Practice · A second solution forces the two lines to coincide · Sensim Math

Variant 1 easy

The system of equations $kx-2y=0$ , $x+y=kx$ has a solution other than $x=0$ , $y=0$ . Find the value of the constant $k$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Both equations are true at $x=0$ , $y=0$ . We want the value of $k$ that gives the system some other solution as well.

Givens

The system is $kx-2y=0$ and $x+y=kx$ .
The system has a solution other than $x=0$ , $y=0$ .

Unknowns

The constant $k$ .

Constraints

$k$ is a constant, the same in both equations.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#1 Draw a Diagram

First reorganize the second equation so both have the form $(\ldots)x + (\ldots)y = 0$ . Picture them as lines: both pass through the origin. Two different lines can share only one point, so a second shared point means the lines are identical, which turns into a proportion of coefficients.

3 · Execute4 carry out the plan

1Move everything to the left side

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

Rewrite the second equation so both equations have

0

on the right.

x + y = kx \;\Rightarrow\; (1-k)x + y = 0

Now both equations have the same shape, which makes them easy to compare.

2Picture the two lines

#1 Draw a Diagram 8.EE.C.8

Each equation is a line through the origin

(0, 0)

, because

x=0

,

y=0

makes both sides

0

. Two different lines meet in at most one point, and these already meet at the origin. A second common solution is possible only if the two equations describe the very same line, which means infinitely many solutions.

One shared point is automatic here; a second one forces everything to be shared.

3Make the coefficients proportional

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

Two equations of the form

(\ldots)x+(\ldots)y=0

describe the same line when their

x

-coefficients and

y

-coefficients are in the same ratio.

\frac{k}{1-k} = \frac{-2}{1}

Same line means one equation is a constant multiple of the other.

4Solve for $k$

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

Cross-multiply and collect the

k

terms.

k = -2(1-k) = -2 + 2k \;\Rightarrow\; -k = -2 \;\Rightarrow\; k = 2

A one-variable linear equation finishes the job.

Answer:

2

4 · Reviewdoes it hold up?

With $k=2$ the equations are $2x-2y=0$ and $x+y=2x$ , and both simplify to $y=x$ . So $x=1$ , $y=1$ (and every point with $y=x$ ) is a solution besides the origin.

Another way: Substitute instead: the first equation gives $y=\frac{k}{2}x$ . Then $x+\frac{k}{2}x=kx$ , so $x\left(1-\frac{k}{2}\right)=0$ . If $x=0$ then $y=0$ too, so a different solution needs $1-\frac{k}{2}=0$ , which again gives $k=2$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognizing that two lines share more than one point only when they are the same line, and solving for $k$.

💡Takeaway. Two lines through the same point share another point only if they are the same line.

Variant 2 easy

The system of equations $kx-2y=0$ , $3x+y=kx$ has a solution other than $x=0$ , $y=0$ . Find the value of the constant $k$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Both equations are true at $x=0$ , $y=0$ . We want the value of $k$ that gives the system some other solution as well.

Givens

The system is $kx-2y=0$ and $3x+y=kx$ .
The system has a solution other than $x=0$ , $y=0$ .

Unknowns

The constant $k$ .

Constraints

$k$ is a constant, the same in both equations.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#1 Draw a Diagram

First reorganize the second equation so both have the form $(\ldots)x + (\ldots)y = 0$ . Picture them as lines: both pass through the origin. Two different lines can share only one point, so a second shared point means the lines are identical, which turns into a proportion of coefficients.

3 · Execute4 carry out the plan

1Move everything to the left side

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

Rewrite the second equation so both equations have

0

on the right.

3x + y = kx \;\Rightarrow\; (3-k)x + y = 0

Now both equations have the same shape, which makes them easy to compare.

2Picture the two lines

#1 Draw a Diagram 8.EE.C.8

Each equation is a line through the origin

(0, 0)

, because

x=0

,

y=0

makes both sides

0

. Two different lines meet in at most one point, and these already meet at the origin. A second common solution is possible only if the two equations describe the very same line, which means infinitely many solutions.

One shared point is automatic here; a second one forces everything to be shared.

3Make the coefficients proportional

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

Two equations of the form

(\ldots)x+(\ldots)y=0

describe the same line when their

x

-coefficients and

y

-coefficients are in the same ratio.

\frac{k}{3-k} = \frac{-2}{1}

Same line means one equation is a constant multiple of the other.

4Solve for $k$

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

Cross-multiply and collect the

k

terms.

k = -2(3-k) = -6 + 2k \;\Rightarrow\; -k = -6 \;\Rightarrow\; k = 6

A one-variable linear equation finishes the job.

Answer:

6

4 · Reviewdoes it hold up?

With $k=6$ the equations are $6x-2y=0$ and $3x+y=6x$ , and both simplify to $y=3x$ . So $x=1$ , $y=3$ (and every point with $y=3x$ ) is a solution besides the origin.

Another way: Substitute instead: the first equation gives $y=\frac{k}{2}x$ . Then $3x+\frac{k}{2}x=kx$ , so $x\left(3-\frac{k}{2}\right)=0$ . If $x=0$ then $y=0$ too, so a different solution needs $3-\frac{k}{2}=0$ , which again gives $k=6$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognizing that two lines share more than one point only when they are the same line, and solving for $k$.

💡Takeaway. Two lines through the same point share another point only if they are the same line.

Variant 3 easy

The system of equations $kx-3y=0$ , $x+2y=kx$ has a solution other than $x=0$ , $y=0$ . Find the value of the constant $k$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Both equations are true at $x=0$ , $y=0$ . We want the value of $k$ that gives the system some other solution as well.

Givens

The system is $kx-3y=0$ and $x+2y=kx$ .
The system has a solution other than $x=0$ , $y=0$ .

Unknowns

The constant $k$ .

Constraints

$k$ is a constant, the same in both equations.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#1 Draw a Diagram

First reorganize the second equation so both have the form $(\ldots)x + (\ldots)y = 0$ . Picture them as lines: both pass through the origin. Two different lines can share only one point, so a second shared point means the lines are identical, which turns into a proportion of coefficients.

3 · Execute4 carry out the plan

1Move everything to the left side

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

Rewrite the second equation so both equations have

0

on the right.

x + 2y = kx \;\Rightarrow\; (1-k)x + 2y = 0

Now both equations have the same shape, which makes them easy to compare.

2Picture the two lines

#1 Draw a Diagram 8.EE.C.8

Each equation is a line through the origin

(0, 0)

, because

x=0

,

y=0

makes both sides

0

. Two different lines meet in at most one point, and these already meet at the origin. A second common solution is possible only if the two equations describe the very same line, which means infinitely many solutions.

One shared point is automatic here; a second one forces everything to be shared.

3Make the coefficients proportional

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

Two equations of the form

(\ldots)x+(\ldots)y=0

describe the same line when their

x

-coefficients and

y

-coefficients are in the same ratio.

\frac{k}{1-k} = \frac{-3}{2}

Same line means one equation is a constant multiple of the other.

4Solve for $k$

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

Cross-multiply and collect the

k

terms.

2k = -3(1-k) = -3 + 3k \;\Rightarrow\; -k = -3 \;\Rightarrow\; k = 3

A one-variable linear equation finishes the job.

Answer:

3

4 · Reviewdoes it hold up?

With $k=3$ the equations are $3x-3y=0$ and $x+2y=3x$ , and both simplify to $y=x$ . So $x=1$ , $y=1$ (and every point with $y=x$ ) is a solution besides the origin.

Another way: Substitute instead: the first equation gives $y=\frac{k}{3}x$ . Then $x+\frac{2k}{3}x=kx$ , so $x\left(1-\frac{k}{3}\right)=0$ . If $x=0$ then $y=0$ too, so a different solution needs $1-\frac{k}{3}=0$ , which again gives $k=3$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognizing that two lines share more than one point only when they are the same line, and solving for $k$.

💡Takeaway. Two lines through the same point share another point only if they are the same line.

Variant 4 easy

The system of equations $kx-3y=0$ , $2x+y=kx$ has a solution other than $x=0$ , $y=0$ . Find the value of the constant $k$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Both equations are true at $x=0$ , $y=0$ . We want the value of $k$ that gives the system some other solution as well.

Givens

The system is $kx-3y=0$ and $2x+y=kx$ .
The system has a solution other than $x=0$ , $y=0$ .

Unknowns

The constant $k$ .

Constraints

$k$ is a constant, the same in both equations.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#1 Draw a Diagram

First reorganize the second equation so both have the form $(\ldots)x + (\ldots)y = 0$ . Picture them as lines: both pass through the origin. Two different lines can share only one point, so a second shared point means the lines are identical, which turns into a proportion of coefficients.

3 · Execute4 carry out the plan

1Move everything to the left side

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

Rewrite the second equation so both equations have

0

on the right.

2x + y = kx \;\Rightarrow\; (2-k)x + y = 0

Now both equations have the same shape, which makes them easy to compare.

2Picture the two lines

#1 Draw a Diagram 8.EE.C.8

Each equation is a line through the origin

(0, 0)

, because

x=0

,

y=0

makes both sides

0

. Two different lines meet in at most one point, and these already meet at the origin. A second common solution is possible only if the two equations describe the very same line, which means infinitely many solutions.

One shared point is automatic here; a second one forces everything to be shared.

3Make the coefficients proportional

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

Two equations of the form

(\ldots)x+(\ldots)y=0

describe the same line when their

x

-coefficients and

y

-coefficients are in the same ratio.

\frac{k}{2-k} = \frac{-3}{1}

Same line means one equation is a constant multiple of the other.

4Solve for $k$

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

Cross-multiply and collect the

k

terms.

k = -3(2-k) = -6 + 3k \;\Rightarrow\; -2k = -6 \;\Rightarrow\; k = 3

A one-variable linear equation finishes the job.

Answer:

3

4 · Reviewdoes it hold up?

With $k=3$ the equations are $3x-3y=0$ and $2x+y=3x$ , and both simplify to $y=x$ . So $x=1$ , $y=1$ (and every point with $y=x$ ) is a solution besides the origin.

Another way: Substitute instead: the first equation gives $y=\frac{k}{3}x$ . Then $2x+\frac{k}{3}x=kx$ , so $x\left(2-\frac{2k}{3}\right)=0$ . If $x=0$ then $y=0$ too, so a different solution needs $2-\frac{2k}{3}=0$ , which again gives $k=3$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognizing that two lines share more than one point only when they are the same line, and solving for $k$.

💡Takeaway. Two lines through the same point share another point only if they are the same line.

Variant 5 medium

The system of equations $kx-4y=0$ , $3x+y=kx$ has a solution other than $x=0$ , $y=0$ . Find the value of the constant $k$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Both equations are true at $x=0$ , $y=0$ . We want the value of $k$ that gives the system some other solution as well.

Givens

The system is $kx-4y=0$ and $3x+y=kx$ .
The system has a solution other than $x=0$ , $y=0$ .

Unknowns

The constant $k$ .

Constraints

$k$ is a constant, the same in both equations.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#1 Draw a Diagram

First reorganize the second equation so both have the form $(\ldots)x + (\ldots)y = 0$ . Picture them as lines: both pass through the origin. Two different lines can share only one point, so a second shared point means the lines are identical, which turns into a proportion of coefficients.

3 · Execute4 carry out the plan

1Move everything to the left side

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

Rewrite the second equation so both equations have

0

on the right.

3x + y = kx \;\Rightarrow\; (3-k)x + y = 0

Now both equations have the same shape, which makes them easy to compare.

2Picture the two lines

#1 Draw a Diagram 8.EE.C.8

Each equation is a line through the origin

(0, 0)

, because

x=0

,

y=0

makes both sides

0

. Two different lines meet in at most one point, and these already meet at the origin. A second common solution is possible only if the two equations describe the very same line, which means infinitely many solutions.

One shared point is automatic here; a second one forces everything to be shared.

3Make the coefficients proportional

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

Two equations of the form

(\ldots)x+(\ldots)y=0

describe the same line when their

x

-coefficients and

y

-coefficients are in the same ratio.

\frac{k}{3-k} = \frac{-4}{1}

Same line means one equation is a constant multiple of the other.

4Solve for $k$

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

Cross-multiply and collect the

k

terms.

k = -4(3-k) = -12 + 4k \;\Rightarrow\; -3k = -12 \;\Rightarrow\; k = 4

A one-variable linear equation finishes the job.

Answer:

4

4 · Reviewdoes it hold up?

With $k=4$ the equations are $4x-4y=0$ and $3x+y=4x$ , and both simplify to $y=x$ . So $x=1$ , $y=1$ (and every point with $y=x$ ) is a solution besides the origin.

Another way: Substitute instead: the first equation gives $y=\frac{k}{4}x$ . Then $3x+\frac{k}{4}x=kx$ , so $x\left(3-\frac{3k}{4}\right)=0$ . If $x=0$ then $y=0$ too, so a different solution needs $3-\frac{3k}{4}=0$ , which again gives $k=4$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognizing that two lines share more than one point only when they are the same line, and solving for $k$.

💡Takeaway. Two lines through the same point share another point only if they are the same line.

Variant 6 medium

The system of equations $kx-4y=0$ , $x+5y=kx$ has a solution other than $x=0$ , $y=0$ . Find the value of the constant $k$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Both equations are true at $x=0$ , $y=0$ . We want the value of $k$ that gives the system some other solution as well.

Givens

The system is $kx-4y=0$ and $x+5y=kx$ .
The system has a solution other than $x=0$ , $y=0$ .

Unknowns

The constant $k$ .

Constraints

$k$ is a constant, the same in both equations.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#1 Draw a Diagram

First reorganize the second equation so both have the form $(\ldots)x + (\ldots)y = 0$ . Picture them as lines: both pass through the origin. Two different lines can share only one point, so a second shared point means the lines are identical, which turns into a proportion of coefficients.

3 · Execute4 carry out the plan

1Move everything to the left side

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

Rewrite the second equation so both equations have

0

on the right.

x + 5y = kx \;\Rightarrow\; (1-k)x + 5y = 0

Now both equations have the same shape, which makes them easy to compare.

2Picture the two lines

#1 Draw a Diagram 8.EE.C.8

Each equation is a line through the origin

(0, 0)

, because

x=0

,

y=0

makes both sides

0

. Two different lines meet in at most one point, and these already meet at the origin. A second common solution is possible only if the two equations describe the very same line, which means infinitely many solutions.

One shared point is automatic here; a second one forces everything to be shared.

3Make the coefficients proportional

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

Two equations of the form

(\ldots)x+(\ldots)y=0

describe the same line when their

x

-coefficients and

y

-coefficients are in the same ratio.

\frac{k}{1-k} = \frac{-4}{5}

Same line means one equation is a constant multiple of the other.

4Solve for $k$

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

Cross-multiply and collect the

k

terms.

5k = -4(1-k) = -4 + 4k \;\Rightarrow\; k = -4

A one-variable linear equation finishes the job.

Answer:

-4

4 · Reviewdoes it hold up?

With $k=-4$ the equations are $-4x-4y=0$ and $x+5y=-4x$ , and both simplify to $y=-x$ . So $x=1$ , $y=-1$ (and every point with $y=-x$ ) is a solution besides the origin.

Another way: Substitute instead: the first equation gives $y=\frac{k}{4}x$ . Then $x+\frac{5k}{4}x=kx$ , so $x\left(1+\frac{k}{4}\right)=0$ . If $x=0$ then $y=0$ too, so a different solution needs $1+\frac{k}{4}=0$ , which again gives $k=-4$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognizing that two lines share more than one point only when they are the same line, and solving for $k$.

💡Takeaway. Two lines through the same point share another point only if they are the same line.

Variant 7 medium

The system of equations $kx-2y=0$ , $5x+y=kx$ has a solution other than $x=0$ , $y=0$ . Find the value of the constant $k$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Both equations are true at $x=0$ , $y=0$ . We want the value of $k$ that gives the system some other solution as well.

Givens

The system is $kx-2y=0$ and $5x+y=kx$ .
The system has a solution other than $x=0$ , $y=0$ .

Unknowns

The constant $k$ .

Constraints

$k$ is a constant, the same in both equations.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#1 Draw a Diagram

First reorganize the second equation so both have the form $(\ldots)x + (\ldots)y = 0$ . Picture them as lines: both pass through the origin. Two different lines can share only one point, so a second shared point means the lines are identical, which turns into a proportion of coefficients.

3 · Execute4 carry out the plan

1Move everything to the left side

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

Rewrite the second equation so both equations have

0

on the right.

5x + y = kx \;\Rightarrow\; (5-k)x + y = 0

Now both equations have the same shape, which makes them easy to compare.

2Picture the two lines

#1 Draw a Diagram 8.EE.C.8

Each equation is a line through the origin

(0, 0)

, because

x=0

,

y=0

makes both sides

0

. Two different lines meet in at most one point, and these already meet at the origin. A second common solution is possible only if the two equations describe the very same line, which means infinitely many solutions.

One shared point is automatic here; a second one forces everything to be shared.

3Make the coefficients proportional

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

Two equations of the form

(\ldots)x+(\ldots)y=0

describe the same line when their

x

-coefficients and

y

-coefficients are in the same ratio.

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

Same line means one equation is a constant multiple of the other.

4Solve for $k$

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

Cross-multiply and collect the

k

terms.

k = -2(5-k) = -10 + 2k \;\Rightarrow\; -k = -10 \;\Rightarrow\; k = 10

A one-variable linear equation finishes the job.

Answer:

10

4 · Reviewdoes it hold up?

With $k=10$ the equations are $10x-2y=0$ and $5x+y=10x$ , and both simplify to $y=5x$ . So $x=1$ , $y=5$ (and every point with $y=5x$ ) is a solution besides the origin.

Another way: Substitute instead: the first equation gives $y=\frac{k}{2}x$ . Then $5x+\frac{k}{2}x=kx$ , so $x\left(5-\frac{k}{2}\right)=0$ . If $x=0$ then $y=0$ too, so a different solution needs $5-\frac{k}{2}=0$ , which again gives $k=10$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognizing that two lines share more than one point only when they are the same line, and solving for $k$.

💡Takeaway. Two lines through the same point share another point only if they are the same line.

Variant 8 medium

The system of equations $kx-6y=0$ , $5x+y=kx$ has a solution other than $x=0$ , $y=0$ . Find the value of the constant $k$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Both equations are true at $x=0$ , $y=0$ . We want the value of $k$ that gives the system some other solution as well.

Givens

The system is $kx-6y=0$ and $5x+y=kx$ .
The system has a solution other than $x=0$ , $y=0$ .

Unknowns

The constant $k$ .

Constraints

$k$ is a constant, the same in both equations.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#1 Draw a Diagram

First reorganize the second equation so both have the form $(\ldots)x + (\ldots)y = 0$ . Picture them as lines: both pass through the origin. Two different lines can share only one point, so a second shared point means the lines are identical, which turns into a proportion of coefficients.

3 · Execute4 carry out the plan

1Move everything to the left side

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

Rewrite the second equation so both equations have

0

on the right.

5x + y = kx \;\Rightarrow\; (5-k)x + y = 0

Now both equations have the same shape, which makes them easy to compare.

2Picture the two lines

#1 Draw a Diagram 8.EE.C.8

Each equation is a line through the origin

(0, 0)

, because

x=0

,

y=0

makes both sides

0

. Two different lines meet in at most one point, and these already meet at the origin. A second common solution is possible only if the two equations describe the very same line, which means infinitely many solutions.

One shared point is automatic here; a second one forces everything to be shared.

3Make the coefficients proportional

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

Two equations of the form

(\ldots)x+(\ldots)y=0

describe the same line when their

x

-coefficients and

y

-coefficients are in the same ratio.

\frac{k}{5-k} = \frac{-6}{1}

Same line means one equation is a constant multiple of the other.

4Solve for $k$

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

Cross-multiply and collect the

k

terms.

k = -6(5-k) = -30 + 6k \;\Rightarrow\; -5k = -30 \;\Rightarrow\; k = 6

A one-variable linear equation finishes the job.

Answer:

6

4 · Reviewdoes it hold up?

With $k=6$ the equations are $6x-6y=0$ and $5x+y=6x$ , and both simplify to $y=x$ . So $x=1$ , $y=1$ (and every point with $y=x$ ) is a solution besides the origin.

Another way: Substitute instead: the first equation gives $y=\frac{k}{6}x$ . Then $5x+\frac{k}{6}x=kx$ , so $x\left(5-\frac{5k}{6}\right)=0$ . If $x=0$ then $y=0$ too, so a different solution needs $5-\frac{5k}{6}=0$ , which again gives $k=6$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognizing that two lines share more than one point only when they are the same line, and solving for $k$.

💡Takeaway. Two lines through the same point share another point only if they are the same line.

Variant 9 hard

The system of equations $kx-6y=0$ , $x+4y=kx$ has a solution other than $x=0$ , $y=0$ . Find the value of the constant $k$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Both equations are true at $x=0$ , $y=0$ . We want the value of $k$ that gives the system some other solution as well.

Givens

The system is $kx-6y=0$ and $x+4y=kx$ .
The system has a solution other than $x=0$ , $y=0$ .

Unknowns

The constant $k$ .

Constraints

$k$ is a constant, the same in both equations.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#1 Draw a Diagram

First reorganize the second equation so both have the form $(\ldots)x + (\ldots)y = 0$ . Picture them as lines: both pass through the origin. Two different lines can share only one point, so a second shared point means the lines are identical, which turns into a proportion of coefficients.

3 · Execute4 carry out the plan

1Move everything to the left side

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

Rewrite the second equation so both equations have

0

on the right.

x + 4y = kx \;\Rightarrow\; (1-k)x + 4y = 0

Now both equations have the same shape, which makes them easy to compare.

2Picture the two lines

#1 Draw a Diagram 8.EE.C.8

Each equation is a line through the origin

(0, 0)

, because

x=0

,

y=0

makes both sides

0

. Two different lines meet in at most one point, and these already meet at the origin. A second common solution is possible only if the two equations describe the very same line, which means infinitely many solutions.

One shared point is automatic here; a second one forces everything to be shared.

3Make the coefficients proportional

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

Two equations of the form

(\ldots)x+(\ldots)y=0

describe the same line when their

x

-coefficients and

y

-coefficients are in the same ratio.

\frac{k}{1-k} = \frac{-6}{4}

Same line means one equation is a constant multiple of the other.

4Solve for $k$

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

Cross-multiply and collect the

k

terms.

4k = -6(1-k) = -6 + 6k \;\Rightarrow\; -2k = -6 \;\Rightarrow\; k = 3

A one-variable linear equation finishes the job.

Answer:

3

4 · Reviewdoes it hold up?

With $k=3$ the equations are $3x-6y=0$ and $x+4y=3x$ , and both simplify to $y=\frac{1}{2}x$ . So $x=2$ , $y=1$ (and every point with $y=\frac{1}{2}x$ ) is a solution besides the origin.

Another way: Substitute instead: the first equation gives $y=\frac{k}{6}x$ . Then $x+\frac{4k}{6}x=kx$ , so $x\left(1-\frac{2k}{6}\right)=0$ . If $x=0$ then $y=0$ too, so a different solution needs $1-\frac{2k}{6}=0$ , which again gives $k=3$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognizing that two lines share more than one point only when they are the same line, and solving for $k$.

💡Takeaway. Two lines through the same point share another point only if they are the same line.

Variant 10 hard

The system of equations $kx-2y=0$ , $2x+3y=kx$ has a solution other than $x=0$ , $y=0$ . Find the value of the constant $k$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Both equations are true at $x=0$ , $y=0$ . We want the value of $k$ that gives the system some other solution as well.

Givens

The system is $kx-2y=0$ and $2x+3y=kx$ .
The system has a solution other than $x=0$ , $y=0$ .

Unknowns

The constant $k$ .

Constraints

$k$ is a constant, the same in both equations.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#1 Draw a Diagram

First reorganize the second equation so both have the form $(\ldots)x + (\ldots)y = 0$ . Picture them as lines: both pass through the origin. Two different lines can share only one point, so a second shared point means the lines are identical, which turns into a proportion of coefficients.

3 · Execute4 carry out the plan

1Move everything to the left side

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

Rewrite the second equation so both equations have

0

on the right.

2x + 3y = kx \;\Rightarrow\; (2-k)x + 3y = 0

Now both equations have the same shape, which makes them easy to compare.

2Picture the two lines

#1 Draw a Diagram 8.EE.C.8

Each equation is a line through the origin

(0, 0)

, because

x=0

,

y=0

makes both sides

0

. Two different lines meet in at most one point, and these already meet at the origin. A second common solution is possible only if the two equations describe the very same line, which means infinitely many solutions.

One shared point is automatic here; a second one forces everything to be shared.

3Make the coefficients proportional

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

Two equations of the form

(\ldots)x+(\ldots)y=0

describe the same line when their

x

-coefficients and

y

-coefficients are in the same ratio.

\frac{k}{2-k} = \frac{-2}{3}

Same line means one equation is a constant multiple of the other.

4Solve for $k$

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

Cross-multiply and collect the

k

terms.

3k = -2(2-k) = -4 + 2k \;\Rightarrow\; k = -4

A one-variable linear equation finishes the job.

Answer:

-4

4 · Reviewdoes it hold up?

With $k=-4$ the equations are $-4x-2y=0$ and $2x+3y=-4x$ , and both simplify to $y=-2x$ . So $x=1$ , $y=-2$ (and every point with $y=-2x$ ) is a solution besides the origin.

Another way: Substitute instead: the first equation gives $y=\frac{k}{2}x$ . Then $2x+\frac{3k}{2}x=kx$ , so $x\left(2+\frac{k}{2}\right)=0$ . If $x=0$ then $y=0$ too, so a different solution needs $2+\frac{k}{2}=0$ , which again gives $k=-4$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognizing that two lines share more than one point only when they are the same line, and solving for $k$.

💡Takeaway. Two lines through the same point share another point only if they are the same line.

Variant 11 hard

The system of equations $kx-3y=0$ , $2x+4y=kx$ has a solution other than $x=0$ , $y=0$ . Find the value of the constant $k$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Both equations are true at $x=0$ , $y=0$ . We want the value of $k$ that gives the system some other solution as well.

Givens

The system is $kx-3y=0$ and $2x+4y=kx$ .
The system has a solution other than $x=0$ , $y=0$ .

Unknowns

The constant $k$ .

Constraints

$k$ is a constant, the same in both equations.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#1 Draw a Diagram

First reorganize the second equation so both have the form $(\ldots)x + (\ldots)y = 0$ . Picture them as lines: both pass through the origin. Two different lines can share only one point, so a second shared point means the lines are identical, which turns into a proportion of coefficients.

3 · Execute4 carry out the plan

1Move everything to the left side

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

Rewrite the second equation so both equations have

0

on the right.

2x + 4y = kx \;\Rightarrow\; (2-k)x + 4y = 0

Now both equations have the same shape, which makes them easy to compare.

2Picture the two lines

#1 Draw a Diagram 8.EE.C.8

Each equation is a line through the origin

(0, 0)

, because

x=0

,

y=0

makes both sides

0

. Two different lines meet in at most one point, and these already meet at the origin. A second common solution is possible only if the two equations describe the very same line, which means infinitely many solutions.

One shared point is automatic here; a second one forces everything to be shared.

3Make the coefficients proportional

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

Two equations of the form

(\ldots)x+(\ldots)y=0

describe the same line when their

x

-coefficients and

y

-coefficients are in the same ratio.

\frac{k}{2-k} = \frac{-3}{4}

Same line means one equation is a constant multiple of the other.

4Solve for $k$

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

Cross-multiply and collect the

k

terms.

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

A one-variable linear equation finishes the job.

Answer:

-6

4 · Reviewdoes it hold up?

With $k=-6$ the equations are $-6x-3y=0$ and $2x+4y=-6x$ , and both simplify to $y=-2x$ . So $x=1$ , $y=-2$ (and every point with $y=-2x$ ) is a solution besides the origin.

Another way: Substitute instead: the first equation gives $y=\frac{k}{3}x$ . Then $2x+\frac{4k}{3}x=kx$ , so $x\left(2+\frac{k}{3}\right)=0$ . If $x=0$ then $y=0$ too, so a different solution needs $2+\frac{k}{3}=0$ , which again gives $k=-6$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognizing that two lines share more than one point only when they are the same line, and solving for $k$.

💡Takeaway. Two lines through the same point share another point only if they are the same line.

Variant 12 hard

The system of equations $kx-5y=0$ , $2x+3y=kx$ has a solution other than $x=0$ , $y=0$ . Find the value of the constant $k$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Both equations are true at $x=0$ , $y=0$ . We want the value of $k$ that gives the system some other solution as well.

Givens

The system is $kx-5y=0$ and $2x+3y=kx$ .
The system has a solution other than $x=0$ , $y=0$ .

Unknowns

The constant $k$ .

Constraints

$k$ is a constant, the same in both equations.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways#1 Draw a Diagram

First reorganize the second equation so both have the form $(\ldots)x + (\ldots)y = 0$ . Picture them as lines: both pass through the origin. Two different lines can share only one point, so a second shared point means the lines are identical, which turns into a proportion of coefficients.

3 · Execute4 carry out the plan

1Move everything to the left side

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

Rewrite the second equation so both equations have

0

on the right.

2x + 3y = kx \;\Rightarrow\; (2-k)x + 3y = 0

Now both equations have the same shape, which makes them easy to compare.

2Picture the two lines

#1 Draw a Diagram 8.EE.C.8

Each equation is a line through the origin

(0, 0)

, because

x=0

,

y=0

makes both sides

0

. Two different lines meet in at most one point, and these already meet at the origin. A second common solution is possible only if the two equations describe the very same line, which means infinitely many solutions.

One shared point is automatic here; a second one forces everything to be shared.

3Make the coefficients proportional

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

Two equations of the form

(\ldots)x+(\ldots)y=0

describe the same line when their

x

-coefficients and

y

-coefficients are in the same ratio.

\frac{k}{2-k} = \frac{-5}{3}

Same line means one equation is a constant multiple of the other.

4Solve for $k$

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

Cross-multiply and collect the

k

terms.

3k = -5(2-k) = -10 + 5k \;\Rightarrow\; -2k = -10 \;\Rightarrow\; k = 5

A one-variable linear equation finishes the job.

Answer:

5

4 · Reviewdoes it hold up?

With $k=5$ the equations are $5x-5y=0$ and $2x+3y=5x$ , and both simplify to $y=x$ . So $x=1$ , $y=1$ (and every point with $y=x$ ) is a solution besides the origin.

Another way: Substitute instead: the first equation gives $y=\frac{k}{5}x$ . Then $2x+\frac{3k}{5}x=kx$ , so $x\left(2-\frac{2k}{5}\right)=0$ . If $x=0$ then $y=0$ too, so a different solution needs $2-\frac{2k}{5}=0$ , which again gives $k=5$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognizing that two lines share more than one point only when they are the same line, and solving for $k$.

💡Takeaway. Two lines through the same point share another point only if they are the same line.

A second solution forces the two lines to coincide · 12 practice problems

Generated variants — 12

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