Practice · Write the solution in terms of the constant · Sensim Math

Variant 1 easy

The values of $x$ and $y$ that satisfy the system of equations $x+y=8$ , $y+z=4$ , $z+x=a$ are both positive. Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three equations link $x$ , $y$ , $z$ and a constant $a$ . We want every value of $a$ for which the solution has $x>0$ and $y>0$ .

Givens

$x+y=8$
$y+z=4$
$z+x=a$
$x$ and $y$ are positive.

Unknowns

The range of values of $a$ .

Constraints

Only $x$ and $y$ must be positive; $z$ has no sign condition.

2 · Planchoose the strategy

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

Each variable appears in exactly two of the three equations, so adding all three gives the total $x+y+z$ at once. Subtracting each original equation from that total isolates one variable in terms of $a$ . The positivity conditions then become two linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

#15 Organize Information in More Ways 8.EE.C.8

Every variable appears twice in the sum.

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

The symmetric pattern makes the total easy to find.

2Isolate $x$ and $y$

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

Subtract

y+z=4

from the total to get

x

, and subtract

z+x=a

to get

y

.

x = 6 + \frac{a}{2} - 4 = 2 + \frac{a}{2}, \quad y = 6 + \frac{a}{2} - a = 6 - \frac{a}{2}

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

3Require $x > 0$

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

Solve the inequality for

a

.

2 + \frac{a}{2} > 0 \;\Rightarrow\; \frac{a}{2} > -2 \;\Rightarrow\; a > -4

Multiplying by the positive number

2

keeps the inequality direction.

4Require $y > 0$ and combine

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

Solve the second inequality, then keep the values of

a

that satisfy both.

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

Both conditions must hold at once, so the answer is the overlap.

Answer:

-4<a<12

4 · Reviewdoes it hold up?

Try $a=0$ , inside the range: $x=2$ , $y=6$ , $z=-2$ , and indeed $2+6=8$ , $6+(-2)=4$ , $-2+2=0$ . At the endpoints $a=-4$ gives $x=0$ and $a=12$ gives $y=0$ , which are not positive, so the endpoints are excluded.

Another way: Solve step by step instead: from $x+y=8$ and $y+z=4$ , subtracting gives $x-z=4$ ; adding $z+x=a$ gives $2x=4+a$ , so $x=2 + \frac{a}{2}$ , and then $y=8-x=6-\frac{a}{2}$ . The same inequalities follow.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the three-equation linear system for $x$ and $y$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning the conditions $x>0$ and $y>0$ into inequalities in $a$ and solving them.

💡Takeaway. Express the solution in terms of the constant first; then any condition on the solution becomes a condition on the constant.

Variant 2 easy

The values of $x$ and $y$ that satisfy the system of equations $x+y=9$ , $y+z=5$ , $z+x=a$ are both positive. Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three equations link $x$ , $y$ , $z$ and a constant $a$ . We want every value of $a$ for which the solution has $x>0$ and $y>0$ .

Givens

$x+y=9$
$y+z=5$
$z+x=a$
$x$ and $y$ are positive.

Unknowns

The range of values of $a$ .

Constraints

Only $x$ and $y$ must be positive; $z$ has no sign condition.

2 · Planchoose the strategy

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

Each variable appears in exactly two of the three equations, so adding all three gives the total $x+y+z$ at once. Subtracting each original equation from that total isolates one variable in terms of $a$ . The positivity conditions then become two linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

#15 Organize Information in More Ways 8.EE.C.8

Every variable appears twice in the sum.

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

The symmetric pattern makes the total easy to find.

2Isolate $x$ and $y$

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

Subtract

y+z=5

from the total to get

x

, and subtract

z+x=a

to get

y

.

x = 7 + \frac{a}{2} - 5 = 2 + \frac{a}{2}, \quad y = 7 + \frac{a}{2} - a = 7 - \frac{a}{2}

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

3Require $x > 0$

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

Solve the inequality for

a

.

2 + \frac{a}{2} > 0 \;\Rightarrow\; \frac{a}{2} > -2 \;\Rightarrow\; a > -4

Multiplying by the positive number

2

keeps the inequality direction.

4Require $y > 0$ and combine

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

Solve the second inequality, then keep the values of

a

that satisfy both.

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

Both conditions must hold at once, so the answer is the overlap.

Answer:

-4<a<14

4 · Reviewdoes it hold up?

Try $a=0$ , inside the range: $x=2$ , $y=7$ , $z=-2$ , and indeed $2+7=9$ , $7+(-2)=5$ , $-2+2=0$ . At the endpoints $a=-4$ gives $x=0$ and $a=14$ gives $y=0$ , which are not positive, so the endpoints are excluded.

Another way: Solve step by step instead: from $x+y=9$ and $y+z=5$ , subtracting gives $x-z=4$ ; adding $z+x=a$ gives $2x=4+a$ , so $x=2 + \frac{a}{2}$ , and then $y=9-x=7-\frac{a}{2}$ . The same inequalities follow.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the three-equation linear system for $x$ and $y$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning the conditions $x>0$ and $y>0$ into inequalities in $a$ and solving them.

💡Takeaway. Express the solution in terms of the constant first; then any condition on the solution becomes a condition on the constant.

Variant 3 easy

The values of $x$ and $y$ that satisfy the system of equations $x+y=5$ , $y+z=9$ , $z+x=a$ are both positive. Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three equations link $x$ , $y$ , $z$ and a constant $a$ . We want every value of $a$ for which the solution has $x>0$ and $y>0$ .

Givens

$x+y=5$
$y+z=9$
$z+x=a$
$x$ and $y$ are positive.

Unknowns

The range of values of $a$ .

Constraints

Only $x$ and $y$ must be positive; $z$ has no sign condition.

2 · Planchoose the strategy

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

Each variable appears in exactly two of the three equations, so adding all three gives the total $x+y+z$ at once. Subtracting each original equation from that total isolates one variable in terms of $a$ . The positivity conditions then become two linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

#15 Organize Information in More Ways 8.EE.C.8

Every variable appears twice in the sum.

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

The symmetric pattern makes the total easy to find.

2Isolate $x$ and $y$

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

Subtract

y+z=9

from the total to get

x

, and subtract

z+x=a

to get

y

.

x = 7 + \frac{a}{2} - 9 = -2 + \frac{a}{2}, \quad y = 7 + \frac{a}{2} - a = 7 - \frac{a}{2}

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

3Require $x > 0$

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

Solve the inequality for

a

.

-2 + \frac{a}{2} > 0 \;\Rightarrow\; \frac{a}{2} > 2 \;\Rightarrow\; a > 4

Multiplying by the positive number

2

keeps the inequality direction.

4Require $y > 0$ and combine

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

Solve the second inequality, then keep the values of

a

that satisfy both.

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

Both conditions must hold at once, so the answer is the overlap.

Answer:

4<a<14

4 · Reviewdoes it hold up?

Try $a=6$ , inside the range: $x=1$ , $y=4$ , $z=5$ , and indeed $1+4=5$ , $4+5=9$ , $5+1=6$ . At the endpoints $a=4$ gives $x=0$ and $a=14$ gives $y=0$ , which are not positive, so the endpoints are excluded.

Another way: Solve step by step instead: from $x+y=5$ and $y+z=9$ , subtracting gives $x-z=-4$ ; adding $z+x=a$ gives $2x=-4+a$ , so $x=-2 + \frac{a}{2}$ , and then $y=5-x=7-\frac{a}{2}$ . The same inequalities follow.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the three-equation linear system for $x$ and $y$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning the conditions $x>0$ and $y>0$ into inequalities in $a$ and solving them.

💡Takeaway. Express the solution in terms of the constant first; then any condition on the solution becomes a condition on the constant.

Variant 4 medium

The values of $x$ and $y$ that satisfy the system of equations $x+y=6$ , $y+z=10$ , $z+x=a$ are both positive. Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three equations link $x$ , $y$ , $z$ and a constant $a$ . We want every value of $a$ for which the solution has $x>0$ and $y>0$ .

Givens

$x+y=6$
$y+z=10$
$z+x=a$
$x$ and $y$ are positive.

Unknowns

The range of values of $a$ .

Constraints

Only $x$ and $y$ must be positive; $z$ has no sign condition.

2 · Planchoose the strategy

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

Each variable appears in exactly two of the three equations, so adding all three gives the total $x+y+z$ at once. Subtracting each original equation from that total isolates one variable in terms of $a$ . The positivity conditions then become two linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

#15 Organize Information in More Ways 8.EE.C.8

Every variable appears twice in the sum.

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

The symmetric pattern makes the total easy to find.

2Isolate $x$ and $y$

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

Subtract

y+z=10

from the total to get

x

, and subtract

z+x=a

to get

y

.

x = 8 + \frac{a}{2} - 10 = -2 + \frac{a}{2}, \quad y = 8 + \frac{a}{2} - a = 8 - \frac{a}{2}

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

3Require $x > 0$

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

Solve the inequality for

a

.

-2 + \frac{a}{2} > 0 \;\Rightarrow\; \frac{a}{2} > 2 \;\Rightarrow\; a > 4

Multiplying by the positive number

2

keeps the inequality direction.

4Require $y > 0$ and combine

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

Solve the second inequality, then keep the values of

a

that satisfy both.

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

Both conditions must hold at once, so the answer is the overlap.

Answer:

4<a<16

4 · Reviewdoes it hold up?

Try $a=6$ , inside the range: $x=1$ , $y=5$ , $z=5$ , and indeed $1+5=6$ , $5+5=10$ , $5+1=6$ . At the endpoints $a=4$ gives $x=0$ and $a=16$ gives $y=0$ , which are not positive, so the endpoints are excluded.

Another way: Solve step by step instead: from $x+y=6$ and $y+z=10$ , subtracting gives $x-z=-4$ ; adding $z+x=a$ gives $2x=-4+a$ , so $x=-2 + \frac{a}{2}$ , and then $y=6-x=8-\frac{a}{2}$ . The same inequalities follow.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the three-equation linear system for $x$ and $y$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning the conditions $x>0$ and $y>0$ into inequalities in $a$ and solving them.

💡Takeaway. Express the solution in terms of the constant first; then any condition on the solution becomes a condition on the constant.

Variant 5 easy

The values of $x$ and $y$ that satisfy the system of equations $x+y=10$ , $y+z=6$ , $z+x=a$ are both positive. Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three equations link $x$ , $y$ , $z$ and a constant $a$ . We want every value of $a$ for which the solution has $x>0$ and $y>0$ .

Givens

$x+y=10$
$y+z=6$
$z+x=a$
$x$ and $y$ are positive.

Unknowns

The range of values of $a$ .

Constraints

Only $x$ and $y$ must be positive; $z$ has no sign condition.

2 · Planchoose the strategy

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

Each variable appears in exactly two of the three equations, so adding all three gives the total $x+y+z$ at once. Subtracting each original equation from that total isolates one variable in terms of $a$ . The positivity conditions then become two linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

#15 Organize Information in More Ways 8.EE.C.8

Every variable appears twice in the sum.

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

The symmetric pattern makes the total easy to find.

2Isolate $x$ and $y$

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

Subtract

y+z=6

from the total to get

x

, and subtract

z+x=a

to get

y

.

x = 8 + \frac{a}{2} - 6 = 2 + \frac{a}{2}, \quad y = 8 + \frac{a}{2} - a = 8 - \frac{a}{2}

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

3Require $x > 0$

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

Solve the inequality for

a

.

2 + \frac{a}{2} > 0 \;\Rightarrow\; \frac{a}{2} > -2 \;\Rightarrow\; a > -4

Multiplying by the positive number

2

keeps the inequality direction.

4Require $y > 0$ and combine

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

Solve the second inequality, then keep the values of

a

that satisfy both.

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

Both conditions must hold at once, so the answer is the overlap.

Answer:

-4<a<16

4 · Reviewdoes it hold up?

Try $a=0$ , inside the range: $x=2$ , $y=8$ , $z=-2$ , and indeed $2+8=10$ , $8+(-2)=6$ , $-2+2=0$ . At the endpoints $a=-4$ gives $x=0$ and $a=16$ gives $y=0$ , which are not positive, so the endpoints are excluded.

Another way: Solve step by step instead: from $x+y=10$ and $y+z=6$ , subtracting gives $x-z=4$ ; adding $z+x=a$ gives $2x=4+a$ , so $x=2 + \frac{a}{2}$ , and then $y=10-x=8-\frac{a}{2}$ . The same inequalities follow.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the three-equation linear system for $x$ and $y$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning the conditions $x>0$ and $y>0$ into inequalities in $a$ and solving them.

💡Takeaway. Express the solution in terms of the constant first; then any condition on the solution becomes a condition on the constant.

Variant 6 medium

The values of $x$ and $y$ that satisfy the system of equations $x+y=11$ , $y+z=3$ , $z+x=a$ are both positive. Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three equations link $x$ , $y$ , $z$ and a constant $a$ . We want every value of $a$ for which the solution has $x>0$ and $y>0$ .

Givens

$x+y=11$
$y+z=3$
$z+x=a$
$x$ and $y$ are positive.

Unknowns

The range of values of $a$ .

Constraints

Only $x$ and $y$ must be positive; $z$ has no sign condition.

2 · Planchoose the strategy

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

Each variable appears in exactly two of the three equations, so adding all three gives the total $x+y+z$ at once. Subtracting each original equation from that total isolates one variable in terms of $a$ . The positivity conditions then become two linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

#15 Organize Information in More Ways 8.EE.C.8

Every variable appears twice in the sum.

2(x+y+z) = 11 + 3 + a \;\Rightarrow\; x + y + z = 7 + \frac{a}{2}

The symmetric pattern makes the total easy to find.

2Isolate $x$ and $y$

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

Subtract

y+z=3

from the total to get

x

, and subtract

z+x=a

to get

y

.

x = 7 + \frac{a}{2} - 3 = 4 + \frac{a}{2}, \quad y = 7 + \frac{a}{2} - a = 7 - \frac{a}{2}

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

3Require $x > 0$

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

Solve the inequality for

a

.

4 + \frac{a}{2} > 0 \;\Rightarrow\; \frac{a}{2} > -4 \;\Rightarrow\; a > -8

Multiplying by the positive number

2

keeps the inequality direction.

4Require $y > 0$ and combine

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

Solve the second inequality, then keep the values of

a

that satisfy both.

7 - \frac{a}{2} > 0 \;\Rightarrow\; a < 14, \quad \text{so } -8 < a < 14

Both conditions must hold at once, so the answer is the overlap.

Answer:

-8<a<14

4 · Reviewdoes it hold up?

Try $a=0$ , inside the range: $x=4$ , $y=7$ , $z=-4$ , and indeed $4+7=11$ , $7+(-4)=3$ , $-4+4=0$ . At the endpoints $a=-8$ gives $x=0$ and $a=14$ gives $y=0$ , which are not positive, so the endpoints are excluded.

Another way: Solve step by step instead: from $x+y=11$ and $y+z=3$ , subtracting gives $x-z=8$ ; adding $z+x=a$ gives $2x=8+a$ , so $x=4 + \frac{a}{2}$ , and then $y=11-x=7-\frac{a}{2}$ . The same inequalities follow.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the three-equation linear system for $x$ and $y$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning the conditions $x>0$ and $y>0$ into inequalities in $a$ and solving them.

💡Takeaway. Express the solution in terms of the constant first; then any condition on the solution becomes a condition on the constant.

Variant 7 medium

The values of $x$ and $y$ that satisfy the system of equations $x+y=4$ , $y+z=12$ , $z+x=a$ are both positive. Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three equations link $x$ , $y$ , $z$ and a constant $a$ . We want every value of $a$ for which the solution has $x>0$ and $y>0$ .

Givens

$x+y=4$
$y+z=12$
$z+x=a$
$x$ and $y$ are positive.

Unknowns

The range of values of $a$ .

Constraints

Only $x$ and $y$ must be positive; $z$ has no sign condition.

2 · Planchoose the strategy

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

Each variable appears in exactly two of the three equations, so adding all three gives the total $x+y+z$ at once. Subtracting each original equation from that total isolates one variable in terms of $a$ . The positivity conditions then become two linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

#15 Organize Information in More Ways 8.EE.C.8

Every variable appears twice in the sum.

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

The symmetric pattern makes the total easy to find.

2Isolate $x$ and $y$

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

Subtract

y+z=12

from the total to get

x

, and subtract

z+x=a

to get

y

.

x = 8 + \frac{a}{2} - 12 = -4 + \frac{a}{2}, \quad y = 8 + \frac{a}{2} - a = 8 - \frac{a}{2}

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

3Require $x > 0$

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

Solve the inequality for

a

.

-4 + \frac{a}{2} > 0 \;\Rightarrow\; \frac{a}{2} > 4 \;\Rightarrow\; a > 8

Multiplying by the positive number

2

keeps the inequality direction.

4Require $y > 0$ and combine

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

Solve the second inequality, then keep the values of

a

that satisfy both.

8 - \frac{a}{2} > 0 \;\Rightarrow\; a < 16, \quad \text{so } 8 < a < 16

Both conditions must hold at once, so the answer is the overlap.

Answer:

8<a<16

4 · Reviewdoes it hold up?

Try $a=10$ , inside the range: $x=1$ , $y=3$ , $z=9$ , and indeed $1+3=4$ , $3+9=12$ , $9+1=10$ . At the endpoints $a=8$ gives $x=0$ and $a=16$ gives $y=0$ , which are not positive, so the endpoints are excluded.

Another way: Solve step by step instead: from $x+y=4$ and $y+z=12$ , subtracting gives $x-z=-8$ ; adding $z+x=a$ gives $2x=-8+a$ , so $x=-4 + \frac{a}{2}$ , and then $y=4-x=8-\frac{a}{2}$ . The same inequalities follow.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the three-equation linear system for $x$ and $y$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning the conditions $x>0$ and $y>0$ into inequalities in $a$ and solving them.

💡Takeaway. Express the solution in terms of the constant first; then any condition on the solution becomes a condition on the constant.

Variant 8 medium

The values of $x$ and $y$ that satisfy the system of equations $x+y=12$ , $y+z=2$ , $z+x=a$ are both positive. Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three equations link $x$ , $y$ , $z$ and a constant $a$ . We want every value of $a$ for which the solution has $x>0$ and $y>0$ .

Givens

$x+y=12$
$y+z=2$
$z+x=a$
$x$ and $y$ are positive.

Unknowns

The range of values of $a$ .

Constraints

Only $x$ and $y$ must be positive; $z$ has no sign condition.

2 · Planchoose the strategy

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

Each variable appears in exactly two of the three equations, so adding all three gives the total $x+y+z$ at once. Subtracting each original equation from that total isolates one variable in terms of $a$ . The positivity conditions then become two linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

#15 Organize Information in More Ways 8.EE.C.8

Every variable appears twice in the sum.

2(x+y+z) = 12 + 2 + a \;\Rightarrow\; x + y + z = 7 + \frac{a}{2}

The symmetric pattern makes the total easy to find.

2Isolate $x$ and $y$

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

Subtract

y+z=2

from the total to get

x

, and subtract

z+x=a

to get

y

.

x = 7 + \frac{a}{2} - 2 = 5 + \frac{a}{2}, \quad y = 7 + \frac{a}{2} - a = 7 - \frac{a}{2}

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

3Require $x > 0$

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

Solve the inequality for

a

.

5 + \frac{a}{2} > 0 \;\Rightarrow\; \frac{a}{2} > -5 \;\Rightarrow\; a > -10

Multiplying by the positive number

2

keeps the inequality direction.

4Require $y > 0$ and combine

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

Solve the second inequality, then keep the values of

a

that satisfy both.

7 - \frac{a}{2} > 0 \;\Rightarrow\; a < 14, \quad \text{so } -10 < a < 14

Both conditions must hold at once, so the answer is the overlap.

Answer:

-10<a<14

4 · Reviewdoes it hold up?

Try $a=0$ , inside the range: $x=5$ , $y=7$ , $z=-5$ , and indeed $5+7=12$ , $7+(-5)=2$ , $-5+5=0$ . At the endpoints $a=-10$ gives $x=0$ and $a=14$ gives $y=0$ , which are not positive, so the endpoints are excluded.

Another way: Solve step by step instead: from $x+y=12$ and $y+z=2$ , subtracting gives $x-z=10$ ; adding $z+x=a$ gives $2x=10+a$ , so $x=5 + \frac{a}{2}$ , and then $y=12-x=7-\frac{a}{2}$ . The same inequalities follow.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the three-equation linear system for $x$ and $y$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning the conditions $x>0$ and $y>0$ into inequalities in $a$ and solving them.

💡Takeaway. Express the solution in terms of the constant first; then any condition on the solution becomes a condition on the constant.

Variant 9 hard

The values of $x$ and $y$ that satisfy the system of equations $x+y=7$ , $y+z=13$ , $z+x=a$ are both positive. Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three equations link $x$ , $y$ , $z$ and a constant $a$ . We want every value of $a$ for which the solution has $x>0$ and $y>0$ .

Givens

$x+y=7$
$y+z=13$
$z+x=a$
$x$ and $y$ are positive.

Unknowns

The range of values of $a$ .

Constraints

Only $x$ and $y$ must be positive; $z$ has no sign condition.

2 · Planchoose the strategy

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

Each variable appears in exactly two of the three equations, so adding all three gives the total $x+y+z$ at once. Subtracting each original equation from that total isolates one variable in terms of $a$ . The positivity conditions then become two linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

#15 Organize Information in More Ways 8.EE.C.8

Every variable appears twice in the sum.

2(x+y+z) = 7 + 13 + a \;\Rightarrow\; x + y + z = 10 + \frac{a}{2}

The symmetric pattern makes the total easy to find.

2Isolate $x$ and $y$

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

Subtract

y+z=13

from the total to get

x

, and subtract

z+x=a

to get

y

.

x = 10 + \frac{a}{2} - 13 = -3 + \frac{a}{2}, \quad y = 10 + \frac{a}{2} - a = 10 - \frac{a}{2}

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

3Require $x > 0$

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

Solve the inequality for

a

.

-3 + \frac{a}{2} > 0 \;\Rightarrow\; \frac{a}{2} > 3 \;\Rightarrow\; a > 6

Multiplying by the positive number

2

keeps the inequality direction.

4Require $y > 0$ and combine

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

Solve the second inequality, then keep the values of

a

that satisfy both.

10 - \frac{a}{2} > 0 \;\Rightarrow\; a < 20, \quad \text{so } 6 < a < 20

Both conditions must hold at once, so the answer is the overlap.

Answer:

6<a<20

4 · Reviewdoes it hold up?

Try $a=8$ , inside the range: $x=1$ , $y=6$ , $z=7$ , and indeed $1+6=7$ , $6+7=13$ , $7+1=8$ . At the endpoints $a=6$ gives $x=0$ and $a=20$ gives $y=0$ , which are not positive, so the endpoints are excluded.

Another way: Solve step by step instead: from $x+y=7$ and $y+z=13$ , subtracting gives $x-z=-6$ ; adding $z+x=a$ gives $2x=-6+a$ , so $x=-3 + \frac{a}{2}$ , and then $y=7-x=10-\frac{a}{2}$ . The same inequalities follow.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the three-equation linear system for $x$ and $y$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning the conditions $x>0$ and $y>0$ into inequalities in $a$ and solving them.

💡Takeaway. Express the solution in terms of the constant first; then any condition on the solution becomes a condition on the constant.

Variant 10 hard

The values of $x$ and $y$ that satisfy the system of equations $x+y=14$ , $y+z=8$ , $z+x=a$ are both positive. Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three equations link $x$ , $y$ , $z$ and a constant $a$ . We want every value of $a$ for which the solution has $x>0$ and $y>0$ .

Givens

$x+y=14$
$y+z=8$
$z+x=a$
$x$ and $y$ are positive.

Unknowns

The range of values of $a$ .

Constraints

Only $x$ and $y$ must be positive; $z$ has no sign condition.

2 · Planchoose the strategy

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

Each variable appears in exactly two of the three equations, so adding all three gives the total $x+y+z$ at once. Subtracting each original equation from that total isolates one variable in terms of $a$ . The positivity conditions then become two linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

#15 Organize Information in More Ways 8.EE.C.8

Every variable appears twice in the sum.

2(x+y+z) = 14 + 8 + a \;\Rightarrow\; x + y + z = 11 + \frac{a}{2}

The symmetric pattern makes the total easy to find.

2Isolate $x$ and $y$

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

Subtract

y+z=8

from the total to get

x

, and subtract

z+x=a

to get

y

.

x = 11 + \frac{a}{2} - 8 = 3 + \frac{a}{2}, \quad y = 11 + \frac{a}{2} - a = 11 - \frac{a}{2}

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

3Require $x > 0$

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

Solve the inequality for

a

.

3 + \frac{a}{2} > 0 \;\Rightarrow\; \frac{a}{2} > -3 \;\Rightarrow\; a > -6

Multiplying by the positive number

2

keeps the inequality direction.

4Require $y > 0$ and combine

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

Solve the second inequality, then keep the values of

a

that satisfy both.

11 - \frac{a}{2} > 0 \;\Rightarrow\; a < 22, \quad \text{so } -6 < a < 22

Both conditions must hold at once, so the answer is the overlap.

Answer:

-6<a<22

4 · Reviewdoes it hold up?

Try $a=0$ , inside the range: $x=3$ , $y=11$ , $z=-3$ , and indeed $3+11=14$ , $11+(-3)=8$ , $-3+3=0$ . At the endpoints $a=-6$ gives $x=0$ and $a=22$ gives $y=0$ , which are not positive, so the endpoints are excluded.

Another way: Solve step by step instead: from $x+y=14$ and $y+z=8$ , subtracting gives $x-z=6$ ; adding $z+x=a$ gives $2x=6+a$ , so $x=3 + \frac{a}{2}$ , and then $y=14-x=11-\frac{a}{2}$ . The same inequalities follow.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the three-equation linear system for $x$ and $y$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning the conditions $x>0$ and $y>0$ into inequalities in $a$ and solving them.

💡Takeaway. Express the solution in terms of the constant first; then any condition on the solution becomes a condition on the constant.

Variant 11 hard

The values of $x$ and $y$ that satisfy the system of equations $x+y=15$ , $y+z=7$ , $z+x=a$ are both positive. Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three equations link $x$ , $y$ , $z$ and a constant $a$ . We want every value of $a$ for which the solution has $x>0$ and $y>0$ .

Givens

$x+y=15$
$y+z=7$
$z+x=a$
$x$ and $y$ are positive.

Unknowns

The range of values of $a$ .

Constraints

Only $x$ and $y$ must be positive; $z$ has no sign condition.

2 · Planchoose the strategy

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

Each variable appears in exactly two of the three equations, so adding all three gives the total $x+y+z$ at once. Subtracting each original equation from that total isolates one variable in terms of $a$ . The positivity conditions then become two linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

#15 Organize Information in More Ways 8.EE.C.8

Every variable appears twice in the sum.

2(x+y+z) = 15 + 7 + a \;\Rightarrow\; x + y + z = 11 + \frac{a}{2}

The symmetric pattern makes the total easy to find.

2Isolate $x$ and $y$

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

Subtract

y+z=7

from the total to get

x

, and subtract

z+x=a

to get

y

.

x = 11 + \frac{a}{2} - 7 = 4 + \frac{a}{2}, \quad y = 11 + \frac{a}{2} - a = 11 - \frac{a}{2}

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

3Require $x > 0$

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

Solve the inequality for

a

.

4 + \frac{a}{2} > 0 \;\Rightarrow\; \frac{a}{2} > -4 \;\Rightarrow\; a > -8

Multiplying by the positive number

2

keeps the inequality direction.

4Require $y > 0$ and combine

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

Solve the second inequality, then keep the values of

a

that satisfy both.

11 - \frac{a}{2} > 0 \;\Rightarrow\; a < 22, \quad \text{so } -8 < a < 22

Both conditions must hold at once, so the answer is the overlap.

Answer:

-8<a<22

4 · Reviewdoes it hold up?

Try $a=0$ , inside the range: $x=4$ , $y=11$ , $z=-4$ , and indeed $4+11=15$ , $11+(-4)=7$ , $-4+4=0$ . At the endpoints $a=-8$ gives $x=0$ and $a=22$ gives $y=0$ , which are not positive, so the endpoints are excluded.

Another way: Solve step by step instead: from $x+y=15$ and $y+z=7$ , subtracting gives $x-z=8$ ; adding $z+x=a$ gives $2x=8+a$ , so $x=4 + \frac{a}{2}$ , and then $y=15-x=11-\frac{a}{2}$ . The same inequalities follow.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the three-equation linear system for $x$ and $y$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning the conditions $x>0$ and $y>0$ into inequalities in $a$ and solving them.

💡Takeaway. Express the solution in terms of the constant first; then any condition on the solution becomes a condition on the constant.

Variant 12 hard

The values of $x$ and $y$ that satisfy the system of equations $x+y=16$ , $y+z=6$ , $z+x=a$ are both positive. Find the range of values of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Three equations link $x$ , $y$ , $z$ and a constant $a$ . We want every value of $a$ for which the solution has $x>0$ and $y>0$ .

Givens

$x+y=16$
$y+z=6$
$z+x=a$
$x$ and $y$ are positive.

Unknowns

The range of values of $a$ .

Constraints

Only $x$ and $y$ must be positive; $z$ has no sign condition.

2 · Planchoose the strategy

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

Each variable appears in exactly two of the three equations, so adding all three gives the total $x+y+z$ at once. Subtracting each original equation from that total isolates one variable in terms of $a$ . The positivity conditions then become two linear inequalities in $a$ .

3 · Execute4 carry out the plan

1Add all three equations

#15 Organize Information in More Ways 8.EE.C.8

Every variable appears twice in the sum.

2(x+y+z) = 16 + 6 + a \;\Rightarrow\; x + y + z = 11 + \frac{a}{2}

The symmetric pattern makes the total easy to find.

2Isolate $x$ and $y$

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

Subtract

y+z=6

from the total to get

x

, and subtract

z+x=a

to get

y

.

x = 11 + \frac{a}{2} - 6 = 5 + \frac{a}{2}, \quad y = 11 + \frac{a}{2} - a = 11 - \frac{a}{2}

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

3Require $x > 0$

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

Solve the inequality for

a

.

5 + \frac{a}{2} > 0 \;\Rightarrow\; \frac{a}{2} > -5 \;\Rightarrow\; a > -10

Multiplying by the positive number

2

keeps the inequality direction.

4Require $y > 0$ and combine

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

Solve the second inequality, then keep the values of

a

that satisfy both.

11 - \frac{a}{2} > 0 \;\Rightarrow\; a < 22, \quad \text{so } -10 < a < 22

Both conditions must hold at once, so the answer is the overlap.

Answer:

-10<a<22

4 · Reviewdoes it hold up?

Try $a=0$ , inside the range: $x=5$ , $y=11$ , $z=-5$ , and indeed $5+11=16$ , $11+(-5)=6$ , $-5+5=0$ . At the endpoints $a=-10$ gives $x=0$ and $a=22$ gives $y=0$ , which are not positive, so the endpoints are excluded.

Another way: Solve step by step instead: from $x+y=16$ and $y+z=6$ , subtracting gives $x-z=10$ ; adding $z+x=a$ gives $2x=10+a$ , so $x=5 + \frac{a}{2}$ , and then $y=16-x=11-\frac{a}{2}$ . The same inequalities follow.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving the three-equation linear system for $x$ and $y$ in terms of $a$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Turning the conditions $x>0$ and $y>0$ into inequalities in $a$ and solving them.

💡Takeaway. Express the solution in terms of the constant first; then any condition on the solution becomes a condition on the constant.

Write the solution in terms of the constant · 12 practice problems

Generated variants — 12

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