Practice · True for every k means an identity in k · Sensim Math

Variant 1 easy

The equation $(2k+3)x-3a=bk+3$ in $x$ always has $x=2$ as a solution, no matter what value $k$ takes. Find the sum $a+b$ of the constants $a$ and $b$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Plugging $x=2$ into the equation must give a true statement for every value of $k$ . Use that to pin down the constants $a$ and $b$ , then add them.

Givens

The equation is $(2k+3)x-3a=bk+3$ .
$x=2$ is a solution for every value of $k$ .
$a$ and $b$ are constants.

Unknowns

The value of $a+b$ .

Constraints

The statement must hold for all $k$ , not just one value of $k$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

After substituting $x=2$ , stop treating $k$ as a fixed number and look at it as the variable: an equation that holds for every $k$ is an identity in $k$ , so both sides must have the same $k$ -part and the same constant part.

3 · Execute4 carry out the plan

1Substitute the solution

#13 Convert to Algebra 7.EE.A.1

Since

x=2

is always a solution, put

x=2

into the equation and expand.

(2k+3)\times 2-3a=bk+3 \;\Rightarrow\; 4k+6-3a=bk+3

Now only

k

,

a

, and

b

are left.

2Read it as an identity in k

#16 Count the Complement 8.EE.C.7

The line

4k+6-3a=bk+3

is true for every

k

. Two expressions in

k

that agree for every

k

must have the same coefficient of

k

.

4k = bk \text{ for all } k \;\Rightarrow\; b=4

If the

k

-parts differed, changing

k

would make the two sides drift apart.

3Match the constant parts

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

With the

k

-parts equal, the parts without

k

must also be equal.

6-3a=3 \;\Rightarrow\; 3a=3 \;\Rightarrow\; a=1

A one-step equation finishes it.

4Add the constants

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

Add the two values.

a+b=1+4=5

Both constants are now known.

Answer:

5

4 · Reviewdoes it hold up?

With $a=1$ and $b=4$ the equation is $(2k+3)x-3=4k+3$ . At $x=2$ the left side is $4k+6-3=4k+3$ , equal to the right side for every $k$ .

Another way: Use two easy values of $k$ (Solve an Easier Related Problem): $k=0$ gives $3\times 2-3a=3$ , so $a=1$ ; $k=1$ gives $5\times 2-3=b+3$ , so $b=4$ . Again $a+b=5$ .

Standardsmin grade 8

8.EE.C.7 Solve linear equations in one variable — Recognizing an equation that holds for every value as an identity.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Expanding $(2k+3)\times 2$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving $6-3a=3$ for $a$.

💡Takeaway. "True no matter what

k

is" is a big clue: match the

k

-parts and the plain-number parts separately.

Variant 2 easy

The equation $(3k+4)x-2a=bk+2$ in $x$ always has $x=3$ as a solution, no matter what value $k$ takes. Find the sum $a+b$ of the constants $a$ and $b$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Plugging $x=3$ into the equation must give a true statement for every value of $k$ . Use that to pin down the constants $a$ and $b$ , then add them.

Givens

The equation is $(3k+4)x-2a=bk+2$ .
$x=3$ is a solution for every value of $k$ .
$a$ and $b$ are constants.

Unknowns

The value of $a+b$ .

Constraints

The statement must hold for all $k$ , not just one value of $k$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

After substituting $x=3$ , stop treating $k$ as a fixed number and look at it as the variable: an equation that holds for every $k$ is an identity in $k$ , so both sides must have the same $k$ -part and the same constant part.

3 · Execute4 carry out the plan

1Substitute the solution

#13 Convert to Algebra 7.EE.A.1

Since

x=3

is always a solution, put

x=3

into the equation and expand.

(3k+4)\times 3-2a=bk+2 \;\Rightarrow\; 9k+12-2a=bk+2

Now only

k

,

a

, and

b

are left.

2Read it as an identity in k

#16 Count the Complement 8.EE.C.7

The line

9k+12-2a=bk+2

is true for every

k

. Two expressions in

k

that agree for every

k

must have the same coefficient of

k

.

9k = bk \text{ for all } k \;\Rightarrow\; b=9

If the

k

-parts differed, changing

k

would make the two sides drift apart.

3Match the constant parts

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

With the

k

-parts equal, the parts without

k

must also be equal.

12-2a=2 \;\Rightarrow\; 2a=10 \;\Rightarrow\; a=5

A one-step equation finishes it.

4Add the constants

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

Add the two values.

a+b=5+9=14

Both constants are now known.

Answer:

14

4 · Reviewdoes it hold up?

With $a=5$ and $b=9$ the equation is $(3k+4)x-10=9k+2$ . At $x=3$ the left side is $9k+12-10=9k+2$ , equal to the right side for every $k$ .

Another way: Use two easy values of $k$ (Solve an Easier Related Problem): $k=0$ gives $4\times 3-2a=2$ , so $a=5$ ; $k=1$ gives $7\times 3-10=b+2$ , so $b=9$ . Again $a+b=14$ .

Standardsmin grade 8

8.EE.C.7 Solve linear equations in one variable — Recognizing an equation that holds for every value as an identity.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Expanding $(3k+4)\times 3$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving $12-2a=2$ for $a$.

💡Takeaway. "True no matter what

k

is" is a big clue: match the

k

-parts and the plain-number parts separately.

Variant 3 medium

The equation $(2k+5)x-4a=bk+3$ in $x$ always has $x=3$ as a solution, no matter what value $k$ takes. Find the sum $a+b$ of the constants $a$ and $b$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Plugging $x=3$ into the equation must give a true statement for every value of $k$ . Use that to pin down the constants $a$ and $b$ , then add them.

Givens

The equation is $(2k+5)x-4a=bk+3$ .
$x=3$ is a solution for every value of $k$ .
$a$ and $b$ are constants.

Unknowns

The value of $a+b$ .

Constraints

The statement must hold for all $k$ , not just one value of $k$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

After substituting $x=3$ , stop treating $k$ as a fixed number and look at it as the variable: an equation that holds for every $k$ is an identity in $k$ , so both sides must have the same $k$ -part and the same constant part.

3 · Execute4 carry out the plan

1Substitute the solution

#13 Convert to Algebra 7.EE.A.1

Since

x=3

is always a solution, put

x=3

into the equation and expand.

(2k+5)\times 3-4a=bk+3 \;\Rightarrow\; 6k+15-4a=bk+3

Now only

k

,

a

, and

b

are left.

2Read it as an identity in k

#16 Count the Complement 8.EE.C.7

The line

6k+15-4a=bk+3

is true for every

k

. Two expressions in

k

that agree for every

k

must have the same coefficient of

k

.

6k = bk \text{ for all } k \;\Rightarrow\; b=6

If the

k

-parts differed, changing

k

would make the two sides drift apart.

3Match the constant parts

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

With the

k

-parts equal, the parts without

k

must also be equal.

15-4a=3 \;\Rightarrow\; 4a=12 \;\Rightarrow\; a=3

A one-step equation finishes it.

4Add the constants

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

Add the two values.

a+b=3+6=9

Both constants are now known.

Answer:

9

4 · Reviewdoes it hold up?

With $a=3$ and $b=6$ the equation is $(2k+5)x-12=6k+3$ . At $x=3$ the left side is $6k+15-12=6k+3$ , equal to the right side for every $k$ .

Another way: Use two easy values of $k$ (Solve an Easier Related Problem): $k=0$ gives $5\times 3-4a=3$ , so $a=3$ ; $k=1$ gives $7\times 3-12=b+3$ , so $b=6$ . Again $a+b=9$ .

Standardsmin grade 8

8.EE.C.7 Solve linear equations in one variable — Recognizing an equation that holds for every value as an identity.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Expanding $(2k+5)\times 3$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving $15-4a=3$ for $a$.

💡Takeaway. "True no matter what

k

is" is a big clue: match the

k

-parts and the plain-number parts separately.

Variant 4 hard

The equation $(3k+2)x-4a=bk+2$ in $x$ always has $x=5$ as a solution, no matter what value $k$ takes. Find the sum $a+b$ of the constants $a$ and $b$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Plugging $x=5$ into the equation must give a true statement for every value of $k$ . Use that to pin down the constants $a$ and $b$ , then add them.

Givens

The equation is $(3k+2)x-4a=bk+2$ .
$x=5$ is a solution for every value of $k$ .
$a$ and $b$ are constants.

Unknowns

The value of $a+b$ .

Constraints

The statement must hold for all $k$ , not just one value of $k$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

After substituting $x=5$ , stop treating $k$ as a fixed number and look at it as the variable: an equation that holds for every $k$ is an identity in $k$ , so both sides must have the same $k$ -part and the same constant part.

3 · Execute4 carry out the plan

1Substitute the solution

#13 Convert to Algebra 7.EE.A.1

Since

x=5

is always a solution, put

x=5

into the equation and expand.

(3k+2)\times 5-4a=bk+2 \;\Rightarrow\; 15k+10-4a=bk+2

Now only

k

,

a

, and

b

are left.

2Read it as an identity in k

#16 Count the Complement 8.EE.C.7

The line

15k+10-4a=bk+2

is true for every

k

. Two expressions in

k

that agree for every

k

must have the same coefficient of

k

.

15k = bk \text{ for all } k \;\Rightarrow\; b=15

If the

k

-parts differed, changing

k

would make the two sides drift apart.

3Match the constant parts

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

With the

k

-parts equal, the parts without

k

must also be equal.

10-4a=2 \;\Rightarrow\; 4a=8 \;\Rightarrow\; a=2

A one-step equation finishes it.

4Add the constants

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

Add the two values.

a+b=2+15=17

Both constants are now known.

Answer:

17

4 · Reviewdoes it hold up?

With $a=2$ and $b=15$ the equation is $(3k+2)x-8=15k+2$ . At $x=5$ the left side is $15k+10-8=15k+2$ , equal to the right side for every $k$ .

Another way: Use two easy values of $k$ (Solve an Easier Related Problem): $k=0$ gives $2\times 5-4a=2$ , so $a=2$ ; $k=1$ gives $5\times 5-8=b+2$ , so $b=15$ . Again $a+b=17$ .

Standardsmin grade 8

8.EE.C.7 Solve linear equations in one variable — Recognizing an equation that holds for every value as an identity.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Expanding $(3k+2)\times 5$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving $10-4a=2$ for $a$.

💡Takeaway. "True no matter what

k

is" is a big clue: match the

k

-parts and the plain-number parts separately.

Variant 5 easy

The equation $(3k+5)x-2a=bk+4$ in $x$ always has $x=2$ as a solution, no matter what value $k$ takes. Find the sum $a+b$ of the constants $a$ and $b$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Plugging $x=2$ into the equation must give a true statement for every value of $k$ . Use that to pin down the constants $a$ and $b$ , then add them.

Givens

The equation is $(3k+5)x-2a=bk+4$ .
$x=2$ is a solution for every value of $k$ .
$a$ and $b$ are constants.

Unknowns

The value of $a+b$ .

Constraints

The statement must hold for all $k$ , not just one value of $k$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

After substituting $x=2$ , stop treating $k$ as a fixed number and look at it as the variable: an equation that holds for every $k$ is an identity in $k$ , so both sides must have the same $k$ -part and the same constant part.

3 · Execute4 carry out the plan

1Substitute the solution

#13 Convert to Algebra 7.EE.A.1

Since

x=2

is always a solution, put

x=2

into the equation and expand.

(3k+5)\times 2-2a=bk+4 \;\Rightarrow\; 6k+10-2a=bk+4

Now only

k

,

a

, and

b

are left.

2Read it as an identity in k

#16 Count the Complement 8.EE.C.7

The line

6k+10-2a=bk+4

is true for every

k

. Two expressions in

k

that agree for every

k

must have the same coefficient of

k

.

6k = bk \text{ for all } k \;\Rightarrow\; b=6

If the

k

-parts differed, changing

k

would make the two sides drift apart.

3Match the constant parts

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

With the

k

-parts equal, the parts without

k

must also be equal.

10-2a=4 \;\Rightarrow\; 2a=6 \;\Rightarrow\; a=3

A one-step equation finishes it.

4Add the constants

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

Add the two values.

a+b=3+6=9

Both constants are now known.

Answer:

9

4 · Reviewdoes it hold up?

With $a=3$ and $b=6$ the equation is $(3k+5)x-6=6k+4$ . At $x=2$ the left side is $6k+10-6=6k+4$ , equal to the right side for every $k$ .

Another way: Use two easy values of $k$ (Solve an Easier Related Problem): $k=0$ gives $5\times 2-2a=4$ , so $a=3$ ; $k=1$ gives $8\times 2-6=b+4$ , so $b=6$ . Again $a+b=9$ .

Standardsmin grade 8

8.EE.C.7 Solve linear equations in one variable — Recognizing an equation that holds for every value as an identity.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Expanding $(3k+5)\times 2$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving $10-2a=4$ for $a$.

💡Takeaway. "True no matter what

k

is" is a big clue: match the

k

-parts and the plain-number parts separately.

Variant 6 medium

The equation $(4k+5)x-3a=bk+5$ in $x$ always has $x=4$ as a solution, no matter what value $k$ takes. Find the sum $a+b$ of the constants $a$ and $b$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Plugging $x=4$ into the equation must give a true statement for every value of $k$ . Use that to pin down the constants $a$ and $b$ , then add them.

Givens

The equation is $(4k+5)x-3a=bk+5$ .
$x=4$ is a solution for every value of $k$ .
$a$ and $b$ are constants.

Unknowns

The value of $a+b$ .

Constraints

The statement must hold for all $k$ , not just one value of $k$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

After substituting $x=4$ , stop treating $k$ as a fixed number and look at it as the variable: an equation that holds for every $k$ is an identity in $k$ , so both sides must have the same $k$ -part and the same constant part.

3 · Execute4 carry out the plan

1Substitute the solution

#13 Convert to Algebra 7.EE.A.1

Since

x=4

is always a solution, put

x=4

into the equation and expand.

(4k+5)\times 4-3a=bk+5 \;\Rightarrow\; 16k+20-3a=bk+5

Now only

k

,

a

, and

b

are left.

2Read it as an identity in k

#16 Count the Complement 8.EE.C.7

The line

16k+20-3a=bk+5

is true for every

k

. Two expressions in

k

that agree for every

k

must have the same coefficient of

k

.

16k = bk \text{ for all } k \;\Rightarrow\; b=16

If the

k

-parts differed, changing

k

would make the two sides drift apart.

3Match the constant parts

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

With the

k

-parts equal, the parts without

k

must also be equal.

20-3a=5 \;\Rightarrow\; 3a=15 \;\Rightarrow\; a=5

A one-step equation finishes it.

4Add the constants

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

Add the two values.

a+b=5+16=21

Both constants are now known.

Answer:

21

4 · Reviewdoes it hold up?

With $a=5$ and $b=16$ the equation is $(4k+5)x-15=16k+5$ . At $x=4$ the left side is $16k+20-15=16k+5$ , equal to the right side for every $k$ .

Another way: Use two easy values of $k$ (Solve an Easier Related Problem): $k=0$ gives $5\times 4-3a=5$ , so $a=5$ ; $k=1$ gives $9\times 4-15=b+5$ , so $b=16$ . Again $a+b=21$ .

Standardsmin grade 8

8.EE.C.7 Solve linear equations in one variable — Recognizing an equation that holds for every value as an identity.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Expanding $(4k+5)\times 4$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving $20-3a=5$ for $a$.

💡Takeaway. "True no matter what

k

is" is a big clue: match the

k

-parts and the plain-number parts separately.

Variant 7 easy

The equation $(2k+4)x-5a=bk+2$ in $x$ always has $x=3$ as a solution, no matter what value $k$ takes. Find the sum $a+b$ of the constants $a$ and $b$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Plugging $x=3$ into the equation must give a true statement for every value of $k$ . Use that to pin down the constants $a$ and $b$ , then add them.

Givens

The equation is $(2k+4)x-5a=bk+2$ .
$x=3$ is a solution for every value of $k$ .
$a$ and $b$ are constants.

Unknowns

The value of $a+b$ .

Constraints

The statement must hold for all $k$ , not just one value of $k$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

After substituting $x=3$ , stop treating $k$ as a fixed number and look at it as the variable: an equation that holds for every $k$ is an identity in $k$ , so both sides must have the same $k$ -part and the same constant part.

3 · Execute4 carry out the plan

1Substitute the solution

#13 Convert to Algebra 7.EE.A.1

Since

x=3

is always a solution, put

x=3

into the equation and expand.

(2k+4)\times 3-5a=bk+2 \;\Rightarrow\; 6k+12-5a=bk+2

Now only

k

,

a

, and

b

are left.

2Read it as an identity in k

#16 Count the Complement 8.EE.C.7

The line

6k+12-5a=bk+2

is true for every

k

. Two expressions in

k

that agree for every

k

must have the same coefficient of

k

.

6k = bk \text{ for all } k \;\Rightarrow\; b=6

If the

k

-parts differed, changing

k

would make the two sides drift apart.

3Match the constant parts

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

With the

k

-parts equal, the parts without

k

must also be equal.

12-5a=2 \;\Rightarrow\; 5a=10 \;\Rightarrow\; a=2

A one-step equation finishes it.

4Add the constants

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

Add the two values.

a+b=2+6=8

Both constants are now known.

Answer:

8

4 · Reviewdoes it hold up?

With $a=2$ and $b=6$ the equation is $(2k+4)x-10=6k+2$ . At $x=3$ the left side is $6k+12-10=6k+2$ , equal to the right side for every $k$ .

Another way: Use two easy values of $k$ (Solve an Easier Related Problem): $k=0$ gives $4\times 3-5a=2$ , so $a=2$ ; $k=1$ gives $6\times 3-10=b+2$ , so $b=6$ . Again $a+b=8$ .

Standardsmin grade 8

8.EE.C.7 Solve linear equations in one variable — Recognizing an equation that holds for every value as an identity.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Expanding $(2k+4)\times 3$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving $12-5a=2$ for $a$.

💡Takeaway. "True no matter what

k

is" is a big clue: match the

k

-parts and the plain-number parts separately.

Variant 8 medium

The equation $(4k+2)x-3a=bk+1$ in $x$ always has $x=5$ as a solution, no matter what value $k$ takes. Find the sum $a+b$ of the constants $a$ and $b$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Plugging $x=5$ into the equation must give a true statement for every value of $k$ . Use that to pin down the constants $a$ and $b$ , then add them.

Givens

The equation is $(4k+2)x-3a=bk+1$ .
$x=5$ is a solution for every value of $k$ .
$a$ and $b$ are constants.

Unknowns

The value of $a+b$ .

Constraints

The statement must hold for all $k$ , not just one value of $k$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

After substituting $x=5$ , stop treating $k$ as a fixed number and look at it as the variable: an equation that holds for every $k$ is an identity in $k$ , so both sides must have the same $k$ -part and the same constant part.

3 · Execute4 carry out the plan

1Substitute the solution

#13 Convert to Algebra 7.EE.A.1

Since

x=5

is always a solution, put

x=5

into the equation and expand.

(4k+2)\times 5-3a=bk+1 \;\Rightarrow\; 20k+10-3a=bk+1

Now only

k

,

a

, and

b

are left.

2Read it as an identity in k

#16 Count the Complement 8.EE.C.7

The line

20k+10-3a=bk+1

is true for every

k

. Two expressions in

k

that agree for every

k

must have the same coefficient of

k

.

20k = bk \text{ for all } k \;\Rightarrow\; b=20

If the

k

-parts differed, changing

k

would make the two sides drift apart.

3Match the constant parts

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

With the

k

-parts equal, the parts without

k

must also be equal.

10-3a=1 \;\Rightarrow\; 3a=9 \;\Rightarrow\; a=3

A one-step equation finishes it.

4Add the constants

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

Add the two values.

a+b=3+20=23

Both constants are now known.

Answer:

23

4 · Reviewdoes it hold up?

With $a=3$ and $b=20$ the equation is $(4k+2)x-9=20k+1$ . At $x=5$ the left side is $20k+10-9=20k+1$ , equal to the right side for every $k$ .

Another way: Use two easy values of $k$ (Solve an Easier Related Problem): $k=0$ gives $2\times 5-3a=1$ , so $a=3$ ; $k=1$ gives $6\times 5-9=b+1$ , so $b=20$ . Again $a+b=23$ .

Standardsmin grade 8

8.EE.C.7 Solve linear equations in one variable — Recognizing an equation that holds for every value as an identity.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Expanding $(4k+2)\times 5$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving $10-3a=1$ for $a$.

💡Takeaway. "True no matter what

k

is" is a big clue: match the

k

-parts and the plain-number parts separately.

Variant 9 medium

The equation $(5k+3)x-2a=bk+2$ in $x$ always has $x=4$ as a solution, no matter what value $k$ takes. Find the sum $a+b$ of the constants $a$ and $b$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Plugging $x=4$ into the equation must give a true statement for every value of $k$ . Use that to pin down the constants $a$ and $b$ , then add them.

Givens

The equation is $(5k+3)x-2a=bk+2$ .
$x=4$ is a solution for every value of $k$ .
$a$ and $b$ are constants.

Unknowns

The value of $a+b$ .

Constraints

The statement must hold for all $k$ , not just one value of $k$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

After substituting $x=4$ , stop treating $k$ as a fixed number and look at it as the variable: an equation that holds for every $k$ is an identity in $k$ , so both sides must have the same $k$ -part and the same constant part.

3 · Execute4 carry out the plan

1Substitute the solution

#13 Convert to Algebra 7.EE.A.1

Since

x=4

is always a solution, put

x=4

into the equation and expand.

(5k+3)\times 4-2a=bk+2 \;\Rightarrow\; 20k+12-2a=bk+2

Now only

k

,

a

, and

b

are left.

2Read it as an identity in k

#16 Count the Complement 8.EE.C.7

The line

20k+12-2a=bk+2

is true for every

k

. Two expressions in

k

that agree for every

k

must have the same coefficient of

k

.

20k = bk \text{ for all } k \;\Rightarrow\; b=20

If the

k

-parts differed, changing

k

would make the two sides drift apart.

3Match the constant parts

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

With the

k

-parts equal, the parts without

k

must also be equal.

12-2a=2 \;\Rightarrow\; 2a=10 \;\Rightarrow\; a=5

A one-step equation finishes it.

4Add the constants

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

Add the two values.

a+b=5+20=25

Both constants are now known.

Answer:

25

4 · Reviewdoes it hold up?

With $a=5$ and $b=20$ the equation is $(5k+3)x-10=20k+2$ . At $x=4$ the left side is $20k+12-10=20k+2$ , equal to the right side for every $k$ .

Another way: Use two easy values of $k$ (Solve an Easier Related Problem): $k=0$ gives $3\times 4-2a=2$ , so $a=5$ ; $k=1$ gives $8\times 4-10=b+2$ , so $b=20$ . Again $a+b=25$ .

Standardsmin grade 8

8.EE.C.7 Solve linear equations in one variable — Recognizing an equation that holds for every value as an identity.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Expanding $(5k+3)\times 4$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving $12-2a=2$ for $a$.

💡Takeaway. "True no matter what

k

is" is a big clue: match the

k

-parts and the plain-number parts separately.

Variant 10 hard

The equation $(2k+6)x-5a=bk+3$ in $x$ always has $x=3$ as a solution, no matter what value $k$ takes. Find the sum $a+b$ of the constants $a$ and $b$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Plugging $x=3$ into the equation must give a true statement for every value of $k$ . Use that to pin down the constants $a$ and $b$ , then add them.

Givens

The equation is $(2k+6)x-5a=bk+3$ .
$x=3$ is a solution for every value of $k$ .
$a$ and $b$ are constants.

Unknowns

The value of $a+b$ .

Constraints

The statement must hold for all $k$ , not just one value of $k$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

After substituting $x=3$ , stop treating $k$ as a fixed number and look at it as the variable: an equation that holds for every $k$ is an identity in $k$ , so both sides must have the same $k$ -part and the same constant part.

3 · Execute4 carry out the plan

1Substitute the solution

#13 Convert to Algebra 7.EE.A.1

Since

x=3

is always a solution, put

x=3

into the equation and expand.

(2k+6)\times 3-5a=bk+3 \;\Rightarrow\; 6k+18-5a=bk+3

Now only

k

,

a

, and

b

are left.

2Read it as an identity in k

#16 Count the Complement 8.EE.C.7

The line

6k+18-5a=bk+3

is true for every

k

. Two expressions in

k

that agree for every

k

must have the same coefficient of

k

.

6k = bk \text{ for all } k \;\Rightarrow\; b=6

If the

k

-parts differed, changing

k

would make the two sides drift apart.

3Match the constant parts

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

With the

k

-parts equal, the parts without

k

must also be equal.

18-5a=3 \;\Rightarrow\; 5a=15 \;\Rightarrow\; a=3

A one-step equation finishes it.

4Add the constants

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

Add the two values.

a+b=3+6=9

Both constants are now known.

Answer:

9

4 · Reviewdoes it hold up?

With $a=3$ and $b=6$ the equation is $(2k+6)x-15=6k+3$ . At $x=3$ the left side is $6k+18-15=6k+3$ , equal to the right side for every $k$ .

Another way: Use two easy values of $k$ (Solve an Easier Related Problem): $k=0$ gives $6\times 3-5a=3$ , so $a=3$ ; $k=1$ gives $8\times 3-15=b+3$ , so $b=6$ . Again $a+b=9$ .

Standardsmin grade 8

8.EE.C.7 Solve linear equations in one variable — Recognizing an equation that holds for every value as an identity.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Expanding $(2k+6)\times 3$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving $18-5a=3$ for $a$.

💡Takeaway. "True no matter what

k

is" is a big clue: match the

k

-parts and the plain-number parts separately.

Variant 11 hard

The equation $(3k+7)x-4a=bk+2$ in $x$ always has $x=2$ as a solution, no matter what value $k$ takes. Find the sum $a+b$ of the constants $a$ and $b$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Plugging $x=2$ into the equation must give a true statement for every value of $k$ . Use that to pin down the constants $a$ and $b$ , then add them.

Givens

The equation is $(3k+7)x-4a=bk+2$ .
$x=2$ is a solution for every value of $k$ .
$a$ and $b$ are constants.

Unknowns

The value of $a+b$ .

Constraints

The statement must hold for all $k$ , not just one value of $k$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

After substituting $x=2$ , stop treating $k$ as a fixed number and look at it as the variable: an equation that holds for every $k$ is an identity in $k$ , so both sides must have the same $k$ -part and the same constant part.

3 · Execute4 carry out the plan

1Substitute the solution

#13 Convert to Algebra 7.EE.A.1

Since

x=2

is always a solution, put

x=2

into the equation and expand.

(3k+7)\times 2-4a=bk+2 \;\Rightarrow\; 6k+14-4a=bk+2

Now only

k

,

a

, and

b

are left.

2Read it as an identity in k

#16 Count the Complement 8.EE.C.7

The line

6k+14-4a=bk+2

is true for every

k

. Two expressions in

k

that agree for every

k

must have the same coefficient of

k

.

6k = bk \text{ for all } k \;\Rightarrow\; b=6

If the

k

-parts differed, changing

k

would make the two sides drift apart.

3Match the constant parts

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

With the

k

-parts equal, the parts without

k

must also be equal.

14-4a=2 \;\Rightarrow\; 4a=12 \;\Rightarrow\; a=3

A one-step equation finishes it.

4Add the constants

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

Add the two values.

a+b=3+6=9

Both constants are now known.

Answer:

9

4 · Reviewdoes it hold up?

With $a=3$ and $b=6$ the equation is $(3k+7)x-12=6k+2$ . At $x=2$ the left side is $6k+14-12=6k+2$ , equal to the right side for every $k$ .

Another way: Use two easy values of $k$ (Solve an Easier Related Problem): $k=0$ gives $7\times 2-4a=2$ , so $a=3$ ; $k=1$ gives $10\times 2-12=b+2$ , so $b=6$ . Again $a+b=9$ .

Standardsmin grade 8

8.EE.C.7 Solve linear equations in one variable — Recognizing an equation that holds for every value as an identity.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Expanding $(3k+7)\times 2$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving $14-4a=2$ for $a$.

💡Takeaway. "True no matter what

k

is" is a big clue: match the

k

-parts and the plain-number parts separately.

Variant 12 hard

The equation $(2k+7)x-3a=bk+5$ in $x$ always has $x=2$ as a solution, no matter what value $k$ takes. Find the sum $a+b$ of the constants $a$ and $b$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Plugging $x=2$ into the equation must give a true statement for every value of $k$ . Use that to pin down the constants $a$ and $b$ , then add them.

Givens

The equation is $(2k+7)x-3a=bk+5$ .
$x=2$ is a solution for every value of $k$ .
$a$ and $b$ are constants.

Unknowns

The value of $a+b$ .

Constraints

The statement must hold for all $k$ , not just one value of $k$ .

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#9 Solve an Easier Related Problem

After substituting $x=2$ , stop treating $k$ as a fixed number and look at it as the variable: an equation that holds for every $k$ is an identity in $k$ , so both sides must have the same $k$ -part and the same constant part.

3 · Execute4 carry out the plan

1Substitute the solution

#13 Convert to Algebra 7.EE.A.1

Since

x=2

is always a solution, put

x=2

into the equation and expand.

(2k+7)\times 2-3a=bk+5 \;\Rightarrow\; 4k+14-3a=bk+5

Now only

k

,

a

, and

b

are left.

2Read it as an identity in k

#16 Count the Complement 8.EE.C.7

The line

4k+14-3a=bk+5

is true for every

k

. Two expressions in

k

that agree for every

k

must have the same coefficient of

k

.

4k = bk \text{ for all } k \;\Rightarrow\; b=4

If the

k

-parts differed, changing

k

would make the two sides drift apart.

3Match the constant parts

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

With the

k

-parts equal, the parts without

k

must also be equal.

14-3a=5 \;\Rightarrow\; 3a=9 \;\Rightarrow\; a=3

A one-step equation finishes it.

4Add the constants

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

Add the two values.

a+b=3+4=7

Both constants are now known.

Answer:

7

4 · Reviewdoes it hold up?

With $a=3$ and $b=4$ the equation is $(2k+7)x-9=4k+5$ . At $x=2$ the left side is $4k+14-9=4k+5$ , equal to the right side for every $k$ .

Another way: Use two easy values of $k$ (Solve an Easier Related Problem): $k=0$ gives $7\times 2-3a=5$ , so $a=3$ ; $k=1$ gives $9\times 2-9=b+5$ , so $b=4$ . Again $a+b=7$ .

Standardsmin grade 8

8.EE.C.7 Solve linear equations in one variable — Recognizing an equation that holds for every value as an identity.
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions — Expanding $(2k+7)\times 2$.
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Solving $14-3a=5$ for $a$.

💡Takeaway. "True no matter what

k

is" is a big clue: match the

k

-parts and the plain-number parts separately.

True for every k means an identity in k · 12 practice problems

Generated variants — 12

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