Practice · A misread answer still fits the untouched equation · Sensim Math

Variant 1 easy

Emma and Noah each solved the system of equations $ax+y=-2$ , $x-by=3$ . Emma misread the constant $a$ and got $x=-3$ , $y=-3$ . Noah misread the constant $b$ and got $x=-1$ , $y=-1$ . Find the solution of the original system.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students solved the same system but each misread a different constant. From their wrong answers we must recover the true constants and then solve the real system.

Givens

The system is $ax+y=-2$ , $x-by=3$ .
Emma misread $a$ and got $x=-3$ , $y=-3$ .
Noah misread $b$ and got $x=-1$ , $y=-1$ .

Unknowns

The solution $(x, y)$ of the original system.

Constraints

Each student copied every number except the one constant they misread.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra

Change focus from what went wrong to what each student got right. Emma copied the second equation correctly, so her answer satisfies it; Noah copied the first equation correctly, so his answer satisfies that one. Each fact pins down one constant, and then the real system is ordinary algebra.

3 · Execute4 carry out the plan

1Use Emma's answer in the equation she copied correctly

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

Emma's mistake was in

a

, which appears only in the first equation. Her pair

x=-3

,

y=-3

therefore satisfies the second equation,

x-by=3

.

(-3) - b \times (-3) = 3 \;\Rightarrow\; -3 + 3b = 3 \;\Rightarrow\; b = 2

A wrong answer is still a correct solution of the equation the student did not change.

2Use Noah's answer in the equation he copied correctly

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

Noah misread

b

, so the first equation

ax+y=-2

was copied correctly, and

x=-1

,

y=-1

satisfies it.

a \times (-1) - 1 = -2 \;\Rightarrow\; -a = -1 \;\Rightarrow\; a = 1

The same idea, used on the other equation, gives the other constant.

3Write the true system and eliminate $y$

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

With

a=1

and

b=2

the system is

x+y=-2

and

x-2y=3

. Multiply the first equation by

-2

so both have

-2y

, then subtract it from the second.

(x-2y)-(-2)(x+y) = 3-(-2) \times (-2) \;\Rightarrow\; 3x = -1 \;\Rightarrow\; x = -\frac{1}{3}

Matching the

y

terms first lets one subtraction cancel them.

4Find $y$

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

Substitute

x=-\frac{1}{3}

into

x+y=-2

.

y = -2 - \left(-\frac{1}{3}\right) = -\frac{6}{3} + \frac{1}{3} = -\frac{5}{3}

Fractions are fine here; nothing says the solution must be whole numbers.

Answer:

x=-\frac{1}{3}

,

y=-\frac{5}{3}

4 · Reviewdoes it hold up?

Check both original equations: $1 \times \left(-\frac{1}{3}\right) + \left(-\frac{5}{3}\right) = -\frac{1}{3} - \frac{5}{3} = -2$ and $1 \times \left(-\frac{1}{3}\right) - 2 \times \left(-\frac{5}{3}\right) = -\frac{1}{3} + \frac{10}{3} = 3$ . Both hold.

Another way: Use substitution instead of elimination: from $x+y=-2$ write $y=-2-x$ , put it into $x-2y=3$ to get $3x+4=3$ , so $x=-\frac{1}{3}$ and $y=-\frac{5}{3}$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using that a solution satisfies an equation to recover $a$ and $b$, then solving the corrected system.

💡Takeaway. A misread constant only spoils its own equation, so a wrong answer still tells you something true about the other one.

Variant 2 easy

Emma and Noah each solved the system of equations $ax+y=-3$ , $x-by=4$ . Emma misread the constant $a$ and got $x=2$ , $y=2$ . Noah misread the constant $b$ and got $x=-2$ , $y=3$ . Find the solution of the original system.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students solved the same system but each misread a different constant. From their wrong answers we must recover the true constants and then solve the real system.

Givens

The system is $ax+y=-3$ , $x-by=4$ .
Emma misread $a$ and got $x=2$ , $y=2$ .
Noah misread $b$ and got $x=-2$ , $y=3$ .

Unknowns

The solution $(x, y)$ of the original system.

Constraints

Each student copied every number except the one constant they misread.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra

Change focus from what went wrong to what each student got right. Emma copied the second equation correctly, so her answer satisfies it; Noah copied the first equation correctly, so his answer satisfies that one. Each fact pins down one constant, and then the real system is ordinary algebra.

3 · Execute4 carry out the plan

1Use Emma's answer in the equation she copied correctly

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

Emma's mistake was in

a

, which appears only in the first equation. Her pair

x=2

,

y=2

therefore satisfies the second equation,

x-by=4

.

2 - b \times 2 = 4 \;\Rightarrow\; 2 - 2b = 4 \;\Rightarrow\; b = -1

A wrong answer is still a correct solution of the equation the student did not change.

2Use Noah's answer in the equation he copied correctly

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

Noah misread

b

, so the first equation

ax+y=-3

was copied correctly, and

x=-2

,

y=3

satisfies it.

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

The same idea, used on the other equation, gives the other constant.

3Write the true system and eliminate $y$

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

With

a=3

and

b=-1

the system is

3x+y=-3

and

x+y=4

. Subtracting the first from the second removes

y

.

(x+y)-(3x+y) = 4-(-3) \;\Rightarrow\; -2x = 7 \;\Rightarrow\; x = -\frac{7}{2}

Both equations have

+y

, so subtraction cancels it at once.

4Find $y$

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

Substitute

x=-\frac{7}{2}

into

x+y=4

.

y = 4 - \left(-\frac{7}{2}\right) = \frac{8}{2} + \frac{7}{2} = \frac{15}{2}

Fractions are fine here; nothing says the solution must be whole numbers.

Answer:

x=-\frac{7}{2}

,

y=\frac{15}{2}

4 · Reviewdoes it hold up?

Check both original equations: $3 \times \left(-\frac{7}{2}\right) + \frac{15}{2} = -\frac{21}{2} + \frac{15}{2} = -3$ and $1 \times \left(-\frac{7}{2}\right) + \frac{15}{2} = -\frac{7}{2} + \frac{15}{2} = 4$ . Both hold.

Another way: Use substitution instead of elimination: from $x+y=4$ write $y=4-x$ , put it into $3x+y=-3$ to get $2x+4=-3$ , so $x=-\frac{7}{2}$ and $y=\frac{15}{2}$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using that a solution satisfies an equation to recover $a$ and $b$, then solving the corrected system.

💡Takeaway. A misread constant only spoils its own equation, so a wrong answer still tells you something true about the other one.

Variant 3 easy

Emma and Noah each solved the system of equations $ax+y=-1$ , $2x-by=3$ . Emma misread the constant $a$ and got $x=3$ , $y=-3$ . Noah misread the constant $b$ and got $x=1$ , $y=2$ . Find the solution of the original system.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students solved the same system but each misread a different constant. From their wrong answers we must recover the true constants and then solve the real system.

Givens

The system is $ax+y=-1$ , $2x-by=3$ .
Emma misread $a$ and got $x=3$ , $y=-3$ .
Noah misread $b$ and got $x=1$ , $y=2$ .

Unknowns

The solution $(x, y)$ of the original system.

Constraints

Each student copied every number except the one constant they misread.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra

Change focus from what went wrong to what each student got right. Emma copied the second equation correctly, so her answer satisfies it; Noah copied the first equation correctly, so his answer satisfies that one. Each fact pins down one constant, and then the real system is ordinary algebra.

3 · Execute4 carry out the plan

1Use Emma's answer in the equation she copied correctly

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

Emma's mistake was in

a

, which appears only in the first equation. Her pair

x=3

,

y=-3

therefore satisfies the second equation,

2x-by=3

.

2 \times 3 - b \times (-3) = 3 \;\Rightarrow\; 6 + 3b = 3 \;\Rightarrow\; b = -1

A wrong answer is still a correct solution of the equation the student did not change.

2Use Noah's answer in the equation he copied correctly

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

Noah misread

b

, so the first equation

ax+y=-1

was copied correctly, and

x=1

,

y=2

satisfies it.

a \times 1 + 2 = -1 \;\Rightarrow\; a = -3

The same idea, used on the other equation, gives the other constant.

3Write the true system and eliminate $y$

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

With

a=-3

and

b=-1

the system is

-3x+y=-1

and

2x+y=3

. Subtracting the first from the second removes

y

.

(2x+y)-(-3x+y) = 3-(-1) \;\Rightarrow\; 5x = 4 \;\Rightarrow\; x = \frac{4}{5}

Both equations have

+y

, so subtraction cancels it at once.

4Find $y$

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

Substitute

x=\frac{4}{5}

into

2x+y=3

.

y = 3 - 2 \times \frac{4}{5} = \frac{15}{5} - \frac{8}{5} = \frac{7}{5}

Fractions are fine here; nothing says the solution must be whole numbers.

Answer:

x=\frac{4}{5}

,

y=\frac{7}{5}

4 · Reviewdoes it hold up?

Check both original equations: $(-3) \times \frac{4}{5} + \frac{7}{5} = -\frac{12}{5} + \frac{7}{5} = -1$ and $2 \times \frac{4}{5} + \frac{7}{5} = \frac{8}{5} + \frac{7}{5} = 3$ . Both hold.

Another way: Use substitution instead of elimination: from $2x+y=3$ write $y=3-2x$ , put it into $-3x+y=-1$ to get $-5x+3=-1$ , so $x=\frac{4}{5}$ and $y=\frac{7}{5}$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using that a solution satisfies an equation to recover $a$ and $b$, then solving the corrected system.

💡Takeaway. A misread constant only spoils its own equation, so a wrong answer still tells you something true about the other one.

Variant 4 easy

Emma and Noah each solved the system of equations $ax+y=-1$ , $3x-by=-2$ . Emma misread the constant $a$ and got $x=-1$ , $y=-1$ . Noah misread the constant $b$ and got $x=-1$ , $y=-3$ . Find the solution of the original system.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students solved the same system but each misread a different constant. From their wrong answers we must recover the true constants and then solve the real system.

Givens

The system is $ax+y=-1$ , $3x-by=-2$ .
Emma misread $a$ and got $x=-1$ , $y=-1$ .
Noah misread $b$ and got $x=-1$ , $y=-3$ .

Unknowns

The solution $(x, y)$ of the original system.

Constraints

Each student copied every number except the one constant they misread.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra

Change focus from what went wrong to what each student got right. Emma copied the second equation correctly, so her answer satisfies it; Noah copied the first equation correctly, so his answer satisfies that one. Each fact pins down one constant, and then the real system is ordinary algebra.

3 · Execute4 carry out the plan

1Use Emma's answer in the equation she copied correctly

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

Emma's mistake was in

a

, which appears only in the first equation. Her pair

x=-1

,

y=-1

therefore satisfies the second equation,

3x-by=-2

.

3 \times (-1) - b \times (-1) = -2 \;\Rightarrow\; -3 + b = -2 \;\Rightarrow\; b = 1

A wrong answer is still a correct solution of the equation the student did not change.

2Use Noah's answer in the equation he copied correctly

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

Noah misread

b

, so the first equation

ax+y=-1

was copied correctly, and

x=-1

,

y=-3

satisfies it.

a \times (-1) - 3 = -1 \;\Rightarrow\; -a = 2 \;\Rightarrow\; a = -2

The same idea, used on the other equation, gives the other constant.

3Write the true system and eliminate $y$

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

With

a=-2

and

b=1

the system is

-2x+y=-1

and

3x-y=-2

. Adding the two equations removes

y

.

(-2x+y)+(3x-y) = -1+(-2) \;\Rightarrow\; x = -3

The

y

terms are opposites, so adding cancels them at once.

4Find $y$

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

Substitute

x=-3

into

-2x+y=-1

.

y = -1 - (-2) \times (-3) = -7

Whole numbers this time, but nothing promised that.

Answer:

x=-3

,

y=-7

4 · Reviewdoes it hold up?

Check both original equations: $(-2) \times (-3) + (-7) = 6 - 7 = -1$ and $3 \times (-3) - (-7) = -9 + 7 = -2$ . Both hold.

Another way: Use substitution instead of elimination: from $-2x+y=-1$ write $y=-1+2x$ , put it into $3x-y=-2$ to get $x+1=-2$ , so $x=-3$ and $y=-7$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using that a solution satisfies an equation to recover $a$ and $b$, then solving the corrected system.

💡Takeaway. A misread constant only spoils its own equation, so a wrong answer still tells you something true about the other one.

Variant 5 medium

Emma and Noah each solved the system of equations $ax+y=-5$ , $3x-by=-3$ . Emma misread the constant $a$ and got $x=1$ , $y=2$ . Noah misread the constant $b$ and got $x=-1$ , $y=-2$ . Find the solution of the original system.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students solved the same system but each misread a different constant. From their wrong answers we must recover the true constants and then solve the real system.

Givens

The system is $ax+y=-5$ , $3x-by=-3$ .
Emma misread $a$ and got $x=1$ , $y=2$ .
Noah misread $b$ and got $x=-1$ , $y=-2$ .

Unknowns

The solution $(x, y)$ of the original system.

Constraints

Each student copied every number except the one constant they misread.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra

Change focus from what went wrong to what each student got right. Emma copied the second equation correctly, so her answer satisfies it; Noah copied the first equation correctly, so his answer satisfies that one. Each fact pins down one constant, and then the real system is ordinary algebra.

3 · Execute4 carry out the plan

1Use Emma's answer in the equation she copied correctly

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

Emma's mistake was in

a

, which appears only in the first equation. Her pair

x=1

,

y=2

therefore satisfies the second equation,

3x-by=-3

.

3 \times 1 - b \times 2 = -3 \;\Rightarrow\; 3 - 2b = -3 \;\Rightarrow\; b = 3

A wrong answer is still a correct solution of the equation the student did not change.

2Use Noah's answer in the equation he copied correctly

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

Noah misread

b

, so the first equation

ax+y=-5

was copied correctly, and

x=-1

,

y=-2

satisfies it.

a \times (-1) - 2 = -5 \;\Rightarrow\; -a = -3 \;\Rightarrow\; a = 3

The same idea, used on the other equation, gives the other constant.

3Write the true system and eliminate $y$

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

With

a=3

and

b=3

the system is

3x+y=-5

and

3x-3y=-3

. Multiply the first equation by

-3

so both have

-3y

, then subtract it from the second.

(3x-3y)-(-3)(3x+y) = -3-(-3) \times (-5) \;\Rightarrow\; 12x = -18 \;\Rightarrow\; x = -\frac{3}{2}

Matching the

y

terms first lets one subtraction cancel them.

4Find $y$

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

Substitute

x=-\frac{3}{2}

into

3x+y=-5

.

y = -5 - 3 \times \left(-\frac{3}{2}\right) = -\frac{10}{2} + \frac{9}{2} = -\frac{1}{2}

Fractions are fine here; nothing says the solution must be whole numbers.

Answer:

x=-\frac{3}{2}

,

y=-\frac{1}{2}

4 · Reviewdoes it hold up?

Check both original equations: $3 \times \left(-\frac{3}{2}\right) + \left(-\frac{1}{2}\right) = -\frac{9}{2} - \frac{1}{2} = -5$ and $3 \times \left(-\frac{3}{2}\right) - 3 \times \left(-\frac{1}{2}\right) = -\frac{9}{2} + \frac{3}{2} = -3$ . Both hold.

Another way: Use substitution instead of elimination: from $3x+y=-5$ write $y=-5-3x$ , put it into $3x-3y=-3$ to get $12x+15=-3$ , so $x=-\frac{3}{2}$ and $y=-\frac{1}{2}$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using that a solution satisfies an equation to recover $a$ and $b$, then solving the corrected system.

💡Takeaway. A misread constant only spoils its own equation, so a wrong answer still tells you something true about the other one.

Variant 6 medium

Emma and Noah each solved the system of equations $ax+y=-2$ , $3x-by=-6$ . Emma misread the constant $a$ and got $x=-4$ , $y=-2$ . Noah misread the constant $b$ and got $x=1$ , $y=1$ . Find the solution of the original system.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students solved the same system but each misread a different constant. From their wrong answers we must recover the true constants and then solve the real system.

Givens

The system is $ax+y=-2$ , $3x-by=-6$ .
Emma misread $a$ and got $x=-4$ , $y=-2$ .
Noah misread $b$ and got $x=1$ , $y=1$ .

Unknowns

The solution $(x, y)$ of the original system.

Constraints

Each student copied every number except the one constant they misread.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra

Change focus from what went wrong to what each student got right. Emma copied the second equation correctly, so her answer satisfies it; Noah copied the first equation correctly, so his answer satisfies that one. Each fact pins down one constant, and then the real system is ordinary algebra.

3 · Execute4 carry out the plan

1Use Emma's answer in the equation she copied correctly

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

Emma's mistake was in

a

, which appears only in the first equation. Her pair

x=-4

,

y=-2

therefore satisfies the second equation,

3x-by=-6

.

3 \times (-4) - b \times (-2) = -6 \;\Rightarrow\; -12 + 2b = -6 \;\Rightarrow\; b = 3

A wrong answer is still a correct solution of the equation the student did not change.

2Use Noah's answer in the equation he copied correctly

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

Noah misread

b

, so the first equation

ax+y=-2

was copied correctly, and

x=1

,

y=1

satisfies it.

a \times 1 + 1 = -2 \;\Rightarrow\; a = -3

The same idea, used on the other equation, gives the other constant.

3Write the true system and eliminate $y$

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

With

a=-3

and

b=3

the system is

-3x+y=-2

and

3x-3y=-6

. Multiply the first equation by

-3

so both have

-3y

, then subtract it from the second.

(3x-3y)-(-3)(-3x+y) = -6-(-3) \times (-2) \;\Rightarrow\; -6x = -12 \;\Rightarrow\; x = 2

Matching the

y

terms first lets one subtraction cancel them.

4Find $y$

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

Substitute

x=2

into

-3x+y=-2

.

y = -2 - (-3) \times 2 = 4

Whole numbers this time, but nothing promised that.

Answer:

x=2

,

y=4

4 · Reviewdoes it hold up?

Check both original equations: $(-3) \times 2 + 4 = -6 + 4 = -2$ and $3 \times 2 - 3 \times 4 = 6 - 12 = -6$ . Both hold.

Another way: Use substitution instead of elimination: from $-3x+y=-2$ write $y=-2+3x$ , put it into $3x-3y=-6$ to get $-6x+6=-6$ , so $x=2$ and $y=4$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using that a solution satisfies an equation to recover $a$ and $b$, then solving the corrected system.

💡Takeaway. A misread constant only spoils its own equation, so a wrong answer still tells you something true about the other one.

Variant 7 medium

Emma and Noah each solved the system of equations $ax+y=-6$ , $2x-by=-7$ . Emma misread the constant $a$ and got $x=-2$ , $y=-3$ . Noah misread the constant $b$ and got $x=-1$ , $y=-2$ . Find the solution of the original system.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students solved the same system but each misread a different constant. From their wrong answers we must recover the true constants and then solve the real system.

Givens

The system is $ax+y=-6$ , $2x-by=-7$ .
Emma misread $a$ and got $x=-2$ , $y=-3$ .
Noah misread $b$ and got $x=-1$ , $y=-2$ .

Unknowns

The solution $(x, y)$ of the original system.

Constraints

Each student copied every number except the one constant they misread.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra

Change focus from what went wrong to what each student got right. Emma copied the second equation correctly, so her answer satisfies it; Noah copied the first equation correctly, so his answer satisfies that one. Each fact pins down one constant, and then the real system is ordinary algebra.

3 · Execute4 carry out the plan

1Use Emma's answer in the equation she copied correctly

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

Emma's mistake was in

a

, which appears only in the first equation. Her pair

x=-2

,

y=-3

therefore satisfies the second equation,

2x-by=-7

.

2 \times (-2) - b \times (-3) = -7 \;\Rightarrow\; -4 + 3b = -7 \;\Rightarrow\; b = -1

A wrong answer is still a correct solution of the equation the student did not change.

2Use Noah's answer in the equation he copied correctly

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

Noah misread

b

, so the first equation

ax+y=-6

was copied correctly, and

x=-1

,

y=-2

satisfies it.

a \times (-1) - 2 = -6 \;\Rightarrow\; -a = -4 \;\Rightarrow\; a = 4

The same idea, used on the other equation, gives the other constant.

3Write the true system and eliminate $y$

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

With

a=4

and

b=-1

the system is

4x+y=-6

and

2x+y=-7

. Subtracting the first from the second removes

y

.

(2x+y)-(4x+y) = -7-(-6) \;\Rightarrow\; -2x = -1 \;\Rightarrow\; x = \frac{1}{2}

Both equations have

+y

, so subtraction cancels it at once.

4Find $y$

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

Substitute

x=\frac{1}{2}

into

2x+y=-7

.

y = -7 - 2 \times \frac{1}{2} = -\frac{14}{2} - \frac{2}{2} = -8

Fractions are fine here; nothing says the solution must be whole numbers.

Answer:

x=\frac{1}{2}

,

y=-8

4 · Reviewdoes it hold up?

Check both original equations: $4 \times \frac{1}{2} + (-8) = 2 - 8 = -6$ and $2 \times \frac{1}{2} + (-8) = 1 - 8 = -7$ . Both hold.

Another way: Use substitution instead of elimination: from $2x+y=-7$ write $y=-7-2x$ , put it into $4x+y=-6$ to get $2x-7=-6$ , so $x=\frac{1}{2}$ and $y=-8$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using that a solution satisfies an equation to recover $a$ and $b$, then solving the corrected system.

💡Takeaway. A misread constant only spoils its own equation, so a wrong answer still tells you something true about the other one.

Variant 8 medium

Emma and Noah each solved the system of equations $ax+y=2$ , $3x-by=-7$ . Emma misread the constant $a$ and got $x=-3$ , $y=-2$ . Noah misread the constant $b$ and got $x=2$ , $y=-2$ . Find the solution of the original system.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students solved the same system but each misread a different constant. From their wrong answers we must recover the true constants and then solve the real system.

Givens

The system is $ax+y=2$ , $3x-by=-7$ .
Emma misread $a$ and got $x=-3$ , $y=-2$ .
Noah misread $b$ and got $x=2$ , $y=-2$ .

Unknowns

The solution $(x, y)$ of the original system.

Constraints

Each student copied every number except the one constant they misread.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra

Change focus from what went wrong to what each student got right. Emma copied the second equation correctly, so her answer satisfies it; Noah copied the first equation correctly, so his answer satisfies that one. Each fact pins down one constant, and then the real system is ordinary algebra.

3 · Execute4 carry out the plan

1Use Emma's answer in the equation she copied correctly

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

Emma's mistake was in

a

, which appears only in the first equation. Her pair

x=-3

,

y=-2

therefore satisfies the second equation,

3x-by=-7

.

3 \times (-3) - b \times (-2) = -7 \;\Rightarrow\; -9 + 2b = -7 \;\Rightarrow\; b = 1

A wrong answer is still a correct solution of the equation the student did not change.

2Use Noah's answer in the equation he copied correctly

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

Noah misread

b

, so the first equation

ax+y=2

was copied correctly, and

x=2

,

y=-2

satisfies it.

a \times 2 - 2 = 2 \;\Rightarrow\; 2a = 4 \;\Rightarrow\; a = 2

The same idea, used on the other equation, gives the other constant.

3Write the true system and eliminate $y$

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

With

a=2

and

b=1

the system is

2x+y=2

and

3x-y=-7

. Adding the two equations removes

y

.

(2x+y)+(3x-y) = 2+(-7) \;\Rightarrow\; 5x = -5 \;\Rightarrow\; x = -1

The

y

terms are opposites, so adding cancels them at once.

4Find $y$

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

Substitute

x=-1

into

2x+y=2

.

y = 2 - 2 \times (-1) = 4

Whole numbers this time, but nothing promised that.

Answer:

x=-1

,

y=4

4 · Reviewdoes it hold up?

Check both original equations: $2 \times (-1) + 4 = -2 + 4 = 2$ and $3 \times (-1) - 4 = -3 - 4 = -7$ . Both hold.

Another way: Use substitution instead of elimination: from $2x+y=2$ write $y=2-2x$ , put it into $3x-y=-7$ to get $5x-2=-7$ , so $x=-1$ and $y=4$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using that a solution satisfies an equation to recover $a$ and $b$, then solving the corrected system.

💡Takeaway. A misread constant only spoils its own equation, so a wrong answer still tells you something true about the other one.

Variant 9 hard

Emma and Noah each solved the system of equations $ax+y=-8$ , $4x-by=9$ . Emma misread the constant $a$ and got $x=3$ , $y=-3$ . Noah misread the constant $b$ and got $x=2$ , $y=-4$ . Find the solution of the original system.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students solved the same system but each misread a different constant. From their wrong answers we must recover the true constants and then solve the real system.

Givens

The system is $ax+y=-8$ , $4x-by=9$ .
Emma misread $a$ and got $x=3$ , $y=-3$ .
Noah misread $b$ and got $x=2$ , $y=-4$ .

Unknowns

The solution $(x, y)$ of the original system.

Constraints

Each student copied every number except the one constant they misread.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra

Change focus from what went wrong to what each student got right. Emma copied the second equation correctly, so her answer satisfies it; Noah copied the first equation correctly, so his answer satisfies that one. Each fact pins down one constant, and then the real system is ordinary algebra.

3 · Execute4 carry out the plan

1Use Emma's answer in the equation she copied correctly

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

Emma's mistake was in

a

, which appears only in the first equation. Her pair

x=3

,

y=-3

therefore satisfies the second equation,

4x-by=9

.

4 \times 3 - b \times (-3) = 9 \;\Rightarrow\; 12 + 3b = 9 \;\Rightarrow\; b = -1

A wrong answer is still a correct solution of the equation the student did not change.

2Use Noah's answer in the equation he copied correctly

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

Noah misread

b

, so the first equation

ax+y=-8

was copied correctly, and

x=2

,

y=-4

satisfies it.

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

The same idea, used on the other equation, gives the other constant.

3Write the true system and eliminate $y$

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

With

a=-2

and

b=-1

the system is

-2x+y=-8

and

4x+y=9

. Subtracting the first from the second removes

y

.

(4x+y)-(-2x+y) = 9-(-8) \;\Rightarrow\; 6x = 17 \;\Rightarrow\; x = \frac{17}{6}

Both equations have

+y

, so subtraction cancels it at once.

4Find $y$

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

Substitute

x=\frac{17}{6}

into

4x+y=9

.

y = 9 - 4 \times \frac{17}{6} = \frac{54}{6} - \frac{68}{6} = -\frac{7}{3}

Fractions are fine here; nothing says the solution must be whole numbers.

Answer:

x=\frac{17}{6}

,

y=-\frac{7}{3}

4 · Reviewdoes it hold up?

Check both original equations: $(-2) \times \frac{17}{6} + \left(-\frac{7}{3}\right) = -\frac{17}{3} - \frac{7}{3} = -8$ and $4 \times \frac{17}{6} + \left(-\frac{7}{3}\right) = \frac{34}{3} - \frac{7}{3} = 9$ . Both hold.

Another way: Use substitution instead of elimination: from $4x+y=9$ write $y=9-4x$ , put it into $-2x+y=-8$ to get $-6x+9=-8$ , so $x=\frac{17}{6}$ and $y=-\frac{7}{3}$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using that a solution satisfies an equation to recover $a$ and $b$, then solving the corrected system.

💡Takeaway. A misread constant only spoils its own equation, so a wrong answer still tells you something true about the other one.

Variant 10 hard

Emma and Noah each solved the system of equations $ax+y=17$ , $x-by=6$ . Emma misread the constant $a$ and got $x=2$ , $y=4$ . Noah misread the constant $b$ and got $x=4$ , $y=1$ . Find the solution of the original system.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students solved the same system but each misread a different constant. From their wrong answers we must recover the true constants and then solve the real system.

Givens

The system is $ax+y=17$ , $x-by=6$ .
Emma misread $a$ and got $x=2$ , $y=4$ .
Noah misread $b$ and got $x=4$ , $y=1$ .

Unknowns

The solution $(x, y)$ of the original system.

Constraints

Each student copied every number except the one constant they misread.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra

Change focus from what went wrong to what each student got right. Emma copied the second equation correctly, so her answer satisfies it; Noah copied the first equation correctly, so his answer satisfies that one. Each fact pins down one constant, and then the real system is ordinary algebra.

3 · Execute4 carry out the plan

1Use Emma's answer in the equation she copied correctly

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

Emma's mistake was in

a

, which appears only in the first equation. Her pair

x=2

,

y=4

therefore satisfies the second equation,

x-by=6

.

2 - b \times 4 = 6 \;\Rightarrow\; 2 - 4b = 6 \;\Rightarrow\; b = -1

A wrong answer is still a correct solution of the equation the student did not change.

2Use Noah's answer in the equation he copied correctly

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

Noah misread

b

, so the first equation

ax+y=17

was copied correctly, and

x=4

,

y=1

satisfies it.

a \times 4 + 1 = 17 \;\Rightarrow\; 4a = 16 \;\Rightarrow\; a = 4

The same idea, used on the other equation, gives the other constant.

3Write the true system and eliminate $y$

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

With

a=4

and

b=-1

the system is

4x+y=17

and

x+y=6

. Subtracting the first from the second removes

y

.

(x+y)-(4x+y) = 6-17 \;\Rightarrow\; -3x = -11 \;\Rightarrow\; x = \frac{11}{3}

Both equations have

+y

, so subtraction cancels it at once.

4Find $y$

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

Substitute

x=\frac{11}{3}

into

x+y=6

.

y = 6 - \frac{11}{3} = \frac{18}{3} - \frac{11}{3} = \frac{7}{3}

Fractions are fine here; nothing says the solution must be whole numbers.

Answer:

x=\frac{11}{3}

,

y=\frac{7}{3}

4 · Reviewdoes it hold up?

Check both original equations: $4 \times \frac{11}{3} + \frac{7}{3} = \frac{44}{3} + \frac{7}{3} = 17$ and $1 \times \frac{11}{3} + \frac{7}{3} = \frac{11}{3} + \frac{7}{3} = 6$ . Both hold.

Another way: Use substitution instead of elimination: from $x+y=6$ write $y=6-x$ , put it into $4x+y=17$ to get $3x+6=17$ , so $x=\frac{11}{3}$ and $y=\frac{7}{3}$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using that a solution satisfies an equation to recover $a$ and $b$, then solving the corrected system.

💡Takeaway. A misread constant only spoils its own equation, so a wrong answer still tells you something true about the other one.

Variant 11 hard

Emma and Noah each solved the system of equations $ax+y=5$ , $3x-by=-10$ . Emma misread the constant $a$ and got $x=-4$ , $y=-2$ . Noah misread the constant $b$ and got $x=2$ , $y=3$ . Find the solution of the original system.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students solved the same system but each misread a different constant. From their wrong answers we must recover the true constants and then solve the real system.

Givens

The system is $ax+y=5$ , $3x-by=-10$ .
Emma misread $a$ and got $x=-4$ , $y=-2$ .
Noah misread $b$ and got $x=2$ , $y=3$ .

Unknowns

The solution $(x, y)$ of the original system.

Constraints

Each student copied every number except the one constant they misread.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra

Change focus from what went wrong to what each student got right. Emma copied the second equation correctly, so her answer satisfies it; Noah copied the first equation correctly, so his answer satisfies that one. Each fact pins down one constant, and then the real system is ordinary algebra.

3 · Execute4 carry out the plan

1Use Emma's answer in the equation she copied correctly

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

Emma's mistake was in

a

, which appears only in the first equation. Her pair

x=-4

,

y=-2

therefore satisfies the second equation,

3x-by=-10

.

3 \times (-4) - b \times (-2) = -10 \;\Rightarrow\; -12 + 2b = -10 \;\Rightarrow\; b = 1

A wrong answer is still a correct solution of the equation the student did not change.

2Use Noah's answer in the equation he copied correctly

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

Noah misread

b

, so the first equation

ax+y=5

was copied correctly, and

x=2

,

y=3

satisfies it.

a \times 2 + 3 = 5 \;\Rightarrow\; 2a = 2 \;\Rightarrow\; a = 1

The same idea, used on the other equation, gives the other constant.

3Write the true system and eliminate $y$

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

With

a=1

and

b=1

the system is

x+y=5

and

3x-y=-10

. Adding the two equations removes

y

.

(x+y)+(3x-y) = 5+(-10) \;\Rightarrow\; 4x = -5 \;\Rightarrow\; x = -\frac{5}{4}

The

y

terms are opposites, so adding cancels them at once.

4Find $y$

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

Substitute

x=-\frac{5}{4}

into

x+y=5

.

y = 5 - \left(-\frac{5}{4}\right) = \frac{20}{4} + \frac{5}{4} = \frac{25}{4}

Fractions are fine here; nothing says the solution must be whole numbers.

Answer:

x=-\frac{5}{4}

,

y=\frac{25}{4}

4 · Reviewdoes it hold up?

Check both original equations: $1 \times \left(-\frac{5}{4}\right) + \frac{25}{4} = -\frac{5}{4} + \frac{25}{4} = 5$ and $3 \times \left(-\frac{5}{4}\right) - \frac{25}{4} = -\frac{15}{4} - \frac{25}{4} = -10$ . Both hold.

Another way: Use substitution instead of elimination: from $x+y=5$ write $y=5-x$ , put it into $3x-y=-10$ to get $4x-5=-10$ , so $x=-\frac{5}{4}$ and $y=\frac{25}{4}$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using that a solution satisfies an equation to recover $a$ and $b$, then solving the corrected system.

💡Takeaway. A misread constant only spoils its own equation, so a wrong answer still tells you something true about the other one.

Variant 12 hard

Emma and Noah each solved the system of equations $ax+y=8$ , $4x-by=-20$ . Emma misread the constant $a$ and got $x=-3$ , $y=-4$ . Noah misread the constant $b$ and got $x=-4$ , $y=4$ . Find the solution of the original system.

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students solved the same system but each misread a different constant. From their wrong answers we must recover the true constants and then solve the real system.

Givens

The system is $ax+y=8$ , $4x-by=-20$ .
Emma misread $a$ and got $x=-3$ , $y=-4$ .
Noah misread $b$ and got $x=-4$ , $y=4$ .

Unknowns

The solution $(x, y)$ of the original system.

Constraints

Each student copied every number except the one constant they misread.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra

Change focus from what went wrong to what each student got right. Emma copied the second equation correctly, so her answer satisfies it; Noah copied the first equation correctly, so his answer satisfies that one. Each fact pins down one constant, and then the real system is ordinary algebra.

3 · Execute4 carry out the plan

1Use Emma's answer in the equation she copied correctly

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

Emma's mistake was in

a

, which appears only in the first equation. Her pair

x=-3

,

y=-4

therefore satisfies the second equation,

4x-by=-20

.

4 \times (-3) - b \times (-4) = -20 \;\Rightarrow\; -12 + 4b = -20 \;\Rightarrow\; b = -2

A wrong answer is still a correct solution of the equation the student did not change.

2Use Noah's answer in the equation he copied correctly

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

Noah misread

b

, so the first equation

ax+y=8

was copied correctly, and

x=-4

,

y=4

satisfies it.

a \times (-4) + 4 = 8 \;\Rightarrow\; -4a = 4 \;\Rightarrow\; a = -1

The same idea, used on the other equation, gives the other constant.

3Write the true system and eliminate $y$

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

With

a=-1

and

b=-2

the system is

-x+y=8

and

4x+2y=-20

. Multiply the first equation by

2

so both have

2y

, then subtract it from the second.

(4x+2y)-2(-x+y) = -20-2 \times 8 \;\Rightarrow\; 6x = -36 \;\Rightarrow\; x = -6

Matching the

y

terms first lets one subtraction cancel them.

4Find $y$

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

Substitute

x=-6

into

-x+y=8

.

y = 8 - (-1) \times (-6) = 2

Whole numbers this time, but nothing promised that.

Answer:

x=-6

,

y=2

4 · Reviewdoes it hold up?

Check both original equations: $(-1) \times (-6) + 2 = 6 + 2 = 8$ and $4 \times (-6) + 2 \times 2 = -24 + 4 = -20$ . Both hold.

Another way: Use substitution instead of elimination: from $-x+y=8$ write $y=8+x$ , put it into $4x+2y=-20$ to get $6x+16=-20$ , so $x=-6$ and $y=2$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Using that a solution satisfies an equation to recover $a$ and $b$, then solving the corrected system.

💡Takeaway. A misread constant only spoils its own equation, so a wrong answer still tells you something true about the other one.

A misread answer still fits the untouched equation · 12 practice problems

Generated variants — 12

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