Practice · Two lines through the origin meeting twice coincide · Sensim Math

Variant 1 easy

The two lines $ax+2y=3x$ and $x-3y+1=ax+1$ meet at some point other than $x=0$ , $y=0$ . Find the value of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The lines $ax+2y=3x$ and $x-3y+1=ax+1$ meet somewhere besides the origin. Find $a$ .

Givens

Line 1: $ax+2y=3x$ .
Line 2: $x-3y+1=ax+1$ .
The lines share a point other than $(0, 0)$ .

Unknowns

The constant $a$ .

Constraints

Two different lines meet in at most one point.

2 · Planchoose the strategy

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

Tidy both equations first. The constant terms vanish, so both lines go through the origin. Two distinct lines share at most one point, and the origin already uses it up; a second common point forces the lines to be the same line.

3 · Execute3 carry out the plan

1Rearrange both equations

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

Move every term to the left side.

(a-3)x+2y=0,\quad (a-1)x+3y=0

Rewriting reveals what the original forms hid: there is no constant term.

2Both lines pass through the origin

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

With no constant term,

x=0

,

y=0

satisfies each equation. Two different lines cross at most once, so if they also meet elsewhere they must coincide, meaning infinitely many common solutions.

Picture two lines through one point: unless they lie on top of each other, they spread apart and never meet again.

3Make the lines identical

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

Two lines through the origin coincide when their slopes are equal. From line 1,

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

; from line 2,

y=-\frac{a-1}{3}x

.

\frac{a-3}{2}=\frac{a-1}{3},\ 3a-9=2a-2,\ a=7

Equal slopes plus a shared point means the same line.

Answer:

a=7

4 · Reviewdoes it hold up?

With $a=7$ the lines are $4x+2y=0$ and $6x+3y=0$ ; both simplify to $y=-2x$ , so they share every point, for example $(1, -2)$ .

Another way: Solve the system directly: eliminating $y$ gives $(3(a-3)-2(a-1))x=0$ , that is $(a-7)x=0$ . Only $x=0$ (and then $y=0$ ) works unless $a-7=0$ , so $a=7$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognising that a system has infinitely many solutions exactly when the two lines coincide.

💡Takeaway. Two lines can share one point or all of them -- never exactly two.

Variant 2 easy

The two lines $ax+2y=3x$ and $x-4y+1=ax+1$ meet at some point other than $x=0$ , $y=0$ . Find the value of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The lines $ax+2y=3x$ and $x-4y+1=ax+1$ meet somewhere besides the origin. Find $a$ .

Givens

Line 1: $ax+2y=3x$ .
Line 2: $x-4y+1=ax+1$ .
The lines share a point other than $(0, 0)$ .

Unknowns

The constant $a$ .

Constraints

Two different lines meet in at most one point.

2 · Planchoose the strategy

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

Tidy both equations first. The constant terms vanish, so both lines go through the origin. Two distinct lines share at most one point, and the origin already uses it up; a second common point forces the lines to be the same line.

3 · Execute3 carry out the plan

1Rearrange both equations

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

Move every term to the left side.

(a-3)x+2y=0,\quad (a-1)x+4y=0

Rewriting reveals what the original forms hid: there is no constant term.

2Both lines pass through the origin

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

With no constant term,

x=0

,

y=0

satisfies each equation. Two different lines cross at most once, so if they also meet elsewhere they must coincide, meaning infinitely many common solutions.

Picture two lines through one point: unless they lie on top of each other, they spread apart and never meet again.

3Make the lines identical

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

Two lines through the origin coincide when their slopes are equal. From line 1,

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

; from line 2,

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

.

\frac{a-3}{2}=\frac{a-1}{4},\ 4a-12=2a-2,\ 2a=10,\ a=5

Equal slopes plus a shared point means the same line.

Answer:

a=5

4 · Reviewdoes it hold up?

With $a=5$ the lines are $2x+2y=0$ and $4x+4y=0$ ; both simplify to $y=-x$ , so they share every point, for example $(1, -1)$ .

Another way: Solve the system directly: eliminating $y$ gives $(4(a-3)-2(a-1))x=0$ , that is $(2a-10)x=0$ . Only $x=0$ (and then $y=0$ ) works unless $2a-10=0$ , so $a=5$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognising that a system has infinitely many solutions exactly when the two lines coincide.

💡Takeaway. Two lines can share one point or all of them -- never exactly two.

Variant 3 medium

The two lines $ax+3y=x$ and $3x-5y+1=ax+1$ meet at some point other than $x=0$ , $y=0$ . Find the value of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The lines $ax+3y=x$ and $3x-5y+1=ax+1$ meet somewhere besides the origin. Find $a$ .

Givens

Line 1: $ax+3y=x$ .
Line 2: $3x-5y+1=ax+1$ .
The lines share a point other than $(0, 0)$ .

Unknowns

The constant $a$ .

Constraints

Two different lines meet in at most one point.

2 · Planchoose the strategy

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

Tidy both equations first. The constant terms vanish, so both lines go through the origin. Two distinct lines share at most one point, and the origin already uses it up; a second common point forces the lines to be the same line.

3 · Execute3 carry out the plan

1Rearrange both equations

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

Move every term to the left side.

(a-1)x+3y=0,\quad (a-3)x+5y=0

Rewriting reveals what the original forms hid: there is no constant term.

2Both lines pass through the origin

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

With no constant term,

x=0

,

y=0

satisfies each equation. Two different lines cross at most once, so if they also meet elsewhere they must coincide, meaning infinitely many common solutions.

Picture two lines through one point: unless they lie on top of each other, they spread apart and never meet again.

3Make the lines identical

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

Two lines through the origin coincide when their slopes are equal. From line 1,

y=-\frac{a-1}{3}x

; from line 2,

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

.

\frac{a-1}{3}=\frac{a-3}{5},\ 5a-5=3a-9,\ 2a=-4,\ a=-2

Equal slopes plus a shared point means the same line.

Answer:

a=-2

4 · Reviewdoes it hold up?

With $a=-2$ the lines are $-3x+3y=0$ and $-5x+5y=0$ ; both simplify to $y=x$ , so they share every point, for example $(1, 1)$ .

Another way: Solve the system directly: eliminating $y$ gives $(5(a-1)-3(a-3))x=0$ , that is $(2a+4)x=0$ . Only $x=0$ (and then $y=0$ ) works unless $2a+4=0$ , so $a=-2$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognising that a system has infinitely many solutions exactly when the two lines coincide.

💡Takeaway. Two lines can share one point or all of them -- never exactly two.

Variant 4 easy

The two lines $ax+2y=5x$ and $x-3y+2=ax+2$ meet at some point other than $x=0$ , $y=0$ . Find the value of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The lines $ax+2y=5x$ and $x-3y+2=ax+2$ meet somewhere besides the origin. Find $a$ .

Givens

Line 1: $ax+2y=5x$ .
Line 2: $x-3y+2=ax+2$ .
The lines share a point other than $(0, 0)$ .

Unknowns

The constant $a$ .

Constraints

Two different lines meet in at most one point.

2 · Planchoose the strategy

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

Tidy both equations first. The constant terms vanish, so both lines go through the origin. Two distinct lines share at most one point, and the origin already uses it up; a second common point forces the lines to be the same line.

3 · Execute3 carry out the plan

1Rearrange both equations

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

Move every term to the left side.

(a-5)x+2y=0,\quad (a-1)x+3y=0

Rewriting reveals what the original forms hid: there is no constant term.

2Both lines pass through the origin

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

With no constant term,

x=0

,

y=0

satisfies each equation. Two different lines cross at most once, so if they also meet elsewhere they must coincide, meaning infinitely many common solutions.

Picture two lines through one point: unless they lie on top of each other, they spread apart and never meet again.

3Make the lines identical

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

Two lines through the origin coincide when their slopes are equal. From line 1,

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

; from line 2,

y=-\frac{a-1}{3}x

.

\frac{a-5}{2}=\frac{a-1}{3},\ 3a-15=2a-2,\ a=13

Equal slopes plus a shared point means the same line.

Answer:

a=13

4 · Reviewdoes it hold up?

With $a=13$ the lines are $8x+2y=0$ and $12x+3y=0$ ; both simplify to $y=-4x$ , so they share every point, for example $(1, -4)$ .

Another way: Solve the system directly: eliminating $y$ gives $(3(a-5)-2(a-1))x=0$ , that is $(a-13)x=0$ . Only $x=0$ (and then $y=0$ ) works unless $a-13=0$ , so $a=13$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognising that a system has infinitely many solutions exactly when the two lines coincide.

💡Takeaway. Two lines can share one point or all of them -- never exactly two.

Variant 5 easy

The two lines $ax+2y=x$ and $4x-3y+5=ax+5$ meet at some point other than $x=0$ , $y=0$ . Find the value of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The lines $ax+2y=x$ and $4x-3y+5=ax+5$ meet somewhere besides the origin. Find $a$ .

Givens

Line 1: $ax+2y=x$ .
Line 2: $4x-3y+5=ax+5$ .
The lines share a point other than $(0, 0)$ .

Unknowns

The constant $a$ .

Constraints

Two different lines meet in at most one point.

2 · Planchoose the strategy

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

Tidy both equations first. The constant terms vanish, so both lines go through the origin. Two distinct lines share at most one point, and the origin already uses it up; a second common point forces the lines to be the same line.

3 · Execute3 carry out the plan

1Rearrange both equations

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

Move every term to the left side.

(a-1)x+2y=0,\quad (a-4)x+3y=0

Rewriting reveals what the original forms hid: there is no constant term.

2Both lines pass through the origin

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

With no constant term,

x=0

,

y=0

satisfies each equation. Two different lines cross at most once, so if they also meet elsewhere they must coincide, meaning infinitely many common solutions.

Picture two lines through one point: unless they lie on top of each other, they spread apart and never meet again.

3Make the lines identical

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

Two lines through the origin coincide when their slopes are equal. From line 1,

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

; from line 2,

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

.

\frac{a-1}{2}=\frac{a-4}{3},\ 3a-3=2a-8,\ a=-5

Equal slopes plus a shared point means the same line.

Answer:

a=-5

4 · Reviewdoes it hold up?

With $a=-5$ the lines are $-6x+2y=0$ and $-9x+3y=0$ ; both simplify to $y=3x$ , so they share every point, for example $(1, 3)$ .

Another way: Solve the system directly: eliminating $y$ gives $(3(a-1)-2(a-4))x=0$ , that is $(a+5)x=0$ . Only $x=0$ (and then $y=0$ ) works unless $a+5=0$ , so $a=-5$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognising that a system has infinitely many solutions exactly when the two lines coincide.

💡Takeaway. Two lines can share one point or all of them -- never exactly two.

Variant 6 medium

The two lines $ax+4y=3x$ and $x-5y+2=ax+2$ meet at some point other than $x=0$ , $y=0$ . Find the value of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The lines $ax+4y=3x$ and $x-5y+2=ax+2$ meet somewhere besides the origin. Find $a$ .

Givens

Line 1: $ax+4y=3x$ .
Line 2: $x-5y+2=ax+2$ .
The lines share a point other than $(0, 0)$ .

Unknowns

The constant $a$ .

Constraints

Two different lines meet in at most one point.

2 · Planchoose the strategy

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

Tidy both equations first. The constant terms vanish, so both lines go through the origin. Two distinct lines share at most one point, and the origin already uses it up; a second common point forces the lines to be the same line.

3 · Execute3 carry out the plan

1Rearrange both equations

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

Move every term to the left side.

(a-3)x+4y=0,\quad (a-1)x+5y=0

Rewriting reveals what the original forms hid: there is no constant term.

2Both lines pass through the origin

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

With no constant term,

x=0

,

y=0

satisfies each equation. Two different lines cross at most once, so if they also meet elsewhere they must coincide, meaning infinitely many common solutions.

Picture two lines through one point: unless they lie on top of each other, they spread apart and never meet again.

3Make the lines identical

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

Two lines through the origin coincide when their slopes are equal. From line 1,

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

; from line 2,

y=-\frac{a-1}{5}x

.

\frac{a-3}{4}=\frac{a-1}{5},\ 5a-15=4a-4,\ a=11

Equal slopes plus a shared point means the same line.

Answer:

a=11

4 · Reviewdoes it hold up?

With $a=11$ the lines are $8x+4y=0$ and $10x+5y=0$ ; both simplify to $y=-2x$ , so they share every point, for example $(1, -2)$ .

Another way: Solve the system directly: eliminating $y$ gives $(5(a-3)-4(a-1))x=0$ , that is $(a-11)x=0$ . Only $x=0$ (and then $y=0$ ) works unless $a-11=0$ , so $a=11$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognising that a system has infinitely many solutions exactly when the two lines coincide.

💡Takeaway. Two lines can share one point or all of them -- never exactly two.

Variant 7 medium

The two lines $ax+3y=4x$ and $2x-4y+1=ax+1$ meet at some point other than $x=0$ , $y=0$ . Find the value of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The lines $ax+3y=4x$ and $2x-4y+1=ax+1$ meet somewhere besides the origin. Find $a$ .

Givens

Line 1: $ax+3y=4x$ .
Line 2: $2x-4y+1=ax+1$ .
The lines share a point other than $(0, 0)$ .

Unknowns

The constant $a$ .

Constraints

Two different lines meet in at most one point.

2 · Planchoose the strategy

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

Tidy both equations first. The constant terms vanish, so both lines go through the origin. Two distinct lines share at most one point, and the origin already uses it up; a second common point forces the lines to be the same line.

3 · Execute3 carry out the plan

1Rearrange both equations

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

Move every term to the left side.

(a-4)x+3y=0,\quad (a-2)x+4y=0

Rewriting reveals what the original forms hid: there is no constant term.

2Both lines pass through the origin

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

With no constant term,

x=0

,

y=0

satisfies each equation. Two different lines cross at most once, so if they also meet elsewhere they must coincide, meaning infinitely many common solutions.

Picture two lines through one point: unless they lie on top of each other, they spread apart and never meet again.

3Make the lines identical

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

Two lines through the origin coincide when their slopes are equal. From line 1,

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

; from line 2,

y=-\frac{a-2}{4}x

.

\frac{a-4}{3}=\frac{a-2}{4},\ 4a-16=3a-6,\ a=10

Equal slopes plus a shared point means the same line.

Answer:

a=10

4 · Reviewdoes it hold up?

With $a=10$ the lines are $6x+3y=0$ and $8x+4y=0$ ; both simplify to $y=-2x$ , so they share every point, for example $(1, -2)$ .

Another way: Solve the system directly: eliminating $y$ gives $(4(a-4)-3(a-2))x=0$ , that is $(a-10)x=0$ . Only $x=0$ (and then $y=0$ ) works unless $a-10=0$ , so $a=10$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognising that a system has infinitely many solutions exactly when the two lines coincide.

💡Takeaway. Two lines can share one point or all of them -- never exactly two.

Variant 8 hard

The two lines $ax+2y=5x$ and $2x-5y+3=ax+3$ meet at some point other than $x=0$ , $y=0$ . Find the value of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The lines $ax+2y=5x$ and $2x-5y+3=ax+3$ meet somewhere besides the origin. Find $a$ .

Givens

Line 1: $ax+2y=5x$ .
Line 2: $2x-5y+3=ax+3$ .
The lines share a point other than $(0, 0)$ .

Unknowns

The constant $a$ .

Constraints

Two different lines meet in at most one point.

2 · Planchoose the strategy

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

Tidy both equations first. The constant terms vanish, so both lines go through the origin. Two distinct lines share at most one point, and the origin already uses it up; a second common point forces the lines to be the same line.

3 · Execute3 carry out the plan

1Rearrange both equations

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

Move every term to the left side.

(a-5)x+2y=0,\quad (a-2)x+5y=0

Rewriting reveals what the original forms hid: there is no constant term.

2Both lines pass through the origin

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

With no constant term,

x=0

,

y=0

satisfies each equation. Two different lines cross at most once, so if they also meet elsewhere they must coincide, meaning infinitely many common solutions.

Picture two lines through one point: unless they lie on top of each other, they spread apart and never meet again.

3Make the lines identical

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

Two lines through the origin coincide when their slopes are equal. From line 1,

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

; from line 2,

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

.

\frac{a-5}{2}=\frac{a-2}{5},\ 5a-25=2a-4,\ 3a=21,\ a=7

Equal slopes plus a shared point means the same line.

Answer:

a=7

4 · Reviewdoes it hold up?

With $a=7$ the lines are $2x+2y=0$ and $5x+5y=0$ ; both simplify to $y=-x$ , so they share every point, for example $(1, -1)$ .

Another way: Solve the system directly: eliminating $y$ gives $(5(a-5)-2(a-2))x=0$ , that is $(3a-21)x=0$ . Only $x=0$ (and then $y=0$ ) works unless $3a-21=0$ , so $a=7$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognising that a system has infinitely many solutions exactly when the two lines coincide.

💡Takeaway. Two lines can share one point or all of them -- never exactly two.

Variant 9 medium

The two lines $ax+3y=2x$ and $5x-4y+2=ax+2$ meet at some point other than $x=0$ , $y=0$ . Find the value of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The lines $ax+3y=2x$ and $5x-4y+2=ax+2$ meet somewhere besides the origin. Find $a$ .

Givens

Line 1: $ax+3y=2x$ .
Line 2: $5x-4y+2=ax+2$ .
The lines share a point other than $(0, 0)$ .

Unknowns

The constant $a$ .

Constraints

Two different lines meet in at most one point.

2 · Planchoose the strategy

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

Tidy both equations first. The constant terms vanish, so both lines go through the origin. Two distinct lines share at most one point, and the origin already uses it up; a second common point forces the lines to be the same line.

3 · Execute3 carry out the plan

1Rearrange both equations

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

Move every term to the left side.

(a-2)x+3y=0,\quad (a-5)x+4y=0

Rewriting reveals what the original forms hid: there is no constant term.

2Both lines pass through the origin

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

With no constant term,

x=0

,

y=0

satisfies each equation. Two different lines cross at most once, so if they also meet elsewhere they must coincide, meaning infinitely many common solutions.

Picture two lines through one point: unless they lie on top of each other, they spread apart and never meet again.

3Make the lines identical

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

Two lines through the origin coincide when their slopes are equal. From line 1,

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

; from line 2,

y=-\frac{a-5}{4}x

.

\frac{a-2}{3}=\frac{a-5}{4},\ 4a-8=3a-15,\ a=-7

Equal slopes plus a shared point means the same line.

Answer:

a=-7

4 · Reviewdoes it hold up?

With $a=-7$ the lines are $-9x+3y=0$ and $-12x+4y=0$ ; both simplify to $y=3x$ , so they share every point, for example $(1, 3)$ .

Another way: Solve the system directly: eliminating $y$ gives $(4(a-2)-3(a-5))x=0$ , that is $(a+7)x=0$ . Only $x=0$ (and then $y=0$ ) works unless $a+7=0$ , so $a=-7$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognising that a system has infinitely many solutions exactly when the two lines coincide.

💡Takeaway. Two lines can share one point or all of them -- never exactly two.

Variant 10 hard

The two lines $ax+2y=2x$ and $5x-3y+3=ax+3$ meet at some point other than $x=0$ , $y=0$ . Find the value of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The lines $ax+2y=2x$ and $5x-3y+3=ax+3$ meet somewhere besides the origin. Find $a$ .

Givens

Line 1: $ax+2y=2x$ .
Line 2: $5x-3y+3=ax+3$ .
The lines share a point other than $(0, 0)$ .

Unknowns

The constant $a$ .

Constraints

Two different lines meet in at most one point.

2 · Planchoose the strategy

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

Tidy both equations first. The constant terms vanish, so both lines go through the origin. Two distinct lines share at most one point, and the origin already uses it up; a second common point forces the lines to be the same line.

3 · Execute3 carry out the plan

1Rearrange both equations

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

Move every term to the left side.

(a-2)x+2y=0,\quad (a-5)x+3y=0

Rewriting reveals what the original forms hid: there is no constant term.

2Both lines pass through the origin

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

With no constant term,

x=0

,

y=0

satisfies each equation. Two different lines cross at most once, so if they also meet elsewhere they must coincide, meaning infinitely many common solutions.

Picture two lines through one point: unless they lie on top of each other, they spread apart and never meet again.

3Make the lines identical

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

Two lines through the origin coincide when their slopes are equal. From line 1,

y=-\frac{a-2}{2}x

; from line 2,

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

.

\frac{a-2}{2}=\frac{a-5}{3},\ 3a-6=2a-10,\ a=-4

Equal slopes plus a shared point means the same line.

Answer:

a=-4

4 · Reviewdoes it hold up?

With $a=-4$ the lines are $-6x+2y=0$ and $-9x+3y=0$ ; both simplify to $y=3x$ , so they share every point, for example $(1, 3)$ .

Another way: Solve the system directly: eliminating $y$ gives $(3(a-2)-2(a-5))x=0$ , that is $(a+4)x=0$ . Only $x=0$ (and then $y=0$ ) works unless $a+4=0$ , so $a=-4$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognising that a system has infinitely many solutions exactly when the two lines coincide.

💡Takeaway. Two lines can share one point or all of them -- never exactly two.

Variant 11 hard

The two lines $ax+3y=5x$ and $2x-6y+4=ax+4$ meet at some point other than $x=0$ , $y=0$ . Find the value of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The lines $ax+3y=5x$ and $2x-6y+4=ax+4$ meet somewhere besides the origin. Find $a$ .

Givens

Line 1: $ax+3y=5x$ .
Line 2: $2x-6y+4=ax+4$ .
The lines share a point other than $(0, 0)$ .

Unknowns

The constant $a$ .

Constraints

Two different lines meet in at most one point.

2 · Planchoose the strategy

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

Tidy both equations first. The constant terms vanish, so both lines go through the origin. Two distinct lines share at most one point, and the origin already uses it up; a second common point forces the lines to be the same line.

3 · Execute3 carry out the plan

1Rearrange both equations

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

Move every term to the left side.

(a-5)x+3y=0,\quad (a-2)x+6y=0

Rewriting reveals what the original forms hid: there is no constant term.

2Both lines pass through the origin

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

With no constant term,

x=0

,

y=0

satisfies each equation. Two different lines cross at most once, so if they also meet elsewhere they must coincide, meaning infinitely many common solutions.

Picture two lines through one point: unless they lie on top of each other, they spread apart and never meet again.

3Make the lines identical

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

Two lines through the origin coincide when their slopes are equal. From line 1,

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

; from line 2,

y=-\frac{a-2}{6}x

.

\frac{a-5}{3}=\frac{a-2}{6},\ 6a-30=3a-6,\ 3a=24,\ a=8

Equal slopes plus a shared point means the same line.

Answer:

a=8

4 · Reviewdoes it hold up?

With $a=8$ the lines are $3x+3y=0$ and $6x+6y=0$ ; both simplify to $y=-x$ , so they share every point, for example $(1, -1)$ .

Another way: Solve the system directly: eliminating $y$ gives $(6(a-5)-3(a-2))x=0$ , that is $(3a-24)x=0$ . Only $x=0$ (and then $y=0$ ) works unless $3a-24=0$ , so $a=8$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognising that a system has infinitely many solutions exactly when the two lines coincide.

💡Takeaway. Two lines can share one point or all of them -- never exactly two.

Variant 12 hard

The two lines $ax+2y=6x$ and $2x-3y+4=ax+4$ meet at some point other than $x=0$ , $y=0$ . Find the value of the constant $a$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

The lines $ax+2y=6x$ and $2x-3y+4=ax+4$ meet somewhere besides the origin. Find $a$ .

Givens

Line 1: $ax+2y=6x$ .
Line 2: $2x-3y+4=ax+4$ .
The lines share a point other than $(0, 0)$ .

Unknowns

The constant $a$ .

Constraints

Two different lines meet in at most one point.

2 · Planchoose the strategy

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

Tidy both equations first. The constant terms vanish, so both lines go through the origin. Two distinct lines share at most one point, and the origin already uses it up; a second common point forces the lines to be the same line.

3 · Execute3 carry out the plan

1Rearrange both equations

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

Move every term to the left side.

(a-6)x+2y=0,\quad (a-2)x+3y=0

Rewriting reveals what the original forms hid: there is no constant term.

2Both lines pass through the origin

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

With no constant term,

x=0

,

y=0

satisfies each equation. Two different lines cross at most once, so if they also meet elsewhere they must coincide, meaning infinitely many common solutions.

Picture two lines through one point: unless they lie on top of each other, they spread apart and never meet again.

3Make the lines identical

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

Two lines through the origin coincide when their slopes are equal. From line 1,

y=-\frac{a-6}{2}x

; from line 2,

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

.

\frac{a-6}{2}=\frac{a-2}{3},\ 3a-18=2a-4,\ a=14

Equal slopes plus a shared point means the same line.

Answer:

a=14

4 · Reviewdoes it hold up?

With $a=14$ the lines are $8x+2y=0$ and $12x+3y=0$ ; both simplify to $y=-4x$ , so they share every point, for example $(1, -4)$ .

Another way: Solve the system directly: eliminating $y$ gives $(3(a-6)-2(a-2))x=0$ , that is $(a-14)x=0$ . Only $x=0$ (and then $y=0$ ) works unless $a-14=0$ , so $a=14$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Recognising that a system has infinitely many solutions exactly when the two lines coincide.

💡Takeaway. Two lines can share one point or all of them -- never exactly two.

Two lines through the origin meeting twice coincide · 12 practice problems

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