<20

.

4Count the integers

#2 Make a Systematic List 7.EE.B.4

The integers with

4<a<20

are

5, 6, \dots, 19

.

19-5+1=15

Both ends are excluded, so

4

and

20

are not counted.

Answer:

15

4 · Reviewdoes it hold up?

Check the extremes: $a=5$ gives $x=\frac{1}{2}$ , $y=7\frac{1}{2}$ , $z=4\frac{1}{2}$ , all positive; $a=4$ gives $x=0$ (not positive), and $a=20$ gives $y=0$ (not positive). So $5$ to $19$ is exactly right.

Another way: Reason without solving: $x=8-y$ and $z=12-y$ , so $a=x+z=20-2y$ . Since $x$ and $z$ are positive, $y<8$ , so $0<y<8$ . As $y$ runs over that range, $a=20-2y$ runs over $4<a<20$ , giving the same $15$ integers.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the system of linear equations for $x$, $y$, $z$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning "positive" into linear inequalities for $a$ and solving them.

💡Takeaway. First let the letter

a

ride along as if it were a number, then ask what it must be.

Variant 11 hard

For the system $\begin{cases} x+y=12 \\ y+z=4 \\ z+x=a \end{cases}$ , the solution values $x$ , $y$ , $z$ are all positive. How many integers $a$ make this possible?

Your answer

Show solution

1 · Understandwhat's really being asked

Solve the system for $x$ , $y$ , $z$ in terms of $a$ . Then require each of them to be positive, which gives inequalities for $a$ , and count the integers that satisfy all of them.

Givens

$x+y=12$ , $y+z=4$ , $z+x=a$
$x$ , $y$ , $z$ are all positive.
$a$ is an integer.

Unknowns

The number of integers $a$ .

Constraints

All three positivity conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#2 Make a Systematic List

The system is symmetric: each variable appears in exactly two equations. Adding all three gives $x+y+z$ , and subtracting each equation from that total isolates one variable. Then positivity is three linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

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

Each of

x

,

y

,

z

appears twice on the left.

2(x+y+z)=12+4+a=16+a \;\Rightarrow\; x+y+z=8+\frac{a}{2}

The total of all three is the easiest thing to find first.

2Isolate each variable

#7 Identify Subproblems 8.EE.C.8

Subtract each original equation from the total.

x=\left(8+\frac{a}{2}\right)-4=4+\frac{a}{2},\; y=\left(8+\frac{a}{2}\right)-a=8-\frac{a}{2},\; z=\left(8+\frac{a}{2}\right)-12=\frac{a}{2}-4

Removing the pair that leaves out a variable leaves exactly that variable.

3Require positivity

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

Set each expression greater than

0

and solve.

4+\frac{a}{2}>0 \Rightarrow a>-8,\quad 8-\frac{a}{2}>0 \Rightarrow a<16,\quad \frac{a}{2}-4>0 \Rightarrow a>8

All three must hold, so the binding conditions are

a>8

and

a<16

:

8<a<16

.

4Count the integers

#2 Make a Systematic List 7.EE.B.4

The integers with

8<a<16

are

9, 10, \dots, 15

.

15-9+1=7

Both ends are excluded, so

8

and

16

are not counted.

Answer:

7

4 · Reviewdoes it hold up?

Check the extremes: $a=9$ gives $x=8\frac{1}{2}$ , $y=3\frac{1}{2}$ , $z=\frac{1}{2}$ , all positive; $a=8$ gives $z=0$ (not positive), and $a=16$ gives $y=0$ (not positive). So $9$ to $15$ is exactly right.

Another way: Reason without solving: $x=12-y$ and $z=4-y$ , so $a=x+z=16-2y$ . Since $x$ and $z$ are positive, $y<4$ , so $0<y<4$ . As $y$ runs over that range, $a=16-2y$ runs over $8<a<16$ , giving the same $7$ integers.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the system of linear equations for $x$, $y$, $z$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning "positive" into linear inequalities for $a$ and solving them.

💡Takeaway. First let the letter

a

ride along as if it were a number, then ask what it must be.

Variant 12 hard

For the system $\begin{cases} x+y=9 \\ y+z=13 \\ z+x=a \end{cases}$ , the solution values $x$ , $y$ , $z$ are all positive. How many integers $a$ make this possible?

Your answer

Show solution

1 · Understandwhat's really being asked

Solve the system for $x$ , $y$ , $z$ in terms of $a$ . Then require each of them to be positive, which gives inequalities for $a$ , and count the integers that satisfy all of them.

Givens

$x+y=9$ , $y+z=13$ , $z+x=a$
$x$ , $y$ , $z$ are all positive.
$a$ is an integer.

Unknowns

The number of integers $a$ .

Constraints

All three positivity conditions must hold at once.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #7 Identify Subproblems#2 Make a Systematic List

The system is symmetric: each variable appears in exactly two equations. Adding all three gives $x+y+z$ , and subtracting each equation from that total isolates one variable. Then positivity is three linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

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

Each of

x

,

y

,

z

appears twice on the left.

2(x+y+z)=9+13+a=22+a \;\Rightarrow\; x+y+z=11+\frac{a}{2}

The total of all three is the easiest thing to find first.

2Isolate each variable

#7 Identify Subproblems 8.EE.C.8

Subtract each original equation from the total.

x=\left(11+\frac{a}{2}\right)-13=\frac{a}{2}-2,\; y=\left(11+\frac{a}{2}\right)-a=11-\frac{a}{2},\; z=\left(11+\frac{a}{2}\right)-9=2+\frac{a}{2}

Removing the pair that leaves out a variable leaves exactly that variable.

3Require positivity

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

Set each expression greater than

0

and solve.

\frac{a}{2}-2>0 \Rightarrow a>4,\quad 11-\frac{a}{2}>0 \Rightarrow a<22,\quad 2+\frac{a}{2}>0 \Rightarrow a>-4

All three must hold, so the binding conditions are

a>4

and

a<22

:

4<a<22

.

4Count the integers

#2 Make a Systematic List 7.EE.B.4

The integers with

4<a<22

are

5, 6, \dots, 21

.

21-5+1=17

Both ends are excluded, so

4

and

22

are not counted.

Answer:

17

4 · Reviewdoes it hold up?

Check the extremes: $a=5$ gives $x=\frac{1}{2}$ , $y=8\frac{1}{2}$ , $z=4\frac{1}{2}$ , all positive; $a=4$ gives $x=0$ (not positive), and $a=22$ gives $y=0$ (not positive). So $5$ to $21$ is exactly right.

Another way: Reason without solving: $x=9-y$ and $z=13-y$ , so $a=x+z=22-2y$ . Since $x$ and $z$ are positive, $y<9$ , so $0<y<9$ . As $y$ runs over that range, $a=22-2y$ runs over $4<a<22$ , giving the same $17$ integers.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the system of linear equations for $x$, $y$, $z$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning "positive" into linear inequalities for $a$ and solving them.

💡Takeaway. First let the letter

a

ride along as if it were a number, then ask what it must be.

Solve in terms of a, then demand positivity · 12 practice problems

Generated variants — 12

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