Practice · A zero product graphs as two lines · Sensim Math

Variant 1 easy

How many ordered pairs $(x, y)$ satisfy both equations $(x-y)(x+y-4)=0$ and $(x-y+2)(y-3)=0$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the points $(x, y)$ that satisfy both $(x-y)(x+y-4)=0$ and $(x-y+2)(y-3)=0$ .

Givens

Equation 1: $(x-y)(x+y-4)=0$ .
Equation 2: $(x-y+2)(y-3)=0$ .

Unknowns

The number of ordered pairs satisfying both.

Constraints

A product is $0$ exactly when at least one factor is $0$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#1 Draw a Diagram

Re-read each equation: a product is zero when one factor is zero, so each equation's graph is a PAIR of lines. A common solution is a crossing between a line from the first pair and a line from the second, so list the four pairings and check which actually meet.

3 · Execute4 carry out the plan

1Split each equation into two lines

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

Equation 1 is

y=x

or

y=-x+4

; equation 2 is

y=x+2

or

y=3

.

Each factored equation draws two lines, not one curve.

2Pair 1 with 1, and 1 with 2

#2 Make a Systematic List 8.EE.C.8

y=x

with

y=x+2

: same slope, different intercepts, so they are parallel and never meet.

y=x

with

y=3

:

x=3

, so

x=3

and the point is

(3, 3)

.

Each crossing is a two-equation system with one solution.

3The other two pairs

#2 Make a Systematic List 8.EE.C.8

y=-x+4

with

y=x+2

:

-x+4=x+2

, so

x=1

and the point is

(1, 3)

.

y=-x+4

with

y=3

:

-x+4=3

, so

x=1

and the point is

(1, 3)

.

Parallel lines contribute nothing.

4Count distinct points

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

Some pairings land on the same point, which counts only once. The distinct points are

(3, 3)

,

(1, 3)

, so there are

2

ordered pairs.

0+1+1+1-1=2

Check for repeats: a point counted twice would make the total too big.

Answer:

2

4 · Reviewdoes it hold up?

Plug in $(1, 3)$ : $1+3-4=0$ makes equation 1 zero, and $3-3=0$ makes equation 2 zero. The other points sit on one line of each pair by construction.

Another way: Draw all four lines on one grid. Only crossings between a line of the first pair and a line of the second count, parallel pairs give none, and lines meeting at one shared point give it once, so the picture shows exactly $2$ of them.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving each pair of linear equations and recognising parallel lines with no solution.

💡Takeaway. A product equal to zero is really two simpler equations wearing one coat.

Variant 2 easy

How many ordered pairs $(x, y)$ satisfy both equations $(x+y)(y-4)=0$ and $(2x-y-3)(x+y-2)=0$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the points $(x, y)$ that satisfy both $(x+y)(y-4)=0$ and $(2x-y-3)(x+y-2)=0$ .

Givens

Equation 1: $(x+y)(y-4)=0$ .
Equation 2: $(2x-y-3)(x+y-2)=0$ .

Unknowns

The number of ordered pairs satisfying both.

Constraints

A product is $0$ exactly when at least one factor is $0$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#1 Draw a Diagram

Re-read each equation: a product is zero when one factor is zero, so each equation's graph is a PAIR of lines. A common solution is a crossing between a line from the first pair and a line from the second, so list the four pairings and check which actually meet.

3 · Execute4 carry out the plan

1Split each equation into two lines

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

Equation 1 is

y=-x

or

y=4

; equation 2 is

y=2x-3

or

y=-x+2

.

Each factored equation draws two lines, not one curve.

2Pair 1 with 1, and 1 with 2

#2 Make a Systematic List 8.EE.C.8

y=-x

with

y=2x-3

:

-x=2x-3

, so

x=1

and the point is

(1, -1)

.

y=-x

with

y=-x+2

: same slope, different intercepts, so they are parallel and never meet.

Each crossing is a two-equation system with one solution.

3The other two pairs

#2 Make a Systematic List 8.EE.C.8

y=4

with

y=2x-3

:

4=2x-3

, so

x=\frac{7}{2}

and the point is

\left(\frac{7}{2}, 4\right)

.

y=4

with

y=-x+2

:

4=-x+2

, so

x=-2

and the point is

(-2, 4)

.

Parallel lines contribute nothing.

4Count distinct points

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

The points

(1, -1)

,

\left(\frac{7}{2}, 4\right)

,

(-2, 4)

are all different, so there are

3

ordered pairs.

1+0+1+1=3

Check for repeats: a point counted twice would make the total too big.

Answer:

3

4 · Reviewdoes it hold up?

Plug in $(-2, 4)$ : $4-4=0$ makes equation 1 zero, and $-2+4-2=0$ makes equation 2 zero. The other points sit on one line of each pair by construction.

Another way: Draw all four lines on one grid. Only crossings between a line of the first pair and a line of the second count, parallel pairs give none, so the picture shows exactly $3$ of them.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving each pair of linear equations and recognising parallel lines with no solution.

💡Takeaway. A product equal to zero is really two simpler equations wearing one coat.

Variant 3 easy

How many ordered pairs $(x, y)$ satisfy both equations $(y-3)(x-y+1)=0$ and $(x+y-5)(x-y-3)=0$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the points $(x, y)$ that satisfy both $(y-3)(x-y+1)=0$ and $(x+y-5)(x-y-3)=0$ .

Givens

Equation 1: $(y-3)(x-y+1)=0$ .
Equation 2: $(x+y-5)(x-y-3)=0$ .

Unknowns

The number of ordered pairs satisfying both.

Constraints

A product is $0$ exactly when at least one factor is $0$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#1 Draw a Diagram

Re-read each equation: a product is zero when one factor is zero, so each equation's graph is a PAIR of lines. A common solution is a crossing between a line from the first pair and a line from the second, so list the four pairings and check which actually meet.

3 · Execute4 carry out the plan

1Split each equation into two lines

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

Equation 1 is

y=3

or

y=x+1

; equation 2 is

y=-x+5

or

y=x-3

.

Each factored equation draws two lines, not one curve.

2Pair 1 with 1, and 1 with 2

#2 Make a Systematic List 8.EE.C.8

y=3

with

y=-x+5

:

3=-x+5

, so

x=2

and the point is

(2, 3)

.

y=3

with

y=x-3

:

3=x-3

, so

x=6

and the point is

(6, 3)

.

Each crossing is a two-equation system with one solution.

3The other two pairs

#2 Make a Systematic List 8.EE.C.8

y=x+1

with

y=-x+5

:

x+1=-x+5

, so

x=2

and the point is

(2, 3)

.

y=x+1

with

y=x-3

: same slope, different intercepts, so they are parallel and never meet.

Parallel lines contribute nothing.

4Count distinct points

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

Some pairings land on the same point, which counts only once. The distinct points are

(2, 3)

,

(6, 3)

, so there are

2

ordered pairs.

1+1+1+0-1=2

Check for repeats: a point counted twice would make the total too big.

Answer:

2

4 · Reviewdoes it hold up?

Plug in $(2, 3)$ : $2-3+1=0$ makes equation 1 zero, and $2+3-5=0$ makes equation 2 zero. The other points sit on one line of each pair by construction.

Another way: Draw all four lines on one grid. Only crossings between a line of the first pair and a line of the second count, parallel pairs give none, and lines meeting at one shared point give it once, so the picture shows exactly $2$ of them.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving each pair of linear equations and recognising parallel lines with no solution.

💡Takeaway. A product equal to zero is really two simpler equations wearing one coat.

Variant 4 easy

How many ordered pairs $(x, y)$ satisfy both equations $(x-y-1)(x+y-3)=0$ and $(y-1)(x-2y+2)=0$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the points $(x, y)$ that satisfy both $(x-y-1)(x+y-3)=0$ and $(y-1)(x-2y+2)=0$ .

Givens

Equation 1: $(x-y-1)(x+y-3)=0$ .
Equation 2: $(y-1)(x-2y+2)=0$ .

Unknowns

The number of ordered pairs satisfying both.

Constraints

A product is $0$ exactly when at least one factor is $0$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#1 Draw a Diagram

Re-read each equation: a product is zero when one factor is zero, so each equation's graph is a PAIR of lines. A common solution is a crossing between a line from the first pair and a line from the second, so list the four pairings and check which actually meet.

3 · Execute4 carry out the plan

1Split each equation into two lines

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

Equation 1 is

y=x-1

or

y=-x+3

; equation 2 is

y=1

or

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

.

Each factored equation draws two lines, not one curve.

2Pair 1 with 1, and 1 with 2

#2 Make a Systematic List 8.EE.C.8

y=x-1

with

y=1

:

x-1=1

, so

x=2

and the point is

(2, 1)

.

y=x-1

with

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

:

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

, so

x=4

and the point is

(4, 3)

.

Each crossing is a two-equation system with one solution.

3The other two pairs

#2 Make a Systematic List 8.EE.C.8

y=-x+3

with

y=1

:

-x+3=1

, so

x=2

and the point is

(2, 1)

.

y=-x+3

with

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

:

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

, so

x=\frac{4}{3}

and the point is

\left(\frac{4}{3}, \frac{5}{3}\right)

.

Watch for a point that has already been found.

4Count distinct points

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

Some pairings land on the same point, which counts only once. The distinct points are

(2, 1)

,

(4, 3)

,

\left(\frac{4}{3}, \frac{5}{3}\right)

, so there are

3

ordered pairs.

1+1+1+1-1=3

Check for repeats: a point counted twice would make the total too big.

Answer:

3

4 · Reviewdoes it hold up?

Plug in $\left(\frac{4}{3}, \frac{5}{3}\right)$ : $\frac{4}{3}+\frac{5}{3}-3=0$ makes equation 1 zero, and $\frac{4}{3}-\frac{10}{3}+2=0$ makes equation 2 zero. The other points sit on one line of each pair by construction.

Another way: Draw all four lines on one grid. Only crossings between a line of the first pair and a line of the second count, and lines meeting at one shared point give it once, so the picture shows exactly $3$ of them.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving each pair of linear equations and recognising parallel lines with no solution.

💡Takeaway. A product equal to zero is really two simpler equations wearing one coat.

Variant 5 medium

How many ordered pairs $(x, y)$ satisfy both equations $(y+2)(3x-y+1)=0$ and $(x-y)(3x-y-4)=0$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the points $(x, y)$ that satisfy both $(y+2)(3x-y+1)=0$ and $(x-y)(3x-y-4)=0$ .

Givens

Equation 1: $(y+2)(3x-y+1)=0$ .
Equation 2: $(x-y)(3x-y-4)=0$ .

Unknowns

The number of ordered pairs satisfying both.

Constraints

A product is $0$ exactly when at least one factor is $0$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#1 Draw a Diagram

Re-read each equation: a product is zero when one factor is zero, so each equation's graph is a PAIR of lines. A common solution is a crossing between a line from the first pair and a line from the second, so list the four pairings and check which actually meet.

3 · Execute4 carry out the plan

1Split each equation into two lines

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

Equation 1 is

y=-2

or

y=3x+1

; equation 2 is

y=x

or

y=3x-4

.

Each factored equation draws two lines, not one curve.

2Pair 1 with 1, and 1 with 2

#2 Make a Systematic List 8.EE.C.8

y=-2

with

y=x

:

-2=x

, so

x=-2

and the point is

(-2, -2)

.

y=-2

with

y=3x-4

:

-2=3x-4

, so

x=\frac{2}{3}

and the point is

\left(\frac{2}{3}, -2\right)

.

Each crossing is a two-equation system with one solution.

3The other two pairs

#2 Make a Systematic List 8.EE.C.8

y=3x+1

with

y=x

:

3x+1=x

, so

x=-\frac{1}{2}

and the point is

\left(-\frac{1}{2}, -\frac{1}{2}\right)

.

y=3x+1

with

y=3x-4

: same slope, different intercepts, so they are parallel and never meet.

Parallel lines contribute nothing.

4Count distinct points

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

The points

(-2, -2)

,

\left(\frac{2}{3}, -2\right)

,

\left(-\frac{1}{2}, -\frac{1}{2}\right)

are all different, so there are

3

ordered pairs.

1+1+1+0=3

Check for repeats: a point counted twice would make the total too big.

Answer:

3

4 · Reviewdoes it hold up?

Plug in $\left(-\frac{1}{2}, -\frac{1}{2}\right)$ : $-\frac{3}{2}+\frac{1}{2}+1=0$ makes equation 1 zero, and $-\frac{1}{2}+\frac{1}{2}=0$ makes equation 2 zero. The other points sit on one line of each pair by construction.

Another way: Draw all four lines on one grid. Only crossings between a line of the first pair and a line of the second count, parallel pairs give none, so the picture shows exactly $3$ of them.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving each pair of linear equations and recognising parallel lines with no solution.

💡Takeaway. A product equal to zero is really two simpler equations wearing one coat.

Variant 6 medium

How many ordered pairs $(x, y)$ satisfy both equations $(2x+y-4)(x-y-1)=0$ and $(y-2)(2x+y)=0$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the points $(x, y)$ that satisfy both $(2x+y-4)(x-y-1)=0$ and $(y-2)(2x+y)=0$ .

Givens

Equation 1: $(2x+y-4)(x-y-1)=0$ .
Equation 2: $(y-2)(2x+y)=0$ .

Unknowns

The number of ordered pairs satisfying both.

Constraints

A product is $0$ exactly when at least one factor is $0$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#1 Draw a Diagram

Re-read each equation: a product is zero when one factor is zero, so each equation's graph is a PAIR of lines. A common solution is a crossing between a line from the first pair and a line from the second, so list the four pairings and check which actually meet.

3 · Execute4 carry out the plan

1Split each equation into two lines

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

Equation 1 is

y=-2x+4

or

y=x-1

; equation 2 is

y=2

or

y=-2x

.

Each factored equation draws two lines, not one curve.

2Pair 1 with 1, and 1 with 2

#2 Make a Systematic List 8.EE.C.8

y=-2x+4

with

y=2

:

-2x+4=2

, so

x=1

and the point is

(1, 2)

.

y=-2x+4

with

y=-2x

: same slope, different intercepts, so they are parallel and never meet.

Each crossing is a two-equation system with one solution.

3The other two pairs

#2 Make a Systematic List 8.EE.C.8

y=x-1

with

y=2

:

x-1=2

, so

x=3

and the point is

(3, 2)

.

y=x-1

with

y=-2x

:

x-1=-2x

, so

x=\frac{1}{3}

and the point is

\left(\frac{1}{3}, -\frac{2}{3}\right)

.

Parallel lines contribute nothing.

4Count distinct points

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

The points

(1, 2)

,

(3, 2)

,

\left(\frac{1}{3}, -\frac{2}{3}\right)

are all different, so there are

3

ordered pairs.

1+0+1+1=3

Check for repeats: a point counted twice would make the total too big.

Answer:

3

4 · Reviewdoes it hold up?

Plug in $\left(\frac{1}{3}, -\frac{2}{3}\right)$ : $\frac{1}{3}+\frac{2}{3}-1=0$ makes equation 1 zero, and $\frac{2}{3}-\frac{2}{3}=0$ makes equation 2 zero. The other points sit on one line of each pair by construction.

Another way: Draw all four lines on one grid. Only crossings between a line of the first pair and a line of the second count, parallel pairs give none, so the picture shows exactly $3$ of them.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving each pair of linear equations and recognising parallel lines with no solution.

💡Takeaway. A product equal to zero is really two simpler equations wearing one coat.

Variant 7 medium

How many ordered pairs $(x, y)$ satisfy both equations $(y-1)(2x-y)=0$ and $(x+y-4)(2x-y+3)=0$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the points $(x, y)$ that satisfy both $(y-1)(2x-y)=0$ and $(x+y-4)(2x-y+3)=0$ .

Givens

Equation 1: $(y-1)(2x-y)=0$ .
Equation 2: $(x+y-4)(2x-y+3)=0$ .

Unknowns

The number of ordered pairs satisfying both.

Constraints

A product is $0$ exactly when at least one factor is $0$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#1 Draw a Diagram

Re-read each equation: a product is zero when one factor is zero, so each equation's graph is a PAIR of lines. A common solution is a crossing between a line from the first pair and a line from the second, so list the four pairings and check which actually meet.

3 · Execute4 carry out the plan

1Split each equation into two lines

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

Equation 1 is

y=1

or

y=2x

; equation 2 is

y=-x+4

or

y=2x+3

.

Each factored equation draws two lines, not one curve.

2Pair 1 with 1, and 1 with 2

#2 Make a Systematic List 8.EE.C.8

y=1

with

y=-x+4

:

1=-x+4

, so

x=3

and the point is

(3, 1)

.

y=1

with

y=2x+3

:

1=2x+3

, so

x=-1

and the point is

(-1, 1)

.

Each crossing is a two-equation system with one solution.

3The other two pairs

#2 Make a Systematic List 8.EE.C.8

y=2x

with

y=-x+4

:

2x=-x+4

, so

x=\frac{4}{3}

and the point is

\left(\frac{4}{3}, \frac{8}{3}\right)

.

y=2x

with

y=2x+3

: same slope, different intercepts, so they are parallel and never meet.

Parallel lines contribute nothing.

4Count distinct points

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

The points

(3, 1)

,

(-1, 1)

,

\left(\frac{4}{3}, \frac{8}{3}\right)

are all different, so there are

3

ordered pairs.

1+1+1+0=3

Check for repeats: a point counted twice would make the total too big.

Answer:

3

4 · Reviewdoes it hold up?

Plug in $\left(\frac{4}{3}, \frac{8}{3}\right)$ : $\frac{8}{3}-\frac{8}{3}=0$ makes equation 1 zero, and $\frac{4}{3}+\frac{8}{3}-4=0$ makes equation 2 zero. The other points sit on one line of each pair by construction.

Another way: Draw all four lines on one grid. Only crossings between a line of the first pair and a line of the second count, parallel pairs give none, so the picture shows exactly $3$ of them.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving each pair of linear equations and recognising parallel lines with no solution.

💡Takeaway. A product equal to zero is really two simpler equations wearing one coat.

Variant 8 medium

How many ordered pairs $(x, y)$ satisfy both equations $(y-1)(x-3y)=0$ and $(x-3y+3)(x+y-5)=0$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the points $(x, y)$ that satisfy both $(y-1)(x-3y)=0$ and $(x-3y+3)(x+y-5)=0$ .

Givens

Equation 1: $(y-1)(x-3y)=0$ .
Equation 2: $(x-3y+3)(x+y-5)=0$ .

Unknowns

The number of ordered pairs satisfying both.

Constraints

A product is $0$ exactly when at least one factor is $0$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#1 Draw a Diagram

Re-read each equation: a product is zero when one factor is zero, so each equation's graph is a PAIR of lines. A common solution is a crossing between a line from the first pair and a line from the second, so list the four pairings and check which actually meet.

3 · Execute4 carry out the plan

1Split each equation into two lines

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

Equation 1 is

y=1

or

y=\frac{1}{3}x

; equation 2 is

y=\frac{1}{3}x+1

or

y=-x+5

.

Each factored equation draws two lines, not one curve.

2Pair 1 with 1, and 1 with 2

#2 Make a Systematic List 8.EE.C.8

y=1

with

y=\frac{1}{3}x+1

:

1=\frac{1}{3}x+1

, so

x=0

and the point is

(0, 1)

.

y=1

with

y=-x+5

:

1=-x+5

, so

x=4

and the point is

(4, 1)

.

Each crossing is a two-equation system with one solution.

3The other two pairs

#2 Make a Systematic List 8.EE.C.8

y=\frac{1}{3}x

with

y=\frac{1}{3}x+1

: same slope, different intercepts, so they are parallel and never meet.

y=\frac{1}{3}x

with

y=-x+5

:

\frac{1}{3}x=-x+5

, so

x=\frac{15}{4}

and the point is

\left(\frac{15}{4}, \frac{5}{4}\right)

.

Parallel lines contribute nothing.

4Count distinct points

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

The points

(0, 1)

,

(4, 1)

,

\left(\frac{15}{4}, \frac{5}{4}\right)

are all different, so there are

3

ordered pairs.

1+1+0+1=3

Check for repeats: a point counted twice would make the total too big.

Answer:

3

4 · Reviewdoes it hold up?

Plug in $\left(\frac{15}{4}, \frac{5}{4}\right)$ : $\frac{15}{4}-\frac{15}{4}=0$ makes equation 1 zero, and $\frac{15}{4}+\frac{5}{4}-5=0$ makes equation 2 zero. The other points sit on one line of each pair by construction.

Another way: Draw all four lines on one grid. Only crossings between a line of the first pair and a line of the second count, parallel pairs give none, so the picture shows exactly $3$ of them.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving each pair of linear equations and recognising parallel lines with no solution.

💡Takeaway. A product equal to zero is really two simpler equations wearing one coat.

Variant 9 hard

How many ordered pairs $(x, y)$ satisfy both equations $(x-y)(x+2y-6)=0$ and $(y-2)(2x-y-2)=0$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the points $(x, y)$ that satisfy both $(x-y)(x+2y-6)=0$ and $(y-2)(2x-y-2)=0$ .

Givens

Equation 1: $(x-y)(x+2y-6)=0$ .
Equation 2: $(y-2)(2x-y-2)=0$ .

Unknowns

The number of ordered pairs satisfying both.

Constraints

A product is $0$ exactly when at least one factor is $0$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#1 Draw a Diagram

Re-read each equation: a product is zero when one factor is zero, so each equation's graph is a PAIR of lines. A common solution is a crossing between a line from the first pair and a line from the second, so list the four pairings and check which actually meet.

3 · Execute4 carry out the plan

1Split each equation into two lines

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

Equation 1 is

y=x

or

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

; equation 2 is

y=2

or

y=2x-2

.

Each factored equation draws two lines, not one curve.

2Pair 1 with 1, and 1 with 2

#2 Make a Systematic List 8.EE.C.8

y=x

with

y=2

:

x=2

, so

x=2

and the point is

(2, 2)

.

y=x

with

y=2x-2

:

x=2x-2

, so

x=2

and the point is

(2, 2)

.

Each crossing is a two-equation system with one solution.

3The other two pairs

#2 Make a Systematic List 8.EE.C.8

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

with

y=2

:

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

, so

x=2

and the point is

(2, 2)

.

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

with

y=2x-2

:

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

, so

x=2

and the point is

(2, 2)

.

Watch for a point that has already been found.

4Count distinct points

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

Every crossing is the same point

(2, 2)

, which counts only once, so there is

1

ordered pair.

1+1+1+1-3=1

Check for repeats: a point counted twice would make the total too big.

Answer:

1

4 · Reviewdoes it hold up?

Plug in $(2, 2)$ : $2+4-6=0$ makes equation 1 zero, and $4-2-2=0$ makes equation 2 zero. The other points sit on one line of each pair by construction.

Another way: Draw all four lines on one grid. Only crossings between a line of the first pair and a line of the second count, and lines meeting at one shared point give it once, so the picture shows exactly $1$ of them.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving each pair of linear equations and recognising parallel lines with no solution.

💡Takeaway. A product equal to zero is really two simpler equations wearing one coat.

Variant 10 hard

How many ordered pairs $(x, y)$ satisfy both equations $(3x-y)(x+2y-4)=0$ and $(3x-y-5)(y+1)=0$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the points $(x, y)$ that satisfy both $(3x-y)(x+2y-4)=0$ and $(3x-y-5)(y+1)=0$ .

Givens

Equation 1: $(3x-y)(x+2y-4)=0$ .
Equation 2: $(3x-y-5)(y+1)=0$ .

Unknowns

The number of ordered pairs satisfying both.

Constraints

A product is $0$ exactly when at least one factor is $0$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#1 Draw a Diagram

Re-read each equation: a product is zero when one factor is zero, so each equation's graph is a PAIR of lines. A common solution is a crossing between a line from the first pair and a line from the second, so list the four pairings and check which actually meet.

3 · Execute4 carry out the plan

1Split each equation into two lines

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

Equation 1 is

y=3x

or

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

; equation 2 is

y=3x-5

or

y=-1

.

Each factored equation draws two lines, not one curve.

2Pair 1 with 1, and 1 with 2

#2 Make a Systematic List 8.EE.C.8

y=3x

with

y=3x-5

: same slope, different intercepts, so they are parallel and never meet.

y=3x

with

y=-1

:

3x=-1

, so

x=-\frac{1}{3}

and the point is

\left(-\frac{1}{3}, -1\right)

.

Each crossing is a two-equation system with one solution.

3The other two pairs

#2 Make a Systematic List 8.EE.C.8

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

with

y=3x-5

:

-\frac{1}{2}x+2=3x-5

, so

x=2

and the point is

(2, 1)

.

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

with

y=-1

:

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

, so

x=6

and the point is

(6, -1)

.

Parallel lines contribute nothing.

4Count distinct points

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

The points

\left(-\frac{1}{3}, -1\right)

,

(2, 1)

,

(6, -1)

are all different, so there are

3

ordered pairs.

0+1+1+1=3

Check for repeats: a point counted twice would make the total too big.

Answer:

3

4 · Reviewdoes it hold up?

Plug in $(6, -1)$ : $6-2-4=0$ makes equation 1 zero, and $-1+1=0$ makes equation 2 zero. The other points sit on one line of each pair by construction.

Another way: Draw all four lines on one grid. Only crossings between a line of the first pair and a line of the second count, parallel pairs give none, so the picture shows exactly $3$ of them.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving each pair of linear equations and recognising parallel lines with no solution.

💡Takeaway. A product equal to zero is really two simpler equations wearing one coat.

Variant 11 hard

How many ordered pairs $(x, y)$ satisfy both equations $(y-2)(x-2y+2)=0$ and $(2x+y-5)(x-2y-2)=0$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the points $(x, y)$ that satisfy both $(y-2)(x-2y+2)=0$ and $(2x+y-5)(x-2y-2)=0$ .

Givens

Equation 1: $(y-2)(x-2y+2)=0$ .
Equation 2: $(2x+y-5)(x-2y-2)=0$ .

Unknowns

The number of ordered pairs satisfying both.

Constraints

A product is $0$ exactly when at least one factor is $0$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#1 Draw a Diagram

Re-read each equation: a product is zero when one factor is zero, so each equation's graph is a PAIR of lines. A common solution is a crossing between a line from the first pair and a line from the second, so list the four pairings and check which actually meet.

3 · Execute4 carry out the plan

1Split each equation into two lines

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

Equation 1 is

y=2

or

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

; equation 2 is

y=-2x+5

or

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

.

Each factored equation draws two lines, not one curve.

2Pair 1 with 1, and 1 with 2

#2 Make a Systematic List 8.EE.C.8

y=2

with

y=-2x+5

:

2=-2x+5

, so

x=\frac{3}{2}

and the point is

\left(\frac{3}{2}, 2\right)

.

y=2

with

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

:

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

, so

x=6

and the point is

(6, 2)

.

Each crossing is a two-equation system with one solution.

3The other two pairs

#2 Make a Systematic List 8.EE.C.8

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

with

y=-2x+5

:

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

, so

x=\frac{8}{5}

and the point is

\left(\frac{8}{5}, \frac{9}{5}\right)

.

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

with

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

: same slope, different intercepts, so they are parallel and never meet.

Parallel lines contribute nothing.

4Count distinct points

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

The points

\left(\frac{3}{2}, 2\right)

,

(6, 2)

,

\left(\frac{8}{5}, \frac{9}{5}\right)

are all different, so there are

3

ordered pairs.

1+1+1+0=3

Check for repeats: a point counted twice would make the total too big.

Answer:

3

4 · Reviewdoes it hold up?

Plug in $\left(\frac{8}{5}, \frac{9}{5}\right)$ : $\frac{8}{5}-\frac{18}{5}+2=0$ makes equation 1 zero, and $\frac{16}{5}+\frac{9}{5}-5=0$ makes equation 2 zero. The other points sit on one line of each pair by construction.

Another way: Draw all four lines on one grid. Only crossings between a line of the first pair and a line of the second count, parallel pairs give none, so the picture shows exactly $3$ of them.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving each pair of linear equations and recognising parallel lines with no solution.

💡Takeaway. A product equal to zero is really two simpler equations wearing one coat.

Variant 12 hard

How many ordered pairs $(x, y)$ satisfy both equations $(x-2y)(3x+y-6)=0$ and $(x-2y+4)(3x+y+3)=0$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Count the points $(x, y)$ that satisfy both $(x-2y)(3x+y-6)=0$ and $(x-2y+4)(3x+y+3)=0$ .

Givens

Equation 1: $(x-2y)(3x+y-6)=0$ .
Equation 2: $(x-2y+4)(3x+y+3)=0$ .

Unknowns

The number of ordered pairs satisfying both.

Constraints

A product is $0$ exactly when at least one factor is $0$ .

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #2 Make a Systematic List#1 Draw a Diagram

Re-read each equation: a product is zero when one factor is zero, so each equation's graph is a PAIR of lines. A common solution is a crossing between a line from the first pair and a line from the second, so list the four pairings and check which actually meet.

3 · Execute4 carry out the plan

1Split each equation into two lines

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

Equation 1 is

y=\frac{1}{2}x

or

y=-3x+6

; equation 2 is

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

or

y=-3x-3

.

Each factored equation draws two lines, not one curve.

2Pair 1 with 1, and 1 with 2

#2 Make a Systematic List 8.EE.C.8

y=\frac{1}{2}x

with

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

: same slope, different intercepts, so they are parallel and never meet.

y=\frac{1}{2}x

with

y=-3x-3

:

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

, so

x=-\frac{6}{7}

and the point is

\left(-\frac{6}{7}, -\frac{3}{7}\right)

.

Each crossing is a two-equation system with one solution.

3The other two pairs

#2 Make a Systematic List 8.EE.C.8

y=-3x+6

with

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

:

-3x+6=\frac{1}{2}x+2

, so

x=\frac{8}{7}

and the point is

\left(\frac{8}{7}, \frac{18}{7}\right)

.

y=-3x+6

with

y=-3x-3

: same slope, different intercepts, so they are parallel and never meet.

Parallel lines contribute nothing.

4Count distinct points

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

The points

\left(-\frac{6}{7}, -\frac{3}{7}\right)

,

\left(\frac{8}{7}, \frac{18}{7}\right)

are all different, so there are

2

ordered pairs.

0+1+1+0=2

Check for repeats: a point counted twice would make the total too big.

Answer:

2

4 · Reviewdoes it hold up?

Plug in $\left(\frac{8}{7}, \frac{18}{7}\right)$ : $\frac{24}{7}+\frac{18}{7}-6=0$ makes equation 1 zero, and $\frac{8}{7}-\frac{36}{7}+4=0$ makes equation 2 zero. The other points sit on one line of each pair by construction.

Another way: Draw all four lines on one grid. Only crossings between a line of the first pair and a line of the second count, parallel pairs give none, so the picture shows exactly $2$ of them.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Solving each pair of linear equations and recognising parallel lines with no solution.

💡Takeaway. A product equal to zero is really two simpler equations wearing one coat.

A zero product graphs as two lines · 12 practice problems

Generated variants — 12

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