Practice · Each wrong solve still reveals one true constant · Sensim Math

Variant 1 easy

Nora solved the equation $\frac{1}{2}x+a=-x+b$ in $x$ , but she misread $a$ as $0$ and got $x=8$ . Dana solved the same equation, but she misread $b$ as $0$ and got $x=-4$ . What is the solution of the original equation when it is solved correctly?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students each misread one constant as $0$ but kept the other one right. Each wrong answer therefore pins down the constant that student read correctly. Recover $a$ and $b$ , then solve the real equation.

Givens

The equation is $\frac{1}{2}x+a=-x+b$ .
With $a$ read as $0$ (and $b$ correct), the solution was $x=8$ .
With $b$ read as $0$ (and $a$ correct), the solution was $x=-4$ .

Unknowns

The solution of the original equation.

Constraints

Nora's $b$ is the true $b$ ; Dana's $a$ is the true $a$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Work backwards from each wrong answer: a solution makes its equation true, so substituting it into the misread equation leaves one unknown, the constant that was read correctly. With both constants found, solve forward.

3 · Execute4 carry out the plan

1Nora's equation gives b

#11 Work Backwards 7.EE.B.4

Nora solved

\frac{1}{2}x=-x+b

and got

x=8

. Substitute.

\frac{1}{2}\times 8=-8+b \;\Rightarrow\; 4=-8+b \;\Rightarrow\; b=12

Her

b

was the real one.

2Dana's equation gives a

#11 Work Backwards 7.EE.B.3

Dana solved

\frac{1}{2}x+a=-x

and got

x=-4

. Substitute.

\frac{1}{2}\times(-4)+a=4 \;\Rightarrow\; -2+a=4 \;\Rightarrow\; a=6

Her

a

was the real one.

3Write the true equation

#7 Identify Subproblems 7.EE.B.4

Put both true constants back.

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

Now nothing is misread.

4Solve it

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

Multiply both sides by

2

, then collect terms.

x+12=-2x+24 \;\Rightarrow\; 3x=12 \;\Rightarrow\; x=4

A standard linear equation.

Answer:

x=4

4 · Reviewdoes it hold up?

At $x=4$ : left side $\frac{1}{2}\times 4+6=8$ ; right side $-4+12=8$ . Equal.

Another way: Notice the true equation is Nora's plus Dana's ideas combined: moving terms gives $\frac{3}{2}x=b-a$ . Nora's run says $\frac{3}{2}\times 8=b$ , so $b=12$ ; Dana's says $\frac{3}{2}\times(-4)=-a$ , so $a=6$ . Then $\frac{3}{2}x=6$ and $x=4$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the linear equations.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing with negative numbers and fractions, such as $\frac{1}{2}\times(-4)$.

💡Takeaway. A wrong answer isn't useless: plug it into the equation that was actually solved and it hands you a true number.

Variant 2 easy

Nora solved the equation $\frac{2}{5}x+a=-x+b$ in $x$ , but she misread $a$ as $0$ and got $x=10$ . Dana solved the same equation, but she misread $b$ as $0$ and got $x=-5$ . What is the solution of the original equation when it is solved correctly?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students each misread one constant as $0$ but kept the other one right. Each wrong answer therefore pins down the constant that student read correctly. Recover $a$ and $b$ , then solve the real equation.

Givens

The equation is $\frac{2}{5}x+a=-x+b$ .
With $a$ read as $0$ (and $b$ correct), the solution was $x=10$ .
With $b$ read as $0$ (and $a$ correct), the solution was $x=-5$ .

Unknowns

The solution of the original equation.

Constraints

Nora's $b$ is the true $b$ ; Dana's $a$ is the true $a$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Work backwards from each wrong answer: a solution makes its equation true, so substituting it into the misread equation leaves one unknown, the constant that was read correctly. With both constants found, solve forward.

3 · Execute4 carry out the plan

1Nora's equation gives b

#11 Work Backwards 7.EE.B.4

Nora solved

\frac{2}{5}x=-x+b

and got

x=10

. Substitute.

\frac{2}{5}\times 10=-10+b \;\Rightarrow\; 4=-10+b \;\Rightarrow\; b=14

Her

b

was the real one.

2Dana's equation gives a

#11 Work Backwards 7.EE.B.3

Dana solved

\frac{2}{5}x+a=-x

and got

x=-5

. Substitute.

\frac{2}{5}\times(-5)+a=5 \;\Rightarrow\; -2+a=5 \;\Rightarrow\; a=7

Her

a

was the real one.

3Write the true equation

#7 Identify Subproblems 7.EE.B.4

Put both true constants back.

\frac{2}{5}x+7=-x+14

Now nothing is misread.

4Solve it

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

Multiply both sides by

5

, then collect terms.

2x+35=-5x+70 \;\Rightarrow\; 7x=35 \;\Rightarrow\; x=5

A standard linear equation.

Answer:

x=5

4 · Reviewdoes it hold up?

At $x=5$ : left side $\frac{2}{5}\times 5+7=9$ ; right side $-5+14=9$ . Equal.

Another way: Notice the true equation is Nora's plus Dana's ideas combined: moving terms gives $\frac{7}{5}x=b-a$ . Nora's run says $\frac{7}{5}\times 10=b$ , so $b=14$ ; Dana's says $\frac{7}{5}\times(-5)=-a$ , so $a=7$ . Then $\frac{7}{5}x=7$ and $x=5$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the linear equations.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing with negative numbers and fractions, such as $\frac{2}{5}\times(-5)$.

💡Takeaway. A wrong answer isn't useless: plug it into the equation that was actually solved and it hands you a true number.

Variant 3 easy

Nora solved the equation $\frac{1}{2}x+a=x+b$ in $x$ , but she misread $a$ as $0$ and got $x=10$ . Dana solved the same equation, but she misread $b$ as $0$ and got $x=-4$ . What is the solution of the original equation when it is solved correctly?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students each misread one constant as $0$ but kept the other one right. Each wrong answer therefore pins down the constant that student read correctly. Recover $a$ and $b$ , then solve the real equation.

Givens

The equation is $\frac{1}{2}x+a=x+b$ .
With $a$ read as $0$ (and $b$ correct), the solution was $x=10$ .
With $b$ read as $0$ (and $a$ correct), the solution was $x=-4$ .

Unknowns

The solution of the original equation.

Constraints

Nora's $b$ is the true $b$ ; Dana's $a$ is the true $a$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Work backwards from each wrong answer: a solution makes its equation true, so substituting it into the misread equation leaves one unknown, the constant that was read correctly. With both constants found, solve forward.

3 · Execute4 carry out the plan

1Nora's equation gives b

#11 Work Backwards 7.EE.B.4

Nora solved

\frac{1}{2}x=x+b

and got

x=10

. Substitute.

\frac{1}{2}\times 10=10+b \;\Rightarrow\; 5=10+b \;\Rightarrow\; b=-5

Her

b

was the real one.

2Dana's equation gives a

#11 Work Backwards 7.EE.B.3

Dana solved

\frac{1}{2}x+a=x

and got

x=-4

. Substitute.

\frac{1}{2}\times(-4)+a=-4 \;\Rightarrow\; -2+a=-4 \;\Rightarrow\; a=-2

Her

a

was the real one.

3Write the true equation

#7 Identify Subproblems 7.EE.B.4

Put both true constants back.

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

Now nothing is misread.

4Solve it

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

Multiply both sides by

2

, then collect terms.

x-4=2x-10 \;\Rightarrow\; -x=-6 \;\Rightarrow\; x=6

A standard linear equation.

Answer:

x=6

4 · Reviewdoes it hold up?

At $x=6$ : left side $\frac{1}{2}\times 6-2=1$ ; right side $6-5=1$ . Equal.

Another way: Notice the true equation is Nora's plus Dana's ideas combined: moving terms gives $-\frac{1}{2}x=b-a$ . Nora's run says $-\frac{1}{2}\times 10=b$ , so $b=-5$ ; Dana's says $-\frac{1}{2}\times(-4)=-a$ , so $a=-2$ . Then $-\frac{1}{2}x=-3$ and $x=6$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the linear equations.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing with negative numbers and fractions, such as $\frac{1}{2}\times(-4)$.

💡Takeaway. A wrong answer isn't useless: plug it into the equation that was actually solved and it hands you a true number.

Variant 4 easy

Nora solved the equation $\frac{3}{2}x+a=-x+b$ in $x$ , but she misread $a$ as $0$ and got $x=4$ . Dana solved the same equation, but she misread $b$ as $0$ and got $x=-10$ . What is the solution of the original equation when it is solved correctly?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students each misread one constant as $0$ but kept the other one right. Each wrong answer therefore pins down the constant that student read correctly. Recover $a$ and $b$ , then solve the real equation.

Givens

The equation is $\frac{3}{2}x+a=-x+b$ .
With $a$ read as $0$ (and $b$ correct), the solution was $x=4$ .
With $b$ read as $0$ (and $a$ correct), the solution was $x=-10$ .

Unknowns

The solution of the original equation.

Constraints

Nora's $b$ is the true $b$ ; Dana's $a$ is the true $a$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Work backwards from each wrong answer: a solution makes its equation true, so substituting it into the misread equation leaves one unknown, the constant that was read correctly. With both constants found, solve forward.

3 · Execute4 carry out the plan

1Nora's equation gives b

#11 Work Backwards 7.EE.B.4

Nora solved

\frac{3}{2}x=-x+b

and got

x=4

. Substitute.

\frac{3}{2}\times 4=-4+b \;\Rightarrow\; 6=-4+b \;\Rightarrow\; b=10

Her

b

was the real one.

2Dana's equation gives a

#11 Work Backwards 7.EE.B.3

Dana solved

\frac{3}{2}x+a=-x

and got

x=-10

. Substitute.

\frac{3}{2}\times(-10)+a=10 \;\Rightarrow\; -15+a=10 \;\Rightarrow\; a=25

Her

a

was the real one.

3Write the true equation

#7 Identify Subproblems 7.EE.B.4

Put both true constants back.

\frac{3}{2}x+25=-x+10

Now nothing is misread.

4Solve it

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

Multiply both sides by

2

, then collect terms.

3x+50=-2x+20 \;\Rightarrow\; 5x=-30 \;\Rightarrow\; x=-6

A standard linear equation.

Answer:

x=-6

4 · Reviewdoes it hold up?

At $x=-6$ : left side $\frac{3}{2}\times(-6)+25=16$ ; right side $6+10=16$ . Equal.

Another way: Notice the true equation is Nora's plus Dana's ideas combined: moving terms gives $\frac{5}{2}x=b-a$ . Nora's run says $\frac{5}{2}\times 4=b$ , so $b=10$ ; Dana's says $\frac{5}{2}\times(-10)=-a$ , so $a=25$ . Then $\frac{5}{2}x=-15$ and $x=-6$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the linear equations.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing with negative numbers and fractions, such as $\frac{3}{2}\times(-10)$.

💡Takeaway. A wrong answer isn't useless: plug it into the equation that was actually solved and it hands you a true number.

Variant 5 medium

Nora solved the equation $\frac{4}{3}x+a=-x+b$ in $x$ , but she misread $a$ as $0$ and got $x=6$ . Dana solved the same equation, but she misread $b$ as $0$ and got $x=-12$ . What is the solution of the original equation when it is solved correctly?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students each misread one constant as $0$ but kept the other one right. Each wrong answer therefore pins down the constant that student read correctly. Recover $a$ and $b$ , then solve the real equation.

Givens

The equation is $\frac{4}{3}x+a=-x+b$ .
With $a$ read as $0$ (and $b$ correct), the solution was $x=6$ .
With $b$ read as $0$ (and $a$ correct), the solution was $x=-12$ .

Unknowns

The solution of the original equation.

Constraints

Nora's $b$ is the true $b$ ; Dana's $a$ is the true $a$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Work backwards from each wrong answer: a solution makes its equation true, so substituting it into the misread equation leaves one unknown, the constant that was read correctly. With both constants found, solve forward.

3 · Execute4 carry out the plan

1Nora's equation gives b

#11 Work Backwards 7.EE.B.4

Nora solved

\frac{4}{3}x=-x+b

and got

x=6

. Substitute.

\frac{4}{3}\times 6=-6+b \;\Rightarrow\; 8=-6+b \;\Rightarrow\; b=14

Her

b

was the real one.

2Dana's equation gives a

#11 Work Backwards 7.EE.B.3

Dana solved

\frac{4}{3}x+a=-x

and got

x=-12

. Substitute.

\frac{4}{3}\times(-12)+a=12 \;\Rightarrow\; -16+a=12 \;\Rightarrow\; a=28

Her

a

was the real one.

3Write the true equation

#7 Identify Subproblems 7.EE.B.4

Put both true constants back.

\frac{4}{3}x+28=-x+14

Now nothing is misread.

4Solve it

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

Multiply both sides by

3

, then collect terms.

4x+84=-3x+42 \;\Rightarrow\; 7x=-42 \;\Rightarrow\; x=-6

A standard linear equation.

Answer:

x=-6

4 · Reviewdoes it hold up?

At $x=-6$ : left side $\frac{4}{3}\times(-6)+28=20$ ; right side $6+14=20$ . Equal.

Another way: Notice the true equation is Nora's plus Dana's ideas combined: moving terms gives $\frac{7}{3}x=b-a$ . Nora's run says $\frac{7}{3}\times 6=b$ , so $b=14$ ; Dana's says $\frac{7}{3}\times(-12)=-a$ , so $a=28$ . Then $\frac{7}{3}x=-14$ and $x=-6$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the linear equations.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing with negative numbers and fractions, such as $\frac{4}{3}\times(-12)$.

💡Takeaway. A wrong answer isn't useless: plug it into the equation that was actually solved and it hands you a true number.

Variant 6 medium

Nora solved the equation $\frac{3}{4}x+a=-x+b$ in $x$ , but she misread $a$ as $0$ and got $x=12$ . Dana solved the same equation, but she misread $b$ as $0$ and got $x=-8$ . What is the solution of the original equation when it is solved correctly?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students each misread one constant as $0$ but kept the other one right. Each wrong answer therefore pins down the constant that student read correctly. Recover $a$ and $b$ , then solve the real equation.

Givens

The equation is $\frac{3}{4}x+a=-x+b$ .
With $a$ read as $0$ (and $b$ correct), the solution was $x=12$ .
With $b$ read as $0$ (and $a$ correct), the solution was $x=-8$ .

Unknowns

The solution of the original equation.

Constraints

Nora's $b$ is the true $b$ ; Dana's $a$ is the true $a$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Work backwards from each wrong answer: a solution makes its equation true, so substituting it into the misread equation leaves one unknown, the constant that was read correctly. With both constants found, solve forward.

3 · Execute4 carry out the plan

1Nora's equation gives b

#11 Work Backwards 7.EE.B.4

Nora solved

\frac{3}{4}x=-x+b

and got

x=12

. Substitute.

\frac{3}{4}\times 12=-12+b \;\Rightarrow\; 9=-12+b \;\Rightarrow\; b=21

Her

b

was the real one.

2Dana's equation gives a

#11 Work Backwards 7.EE.B.3

Dana solved

\frac{3}{4}x+a=-x

and got

x=-8

. Substitute.

\frac{3}{4}\times(-8)+a=8 \;\Rightarrow\; -6+a=8 \;\Rightarrow\; a=14

Her

a

was the real one.

3Write the true equation

#7 Identify Subproblems 7.EE.B.4

Put both true constants back.

\frac{3}{4}x+14=-x+21

Now nothing is misread.

4Solve it

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

Multiply both sides by

4

, then collect terms.

3x+56=-4x+84 \;\Rightarrow\; 7x=28 \;\Rightarrow\; x=4

A standard linear equation.

Answer:

x=4

4 · Reviewdoes it hold up?

At $x=4$ : left side $\frac{3}{4}\times 4+14=17$ ; right side $-4+21=17$ . Equal.

Another way: Notice the true equation is Nora's plus Dana's ideas combined: moving terms gives $\frac{7}{4}x=b-a$ . Nora's run says $\frac{7}{4}\times 12=b$ , so $b=21$ ; Dana's says $\frac{7}{4}\times(-8)=-a$ , so $a=14$ . Then $\frac{7}{4}x=7$ and $x=4$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the linear equations.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing with negative numbers and fractions, such as $\frac{3}{4}\times(-8)$.

💡Takeaway. A wrong answer isn't useless: plug it into the equation that was actually solved and it hands you a true number.

Variant 7 medium

Nora solved the equation $\frac{5}{6}x+a=-x+b$ in $x$ , but she misread $a$ as $0$ and got $x=12$ . Dana solved the same equation, but she misread $b$ as $0$ and got $x=-6$ . What is the solution of the original equation when it is solved correctly?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students each misread one constant as $0$ but kept the other one right. Each wrong answer therefore pins down the constant that student read correctly. Recover $a$ and $b$ , then solve the real equation.

Givens

The equation is $\frac{5}{6}x+a=-x+b$ .
With $a$ read as $0$ (and $b$ correct), the solution was $x=12$ .
With $b$ read as $0$ (and $a$ correct), the solution was $x=-6$ .

Unknowns

The solution of the original equation.

Constraints

Nora's $b$ is the true $b$ ; Dana's $a$ is the true $a$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Work backwards from each wrong answer: a solution makes its equation true, so substituting it into the misread equation leaves one unknown, the constant that was read correctly. With both constants found, solve forward.

3 · Execute4 carry out the plan

1Nora's equation gives b

#11 Work Backwards 7.EE.B.4

Nora solved

\frac{5}{6}x=-x+b

and got

x=12

. Substitute.

\frac{5}{6}\times 12=-12+b \;\Rightarrow\; 10=-12+b \;\Rightarrow\; b=22

Her

b

was the real one.

2Dana's equation gives a

#11 Work Backwards 7.EE.B.3

Dana solved

\frac{5}{6}x+a=-x

and got

x=-6

. Substitute.

\frac{5}{6}\times(-6)+a=6 \;\Rightarrow\; -5+a=6 \;\Rightarrow\; a=11

Her

a

was the real one.

3Write the true equation

#7 Identify Subproblems 7.EE.B.4

Put both true constants back.

\frac{5}{6}x+11=-x+22

Now nothing is misread.

4Solve it

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

Multiply both sides by

6

, then collect terms.

5x+66=-6x+132 \;\Rightarrow\; 11x=66 \;\Rightarrow\; x=6

A standard linear equation.

Answer:

x=6

4 · Reviewdoes it hold up?

At $x=6$ : left side $\frac{5}{6}\times 6+11=16$ ; right side $-6+22=16$ . Equal.

Another way: Notice the true equation is Nora's plus Dana's ideas combined: moving terms gives $\frac{11}{6}x=b-a$ . Nora's run says $\frac{11}{6}\times 12=b$ , so $b=22$ ; Dana's says $\frac{11}{6}\times(-6)=-a$ , so $a=11$ . Then $\frac{11}{6}x=11$ and $x=6$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the linear equations.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing with negative numbers and fractions, such as $\frac{5}{6}\times(-6)$.

💡Takeaway. A wrong answer isn't useless: plug it into the equation that was actually solved and it hands you a true number.

Variant 8 medium

Nora solved the equation $\frac{3}{5}x+a=-x+b$ in $x$ , but she misread $a$ as $0$ and got $x=15$ . Dana solved the same equation, but she misread $b$ as $0$ and got $x=5$ . What is the solution of the original equation when it is solved correctly?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students each misread one constant as $0$ but kept the other one right. Each wrong answer therefore pins down the constant that student read correctly. Recover $a$ and $b$ , then solve the real equation.

Givens

The equation is $\frac{3}{5}x+a=-x+b$ .
With $a$ read as $0$ (and $b$ correct), the solution was $x=15$ .
With $b$ read as $0$ (and $a$ correct), the solution was $x=5$ .

Unknowns

The solution of the original equation.

Constraints

Nora's $b$ is the true $b$ ; Dana's $a$ is the true $a$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Work backwards from each wrong answer: a solution makes its equation true, so substituting it into the misread equation leaves one unknown, the constant that was read correctly. With both constants found, solve forward.

3 · Execute4 carry out the plan

1Nora's equation gives b

#11 Work Backwards 7.EE.B.4

Nora solved

\frac{3}{5}x=-x+b

and got

x=15

. Substitute.

\frac{3}{5}\times 15=-15+b \;\Rightarrow\; 9=-15+b \;\Rightarrow\; b=24

Her

b

was the real one.

2Dana's equation gives a

#11 Work Backwards 7.EE.B.3

Dana solved

\frac{3}{5}x+a=-x

and got

x=5

. Substitute.

\frac{3}{5}\times 5+a=-5 \;\Rightarrow\; 3+a=-5 \;\Rightarrow\; a=-8

Her

a

was the real one.

3Write the true equation

#7 Identify Subproblems 7.EE.B.4

Put both true constants back.

\frac{3}{5}x-8=-x+24

Now nothing is misread.

4Solve it

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

Multiply both sides by

5

, then collect terms.

3x-40=-5x+120 \;\Rightarrow\; 8x=160 \;\Rightarrow\; x=20

A standard linear equation.

Answer:

x=20

4 · Reviewdoes it hold up?

At $x=20$ : left side $\frac{3}{5}\times 20-8=4$ ; right side $-20+24=4$ . Equal.

Another way: Notice the true equation is Nora's plus Dana's ideas combined: moving terms gives $\frac{8}{5}x=b-a$ . Nora's run says $\frac{8}{5}\times 15=b$ , so $b=24$ ; Dana's says $\frac{8}{5}\times 5=-a$ , so $a=-8$ . Then $\frac{8}{5}x=32$ and $x=20$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the linear equations.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing with negative numbers and fractions, such as $\frac{3}{5}\times 5$.

💡Takeaway. A wrong answer isn't useless: plug it into the equation that was actually solved and it hands you a true number.

Variant 9 hard

Nora solved the equation $\frac{1}{3}x+a=-2x+b$ in $x$ , but she misread $a$ as $0$ and got $x=9$ . Dana solved the same equation, but she misread $b$ as $0$ and got $x=-3$ . What is the solution of the original equation when it is solved correctly?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students each misread one constant as $0$ but kept the other one right. Each wrong answer therefore pins down the constant that student read correctly. Recover $a$ and $b$ , then solve the real equation.

Givens

The equation is $\frac{1}{3}x+a=-2x+b$ .
With $a$ read as $0$ (and $b$ correct), the solution was $x=9$ .
With $b$ read as $0$ (and $a$ correct), the solution was $x=-3$ .

Unknowns

The solution of the original equation.

Constraints

Nora's $b$ is the true $b$ ; Dana's $a$ is the true $a$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Work backwards from each wrong answer: a solution makes its equation true, so substituting it into the misread equation leaves one unknown, the constant that was read correctly. With both constants found, solve forward.

3 · Execute4 carry out the plan

1Nora's equation gives b

#11 Work Backwards 7.EE.B.4

Nora solved

\frac{1}{3}x=-2x+b

and got

x=9

. Substitute.

\frac{1}{3}\times 9=-18+b \;\Rightarrow\; 3=-18+b \;\Rightarrow\; b=21

Her

b

was the real one.

2Dana's equation gives a

#11 Work Backwards 7.EE.B.3

Dana solved

\frac{1}{3}x+a=-2x

and got

x=-3

. Substitute.

\frac{1}{3}\times(-3)+a=6 \;\Rightarrow\; -1+a=6 \;\Rightarrow\; a=7

Her

a

was the real one.

3Write the true equation

#7 Identify Subproblems 7.EE.B.4

Put both true constants back.

\frac{1}{3}x+7=-2x+21

Now nothing is misread.

4Solve it

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

Multiply both sides by

3

, then collect terms.

x+21=-6x+63 \;\Rightarrow\; 7x=42 \;\Rightarrow\; x=6

A standard linear equation.

Answer:

x=6

4 · Reviewdoes it hold up?

At $x=6$ : left side $\frac{1}{3}\times 6+7=9$ ; right side $-12+21=9$ . Equal.

Another way: Notice the true equation is Nora's plus Dana's ideas combined: moving terms gives $\frac{7}{3}x=b-a$ . Nora's run says $\frac{7}{3}\times 9=b$ , so $b=21$ ; Dana's says $\frac{7}{3}\times(-3)=-a$ , so $a=7$ . Then $\frac{7}{3}x=14$ and $x=6$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the linear equations.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing with negative numbers and fractions, such as $\frac{1}{3}\times(-3)$.

💡Takeaway. A wrong answer isn't useless: plug it into the equation that was actually solved and it hands you a true number.

Variant 10 hard

Nora solved the equation $\frac{2}{3}x+a=-2x+b$ in $x$ , but she misread $a$ as $0$ and got $x=9$ . Dana solved the same equation, but she misread $b$ as $0$ and got $x=-3$ . What is the solution of the original equation when it is solved correctly?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students each misread one constant as $0$ but kept the other one right. Each wrong answer therefore pins down the constant that student read correctly. Recover $a$ and $b$ , then solve the real equation.

Givens

The equation is $\frac{2}{3}x+a=-2x+b$ .
With $a$ read as $0$ (and $b$ correct), the solution was $x=9$ .
With $b$ read as $0$ (and $a$ correct), the solution was $x=-3$ .

Unknowns

The solution of the original equation.

Constraints

Nora's $b$ is the true $b$ ; Dana's $a$ is the true $a$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Work backwards from each wrong answer: a solution makes its equation true, so substituting it into the misread equation leaves one unknown, the constant that was read correctly. With both constants found, solve forward.

3 · Execute4 carry out the plan

1Nora's equation gives b

#11 Work Backwards 7.EE.B.4

Nora solved

\frac{2}{3}x=-2x+b

and got

x=9

. Substitute.

\frac{2}{3}\times 9=-18+b \;\Rightarrow\; 6=-18+b \;\Rightarrow\; b=24

Her

b

was the real one.

2Dana's equation gives a

#11 Work Backwards 7.EE.B.3

Dana solved

\frac{2}{3}x+a=-2x

and got

x=-3

. Substitute.

\frac{2}{3}\times(-3)+a=6 \;\Rightarrow\; -2+a=6 \;\Rightarrow\; a=8

Her

a

was the real one.

3Write the true equation

#7 Identify Subproblems 7.EE.B.4

Put both true constants back.

\frac{2}{3}x+8=-2x+24

Now nothing is misread.

4Solve it

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

Multiply both sides by

3

, then collect terms.

2x+24=-6x+72 \;\Rightarrow\; 8x=48 \;\Rightarrow\; x=6

A standard linear equation.

Answer:

x=6

4 · Reviewdoes it hold up?

At $x=6$ : left side $\frac{2}{3}\times 6+8=12$ ; right side $-12+24=12$ . Equal.

Another way: Notice the true equation is Nora's plus Dana's ideas combined: moving terms gives $\frac{8}{3}x=b-a$ . Nora's run says $\frac{8}{3}\times 9=b$ , so $b=24$ ; Dana's says $\frac{8}{3}\times(-3)=-a$ , so $a=8$ . Then $\frac{8}{3}x=16$ and $x=6$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the linear equations.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing with negative numbers and fractions, such as $\frac{2}{3}\times(-3)$.

💡Takeaway. A wrong answer isn't useless: plug it into the equation that was actually solved and it hands you a true number.

Variant 11 hard

Nora solved the equation $\frac{2}{3}x+a=-x+b$ in $x$ , but she misread $a$ as $0$ and got $x=18$ . Dana solved the same equation, but she misread $b$ as $0$ and got $x=-6$ . What is the solution of the original equation when it is solved correctly?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students each misread one constant as $0$ but kept the other one right. Each wrong answer therefore pins down the constant that student read correctly. Recover $a$ and $b$ , then solve the real equation.

Givens

The equation is $\frac{2}{3}x+a=-x+b$ .
With $a$ read as $0$ (and $b$ correct), the solution was $x=18$ .
With $b$ read as $0$ (and $a$ correct), the solution was $x=-6$ .

Unknowns

The solution of the original equation.

Constraints

Nora's $b$ is the true $b$ ; Dana's $a$ is the true $a$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Work backwards from each wrong answer: a solution makes its equation true, so substituting it into the misread equation leaves one unknown, the constant that was read correctly. With both constants found, solve forward.

3 · Execute4 carry out the plan

1Nora's equation gives b

#11 Work Backwards 7.EE.B.4

Nora solved

\frac{2}{3}x=-x+b

and got

x=18

. Substitute.

\frac{2}{3}\times 18=-18+b \;\Rightarrow\; 12=-18+b \;\Rightarrow\; b=30

Her

b

was the real one.

2Dana's equation gives a

#11 Work Backwards 7.EE.B.3

Dana solved

\frac{2}{3}x+a=-x

and got

x=-6

. Substitute.

\frac{2}{3}\times(-6)+a=6 \;\Rightarrow\; -4+a=6 \;\Rightarrow\; a=10

Her

a

was the real one.

3Write the true equation

#7 Identify Subproblems 7.EE.B.4

Put both true constants back.

\frac{2}{3}x+10=-x+30

Now nothing is misread.

4Solve it

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

Multiply both sides by

3

, then collect terms.

2x+30=-3x+90 \;\Rightarrow\; 5x=60 \;\Rightarrow\; x=12

A standard linear equation.

Answer:

x=12

4 · Reviewdoes it hold up?

At $x=12$ : left side $\frac{2}{3}\times 12+10=18$ ; right side $-12+30=18$ . Equal.

Another way: Notice the true equation is Nora's plus Dana's ideas combined: moving terms gives $\frac{5}{3}x=b-a$ . Nora's run says $\frac{5}{3}\times 18=b$ , so $b=30$ ; Dana's says $\frac{5}{3}\times(-6)=-a$ , so $a=10$ . Then $\frac{5}{3}x=20$ and $x=12$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the linear equations.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing with negative numbers and fractions, such as $\frac{2}{3}\times(-6)$.

💡Takeaway. A wrong answer isn't useless: plug it into the equation that was actually solved and it hands you a true number.

Variant 12 hard

Nora solved the equation $\frac{1}{4}x+a=-x+b$ in $x$ , but she misread $a$ as $0$ and got $x=20$ . Dana solved the same equation, but she misread $b$ as $0$ and got $x=-8$ . What is the solution of the original equation when it is solved correctly?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Two students each misread one constant as $0$ but kept the other one right. Each wrong answer therefore pins down the constant that student read correctly. Recover $a$ and $b$ , then solve the real equation.

Givens

The equation is $\frac{1}{4}x+a=-x+b$ .
With $a$ read as $0$ (and $b$ correct), the solution was $x=20$ .
With $b$ read as $0$ (and $a$ correct), the solution was $x=-8$ .

Unknowns

The solution of the original equation.

Constraints

Nora's $b$ is the true $b$ ; Dana's $a$ is the true $a$ .

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems#13 Convert to Algebra

Work backwards from each wrong answer: a solution makes its equation true, so substituting it into the misread equation leaves one unknown, the constant that was read correctly. With both constants found, solve forward.

3 · Execute4 carry out the plan

1Nora's equation gives b

#11 Work Backwards 7.EE.B.4

Nora solved

\frac{1}{4}x=-x+b

and got

x=20

. Substitute.

\frac{1}{4}\times 20=-20+b \;\Rightarrow\; 5=-20+b \;\Rightarrow\; b=25

Her

b

was the real one.

2Dana's equation gives a

#11 Work Backwards 7.EE.B.3

Dana solved

\frac{1}{4}x+a=-x

and got

x=-8

. Substitute.

\frac{1}{4}\times(-8)+a=8 \;\Rightarrow\; -2+a=8 \;\Rightarrow\; a=10

Her

a

was the real one.

3Write the true equation

#7 Identify Subproblems 7.EE.B.4

Put both true constants back.

\frac{1}{4}x+10=-x+25

Now nothing is misread.

4Solve it

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

Multiply both sides by

4

, then collect terms.

x+40=-4x+100 \;\Rightarrow\; 5x=60 \;\Rightarrow\; x=12

A standard linear equation.

Answer:

x=12

4 · Reviewdoes it hold up?

At $x=12$ : left side $\frac{1}{4}\times 12+10=13$ ; right side $-12+25=13$ . Equal.

Another way: Notice the true equation is Nora's plus Dana's ideas combined: moving terms gives $\frac{5}{4}x=b-a$ . Nora's run says $\frac{5}{4}\times 20=b$ , so $b=25$ ; Dana's says $\frac{5}{4}\times(-8)=-a$ , so $a=10$ . Then $\frac{5}{4}x=15$ and $x=12$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing and solving the linear equations.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing with negative numbers and fractions, such as $\frac{1}{4}\times(-8)$.

💡Takeaway. A wrong answer isn't useless: plug it into the equation that was actually solved and it hands you a true number.

Each wrong solve still reveals one true constant · 12 practice problems

Generated variants — 12

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