Practice · Find the common point from the known lines · Sensim Math

Variant 1 easy

The four lines $5x+y=12$ , $ax+by=6$ , $x+5y=12$ , and $-ax+by=2$ all meet at one point. Find the values of the constants $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four lines pass through one common point. Two of the lines contain the unknown constants $a$ and $b$ ; the other two are fully known. We need $a$ and $b$ .

Givens

The lines $5x+y=12$ , $ax+by=6$ , $x+5y=12$ , $-ax+by=2$ .
All four lines meet at one point.

Unknowns

The constants $a$ and $b$ .

Constraints

The common point lies on every one of the four lines.

2 · Planchoose the strategy

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

Change focus from the unknown constants to the common point. The two lines with no constants already determine that point. Once it is known, plugging it into the other two lines gives a system whose unknowns are $a$ and $b$ .

3 · Execute4 carry out the plan

1Find the point from the two known lines

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

Solve

5x+y=12

and

x+5y=12

. Adding them gives

6x+6y=24

, so

x+y=4

; subtracting gives

4x-4y=0

, so

x=y

.

x = y = 2

Two different lines meet in exactly one point, so this is the shared point of all four.

2Put the point into the lines with constants

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

The point

(2, 2)

is on

ax+by=6

and on

-ax+by=2

.

2a + 2b = 6, \quad -2a + 2b = 2

Now

a

and

b

are the unknowns of a new linear system.

3Simplify both equations

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

Divide the first by

2

and the second by

2

.

a + b = 3, \quad -a + b = 1

Smaller coefficients mean fewer arithmetic slips.

4Solve for $a$ and $b$

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

Substitute

a = 3 - b

into the second equation.

-(3 - b) + b = 1 \;\Rightarrow\; -3 + 2b = 1 \;\Rightarrow\; b = 2, \quad a = 3 - 2 = 1

One substitution finishes the system.

Answer:

a=1

,

b=2

4 · Reviewdoes it hold up?

With $a=1$ , $b=2$ the lines are $x+2y=6$ and $-x+2y=2$ . At $(2, 2)$ : $2+4=6$ and $-2+4=2$ . Both pass through $(2, 2)$ , as do $5x+y=12$ and $x+5y=12$ .

Another way: Eliminate instead: adding $a+b=3$ and $-a+b=1$ gives $2b=4$ , so $b=2$ and $a=1$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Finding the intersection point of two lines and solving the system it produces for $a$ and $b$.

💡Takeaway. When several lines share a point, let the fully known lines find the point first.

Variant 2 easy

The four lines $x+4y=9$ , $ax+by=-7$ , $4x+y=6$ , and $-2ax+by=-4$ all meet at one point. Find the values of the constants $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four lines pass through one common point. Two of the lines contain the unknown constants $a$ and $b$ ; the other two are fully known. We need $a$ and $b$ .

Givens

The lines $x+4y=9$ , $ax+by=-7$ , $4x+y=6$ , $-2ax+by=-4$ .
All four lines meet at one point.

Unknowns

The constants $a$ and $b$ .

Constraints

The common point lies on every one of the four lines.

2 · Planchoose the strategy

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

Change focus from the unknown constants to the common point. The two lines with no constants already determine that point. Once it is known, plugging it into the other two lines gives a system whose unknowns are $a$ and $b$ .

3 · Execute4 carry out the plan

1Find the point from the two known lines

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

Solve

x+4y=9

and

4x+y=6

. Adding them gives

5x+5y=15

, so

x+y=3

; subtracting gives

-3x+3y=3

, so

x-y=-1

.

x = 1, \quad y = 2

Two different lines meet in exactly one point, so this is the shared point of all four.

2Put the point into the lines with constants

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

The point

(1, 2)

is on

ax+by=-7

and on

-2ax+by=-4

.

a + 2b = -7, \quad -2a + 2b = -4

Now

a

and

b

are the unknowns of a new linear system.

3Simplify both equations

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

The first is already as simple as it gets; divide the second by

2

.

a + 2b = -7, \quad -a + b = -2

Smaller coefficients mean fewer arithmetic slips.

4Solve for $a$ and $b$

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

Substitute

a = -7 - 2b

into the second equation.

-(-7 - 2b) + b = -2 \;\Rightarrow\; 7 + 3b = -2 \;\Rightarrow\; b = -3, \quad a = -7 - 2 \times (-3) = -1

One substitution finishes the system.

Answer:

a=-1

,

b=-3

4 · Reviewdoes it hold up?

With $a=-1$ , $b=-3$ the lines are $-x-3y=-7$ and $2x-3y=-4$ . At $(1, 2)$ : $-1-6=-7$ and $2-6=-4$ . Both pass through $(1, 2)$ , as do $x+4y=9$ and $4x+y=6$ .

Another way: Eliminate instead: adding $a+2b=-7$ and $-a+b=-2$ gives $3b=-9$ , so $b=-3$ and $a=-1$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Finding the intersection point of two lines and solving the system it produces for $a$ and $b$.

💡Takeaway. When several lines share a point, let the fully known lines find the point first.

Variant 3 easy

The four lines $2x+y=2$ , $ax+by=-8$ , $x+2y=-2$ , and $2ax-3by=14$ all meet at one point. Find the values of the constants $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four lines pass through one common point. Two of the lines contain the unknown constants $a$ and $b$ ; the other two are fully known. We need $a$ and $b$ .

Givens

The lines $2x+y=2$ , $ax+by=-8$ , $x+2y=-2$ , $2ax-3by=14$ .
All four lines meet at one point.

Unknowns

The constants $a$ and $b$ .

Constraints

The common point lies on every one of the four lines.

2 · Planchoose the strategy

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

Change focus from the unknown constants to the common point. The two lines with no constants already determine that point. Once it is known, plugging it into the other two lines gives a system whose unknowns are $a$ and $b$ .

3 · Execute4 carry out the plan

1Find the point from the two known lines

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

Solve

2x+y=2

and

x+2y=-2

. Adding them gives

3x+3y=0

, so

x+y=0

; subtracting gives

x-y=4

.

x = 2, \quad y = -2

Two different lines meet in exactly one point, so this is the shared point of all four.

2Put the point into the lines with constants

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

The point

(2, -2)

is on

ax+by=-8

and on

2ax-3by=14

.

2a - 2b = -8, \quad 4a + 6b = 14

Now

a

and

b

are the unknowns of a new linear system.

3Simplify both equations

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

Divide the first by

2

and the second by

2

.

a - b = -4, \quad 2a + 3b = 7

Smaller coefficients mean fewer arithmetic slips.

4Solve for $a$ and $b$

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

Substitute

a = -4 + b

into the second equation.

2(-4 + b) + 3b = 7 \;\Rightarrow\; -8 + 5b = 7 \;\Rightarrow\; b = 3, \quad a = -4 + 3 = -1

One substitution finishes the system.

Answer:

a=-1

,

b=3

4 · Reviewdoes it hold up?

With $a=-1$ , $b=3$ the lines are $-x+3y=-8$ and $-2x-9y=14$ . At $(2, -2)$ : $-2-6=-8$ and $-4+18=14$ . Both pass through $(2, -2)$ , as do $2x+y=2$ and $x+2y=-2$ .

Another way: Eliminate instead: doubling $a-b=-4$ gives $2a-2b=-8$ ; subtracting $2a+3b=7$ gives $-5b=-15$ , so $b=3$ and $a=-1$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Finding the intersection point of two lines and solving the system it produces for $a$ and $b$.

💡Takeaway. When several lines share a point, let the fully known lines find the point first.

Variant 4 easy

The four lines $2x+y=0$ , $ax+by=8$ , $x+2y=-3$ , and $-2ax-3by=-20$ all meet at one point. Find the values of the constants $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four lines pass through one common point. Two of the lines contain the unknown constants $a$ and $b$ ; the other two are fully known. We need $a$ and $b$ .

Givens

The lines $2x+y=0$ , $ax+by=8$ , $x+2y=-3$ , $-2ax-3by=-20$ .
All four lines meet at one point.

Unknowns

The constants $a$ and $b$ .

Constraints

The common point lies on every one of the four lines.

2 · Planchoose the strategy

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

Change focus from the unknown constants to the common point. The two lines with no constants already determine that point. Once it is known, plugging it into the other two lines gives a system whose unknowns are $a$ and $b$ .

3 · Execute4 carry out the plan

1Find the point from the two known lines

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

Solve

2x+y=0

and

x+2y=-3

. Adding them gives

3x+3y=-3

, so

x+y=-1

; subtracting gives

x-y=3

.

x = 1, \quad y = -2

Two different lines meet in exactly one point, so this is the shared point of all four.

2Put the point into the lines with constants

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

The point

(1, -2)

is on

ax+by=8

and on

-2ax-3by=-20

.

a - 2b = 8, \quad -2a + 6b = -20

Now

a

and

b

are the unknowns of a new linear system.

3Simplify both equations

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

The first is already as simple as it gets; divide the second by

2

.

a - 2b = 8, \quad -a + 3b = -10

Smaller coefficients mean fewer arithmetic slips.

4Solve for $a$ and $b$

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

Substitute

a = 8 + 2b

into the second equation.

-(8 + 2b) + 3b = -10 \;\Rightarrow\; -8 + b = -10 \;\Rightarrow\; b = -2, \quad a = 8 + 2 \times (-2) = 4

One substitution finishes the system.

Answer:

a=4

,

b=-2

4 · Reviewdoes it hold up?

With $a=4$ , $b=-2$ the lines are $4x-2y=8$ and $-8x+6y=-20$ . At $(1, -2)$ : $4+4=8$ and $-8-12=-20$ . Both pass through $(1, -2)$ , as do $2x+y=0$ and $x+2y=-3$ .

Another way: Eliminate instead: adding $a-2b=8$ and $-a+3b=-10$ gives $b=-2$ , then $a=4$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Finding the intersection point of two lines and solving the system it produces for $a$ and $b$.

💡Takeaway. When several lines share a point, let the fully known lines find the point first.

Variant 5 medium

The four lines $3x+y=12$ , $ax+by=-10$ , $x+3y=20$ , and $-3ax-2by=18$ all meet at one point. Find the values of the constants $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four lines pass through one common point. Two of the lines contain the unknown constants $a$ and $b$ ; the other two are fully known. We need $a$ and $b$ .

Givens

The lines $3x+y=12$ , $ax+by=-10$ , $x+3y=20$ , $-3ax-2by=18$ .
All four lines meet at one point.

Unknowns

The constants $a$ and $b$ .

Constraints

The common point lies on every one of the four lines.

2 · Planchoose the strategy

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

Change focus from the unknown constants to the common point. The two lines with no constants already determine that point. Once it is known, plugging it into the other two lines gives a system whose unknowns are $a$ and $b$ .

3 · Execute4 carry out the plan

1Find the point from the two known lines

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

Solve

3x+y=12

and

x+3y=20

. Adding them gives

4x+4y=32

, so

x+y=8

; subtracting gives

2x-2y=-8

, so

x-y=-4

.

x = 2, \quad y = 6

Two different lines meet in exactly one point, so this is the shared point of all four.

2Put the point into the lines with constants

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

The point

(2, 6)

is on

ax+by=-10

and on

-3ax-2by=18

.

2a + 6b = -10, \quad -6a - 12b = 18

Now

a

and

b

are the unknowns of a new linear system.

3Simplify both equations

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

Divide the first by

2

and the second by

6

.

a + 3b = -5, \quad -a - 2b = 3

Smaller coefficients mean fewer arithmetic slips.

4Solve for $a$ and $b$

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

Substitute

a = -5 - 3b

into the second equation.

-(-5 - 3b) - 2b = 3 \;\Rightarrow\; 5 + b = 3 \;\Rightarrow\; b = -2, \quad a = -5 - 3 \times (-2) = 1

One substitution finishes the system.

Answer:

a=1

,

b=-2

4 · Reviewdoes it hold up?

With $a=1$ , $b=-2$ the lines are $x-2y=-10$ and $-3x+4y=18$ . At $(2, 6)$ : $2-12=-10$ and $-6+24=18$ . Both pass through $(2, 6)$ , as do $3x+y=12$ and $x+3y=20$ .

Another way: Eliminate instead: adding $a+3b=-5$ and $-a-2b=3$ gives $b=-2$ , then $a=1$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Finding the intersection point of two lines and solving the system it produces for $a$ and $b$.

💡Takeaway. When several lines share a point, let the fully known lines find the point first.

Variant 6 medium

The four lines $3x+y=8$ , $ax+by=4$ , $x+3y=8$ , and $2ax-3by=-22$ all meet at one point. Find the values of the constants $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four lines pass through one common point. Two of the lines contain the unknown constants $a$ and $b$ ; the other two are fully known. We need $a$ and $b$ .

Givens

The lines $3x+y=8$ , $ax+by=4$ , $x+3y=8$ , $2ax-3by=-22$ .
All four lines meet at one point.

Unknowns

The constants $a$ and $b$ .

Constraints

The common point lies on every one of the four lines.

2 · Planchoose the strategy

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

Change focus from the unknown constants to the common point. The two lines with no constants already determine that point. Once it is known, plugging it into the other two lines gives a system whose unknowns are $a$ and $b$ .

3 · Execute4 carry out the plan

1Find the point from the two known lines

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

Solve

3x+y=8

and

x+3y=8

. Adding them gives

4x+4y=16

, so

x+y=4

; subtracting gives

2x-2y=0

, so

x=y

.

x = y = 2

Two different lines meet in exactly one point, so this is the shared point of all four.

2Put the point into the lines with constants

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

The point

(2, 2)

is on

ax+by=4

and on

2ax-3by=-22

.

2a + 2b = 4, \quad 4a - 6b = -22

Now

a

and

b

are the unknowns of a new linear system.

3Simplify both equations

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

Divide the first by

2

and the second by

2

.

a + b = 2, \quad 2a - 3b = -11

Smaller coefficients mean fewer arithmetic slips.

4Solve for $a$ and $b$

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

Substitute

a = 2 - b

into the second equation.

2(2 - b) - 3b = -11 \;\Rightarrow\; 4 - 5b = -11 \;\Rightarrow\; b = 3, \quad a = 2 - 3 = -1

One substitution finishes the system.

Answer:

a=-1

,

b=3

4 · Reviewdoes it hold up?

With $a=-1$ , $b=3$ the lines are $-x+3y=4$ and $-2x-9y=-22$ . At $(2, 2)$ : $-2+6=4$ and $-4-18=-22$ . Both pass through $(2, 2)$ , as do $3x+y=8$ and $x+3y=8$ .

Another way: Eliminate instead: doubling $a+b=2$ gives $2a+2b=4$ ; subtracting $2a-3b=-11$ gives $5b=15$ , so $b=3$ and $a=-1$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Finding the intersection point of two lines and solving the system it produces for $a$ and $b$.

💡Takeaway. When several lines share a point, let the fully known lines find the point first.

Variant 7 medium

The four lines $5x+2y=9$ , $ax+by=5$ , $2x+5y=12$ , and $-2ax+by=-4$ all meet at one point. Find the values of the constants $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four lines pass through one common point. Two of the lines contain the unknown constants $a$ and $b$ ; the other two are fully known. We need $a$ and $b$ .

Givens

The lines $5x+2y=9$ , $ax+by=5$ , $2x+5y=12$ , $-2ax+by=-4$ .
All four lines meet at one point.

Unknowns

The constants $a$ and $b$ .

Constraints

The common point lies on every one of the four lines.

2 · Planchoose the strategy

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

Change focus from the unknown constants to the common point. The two lines with no constants already determine that point. Once it is known, plugging it into the other two lines gives a system whose unknowns are $a$ and $b$ .

3 · Execute4 carry out the plan

1Find the point from the two known lines

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

Solve

5x+2y=9

and

2x+5y=12

. Adding them gives

7x+7y=21

, so

x+y=3

; subtracting gives

3x-3y=-3

, so

x-y=-1

.

x = 1, \quad y = 2

Two different lines meet in exactly one point, so this is the shared point of all four.

2Put the point into the lines with constants

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

The point

(1, 2)

is on

ax+by=5

and on

-2ax+by=-4

.

a + 2b = 5, \quad -2a + 2b = -4

Now

a

and

b

are the unknowns of a new linear system.

3Simplify both equations

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

The first is already as simple as it gets; divide the second by

2

.

a + 2b = 5, \quad -a + b = -2

Smaller coefficients mean fewer arithmetic slips.

4Solve for $a$ and $b$

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

Substitute

a = 5 - 2b

into the second equation.

-(5 - 2b) + b = -2 \;\Rightarrow\; -5 + 3b = -2 \;\Rightarrow\; b = 1, \quad a = 5 - 2 \times 1 = 3

One substitution finishes the system.

Answer:

a=3

,

b=1

4 · Reviewdoes it hold up?

With $a=3$ , $b=1$ the lines are $3x+y=5$ and $-6x+y=-4$ . At $(1, 2)$ : $3+2=5$ and $-6+2=-4$ . Both pass through $(1, 2)$ , as do $5x+2y=9$ and $2x+5y=12$ .

Another way: Eliminate instead: adding $a+2b=5$ and $-a+b=-2$ gives $3b=3$ , so $b=1$ and $a=3$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Finding the intersection point of two lines and solving the system it produces for $a$ and $b$.

💡Takeaway. When several lines share a point, let the fully known lines find the point first.

Variant 8 medium

The four lines $3x+5y=-4$ , $ax+by=-10$ , $5x+3y=4$ , and $-ax-3by=18$ all meet at one point. Find the values of the constants $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four lines pass through one common point. Two of the lines contain the unknown constants $a$ and $b$ ; the other two are fully known. We need $a$ and $b$ .

Givens

The lines $3x+5y=-4$ , $ax+by=-10$ , $5x+3y=4$ , $-ax-3by=18$ .
All four lines meet at one point.

Unknowns

The constants $a$ and $b$ .

Constraints

The common point lies on every one of the four lines.

2 · Planchoose the strategy

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

Change focus from the unknown constants to the common point. The two lines with no constants already determine that point. Once it is known, plugging it into the other two lines gives a system whose unknowns are $a$ and $b$ .

3 · Execute4 carry out the plan

1Find the point from the two known lines

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

Solve

3x+5y=-4

and

5x+3y=4

. Adding them gives

8x+8y=0

, so

x+y=0

; subtracting gives

-2x+2y=-8

, so

x-y=4

.

x = 2, \quad y = -2

Two different lines meet in exactly one point, so this is the shared point of all four.

2Put the point into the lines with constants

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

The point

(2, -2)

is on

ax+by=-10

and on

-ax-3by=18

.

2a - 2b = -10, \quad -2a + 6b = 18

Now

a

and

b

are the unknowns of a new linear system.

3Simplify both equations

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

Divide the first by

2

and the second by

2

.

a - b = -5, \quad -a + 3b = 9

Smaller coefficients mean fewer arithmetic slips.

4Solve for $a$ and $b$

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

Substitute

a = -5 + b

into the second equation.

-(-5 + b) + 3b = 9 \;\Rightarrow\; 5 + 2b = 9 \;\Rightarrow\; b = 2, \quad a = -5 + 2 = -3

One substitution finishes the system.

Answer:

a=-3

,

b=2

4 · Reviewdoes it hold up?

With $a=-3$ , $b=2$ the lines are $-3x+2y=-10$ and $3x-6y=18$ . At $(2, -2)$ : $-6-4=-10$ and $6+12=18$ . Both pass through $(2, -2)$ , as do $3x+5y=-4$ and $5x+3y=4$ .

Another way: Eliminate instead: adding $a-b=-5$ and $-a+3b=9$ gives $2b=4$ , so $b=2$ and $a=-3$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Finding the intersection point of two lines and solving the system it produces for $a$ and $b$.

💡Takeaway. When several lines share a point, let the fully known lines find the point first.

Variant 9 hard

The four lines $4x+3y=26$ , $ax+by=14$ , $3x+4y=30$ , and $2ax-by=10$ all meet at one point. Find the values of the constants $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four lines pass through one common point. Two of the lines contain the unknown constants $a$ and $b$ ; the other two are fully known. We need $a$ and $b$ .

Givens

The lines $4x+3y=26$ , $ax+by=14$ , $3x+4y=30$ , $2ax-by=10$ .
All four lines meet at one point.

Unknowns

The constants $a$ and $b$ .

Constraints

The common point lies on every one of the four lines.

2 · Planchoose the strategy

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

Change focus from the unknown constants to the common point. The two lines with no constants already determine that point. Once it is known, plugging it into the other two lines gives a system whose unknowns are $a$ and $b$ .

3 · Execute4 carry out the plan

1Find the point from the two known lines

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

Solve

4x+3y=26

and

3x+4y=30

. Adding them gives

7x+7y=56

, so

x+y=8

; subtracting gives

x-y=-4

.

x = 2, \quad y = 6

Two different lines meet in exactly one point, so this is the shared point of all four.

2Put the point into the lines with constants

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

The point

(2, 6)

is on

ax+by=14

and on

2ax-by=10

.

2a + 6b = 14, \quad 4a - 6b = 10

Now

a

and

b

are the unknowns of a new linear system.

3Simplify both equations

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

Divide the first by

2

and the second by

2

.

a + 3b = 7, \quad 2a - 3b = 5

Smaller coefficients mean fewer arithmetic slips.

4Solve for $a$ and $b$

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

Substitute

a = 7 - 3b

into the second equation.

2(7 - 3b) - 3b = 5 \;\Rightarrow\; 14 - 9b = 5 \;\Rightarrow\; b = 1, \quad a = 7 - 3 \times 1 = 4

One substitution finishes the system.

Answer:

a=4

,

b=1

4 · Reviewdoes it hold up?

With $a=4$ , $b=1$ the lines are $4x+y=14$ and $8x-y=10$ . At $(2, 6)$ : $8+6=14$ and $16-6=10$ . Both pass through $(2, 6)$ , as do $4x+3y=26$ and $3x+4y=30$ .

Another way: Eliminate instead: doubling $a+3b=7$ gives $2a+6b=14$ ; subtracting $2a-3b=5$ gives $9b=9$ , so $b=1$ and $a=4$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Finding the intersection point of two lines and solving the system it produces for $a$ and $b$.

💡Takeaway. When several lines share a point, let the fully known lines find the point first.

Variant 10 hard

The four lines $3x+2y=3$ , $ax+by=6$ , $2x+3y=-3$ , and $-ax-3by=-30$ all meet at one point. Find the values of the constants $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four lines pass through one common point. Two of the lines contain the unknown constants $a$ and $b$ ; the other two are fully known. We need $a$ and $b$ .

Givens

The lines $3x+2y=3$ , $ax+by=6$ , $2x+3y=-3$ , $-ax-3by=-30$ .
All four lines meet at one point.

Unknowns

The constants $a$ and $b$ .

Constraints

The common point lies on every one of the four lines.

2 · Planchoose the strategy

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

Change focus from the unknown constants to the common point. The two lines with no constants already determine that point. Once it is known, plugging it into the other two lines gives a system whose unknowns are $a$ and $b$ .

3 · Execute4 carry out the plan

1Find the point from the two known lines

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

Solve

3x+2y=3

and

2x+3y=-3

. Adding them gives

5x+5y=0

, so

x+y=0

; subtracting gives

x-y=6

.

x = 3, \quad y = -3

Two different lines meet in exactly one point, so this is the shared point of all four.

2Put the point into the lines with constants

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

The point

(3, -3)

is on

ax+by=6

and on

-ax-3by=-30

.

3a - 3b = 6, \quad -3a + 9b = -30

Now

a

and

b

are the unknowns of a new linear system.

3Simplify both equations

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

Divide the first by

3

and the second by

3

.

a - b = 2, \quad -a + 3b = -10

Smaller coefficients mean fewer arithmetic slips.

4Solve for $a$ and $b$

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

Substitute

a = 2 + b

into the second equation.

-(2 + b) + 3b = -10 \;\Rightarrow\; -2 + 2b = -10 \;\Rightarrow\; b = -4, \quad a = 2 + (-4) = -2

One substitution finishes the system.

Answer:

a=-2

,

b=-4

4 · Reviewdoes it hold up?

With $a=-2$ , $b=-4$ the lines are $-2x-4y=6$ and $2x+12y=-30$ . At $(3, -3)$ : $-6+12=6$ and $6-36=-30$ . Both pass through $(3, -3)$ , as do $3x+2y=3$ and $2x+3y=-3$ .

Another way: Eliminate instead: adding $a-b=2$ and $-a+3b=-10$ gives $2b=-8$ , so $b=-4$ and $a=-2$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Finding the intersection point of two lines and solving the system it produces for $a$ and $b$.

💡Takeaway. When several lines share a point, let the fully known lines find the point first.

Variant 11 hard

The four lines $5x+2y=9$ , $ax+by=-6$ , $2x+5y=-9$ , and $-ax+3by=-42$ all meet at one point. Find the values of the constants $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four lines pass through one common point. Two of the lines contain the unknown constants $a$ and $b$ ; the other two are fully known. We need $a$ and $b$ .

Givens

The lines $5x+2y=9$ , $ax+by=-6$ , $2x+5y=-9$ , $-ax+3by=-42$ .
All four lines meet at one point.

Unknowns

The constants $a$ and $b$ .

Constraints

The common point lies on every one of the four lines.

2 · Planchoose the strategy

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

Change focus from the unknown constants to the common point. The two lines with no constants already determine that point. Once it is known, plugging it into the other two lines gives a system whose unknowns are $a$ and $b$ .

3 · Execute4 carry out the plan

1Find the point from the two known lines

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

Solve

5x+2y=9

and

2x+5y=-9

. Adding them gives

7x+7y=0

, so

x+y=0

; subtracting gives

3x-3y=18

, so

x-y=6

.

x = 3, \quad y = -3

Two different lines meet in exactly one point, so this is the shared point of all four.

2Put the point into the lines with constants

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

The point

(3, -3)

is on

ax+by=-6

and on

-ax+3by=-42

.

3a - 3b = -6, \quad -3a - 9b = -42

Now

a

and

b

are the unknowns of a new linear system.

3Simplify both equations

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

Divide the first by

3

and the second by

3

.

a - b = -2, \quad -a - 3b = -14

Smaller coefficients mean fewer arithmetic slips.

4Solve for $a$ and $b$

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

Substitute

a = -2 + b

into the second equation.

-(-2 + b) - 3b = -14 \;\Rightarrow\; 2 - 4b = -14 \;\Rightarrow\; b = 4, \quad a = -2 + 4 = 2

One substitution finishes the system.

Answer:

a=2

,

b=4

4 · Reviewdoes it hold up?

With $a=2$ , $b=4$ the lines are $2x+4y=-6$ and $-2x+12y=-42$ . At $(3, -3)$ : $6-12=-6$ and $-6-36=-42$ . Both pass through $(3, -3)$ , as do $5x+2y=9$ and $2x+5y=-9$ .

Another way: Eliminate instead: adding $a-b=-2$ and $-a-3b=-14$ gives $-4b=-16$ , so $b=4$ and $a=2$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Finding the intersection point of two lines and solving the system it produces for $a$ and $b$.

💡Takeaway. When several lines share a point, let the fully known lines find the point first.

Variant 12 hard

The four lines $3x+4y=30$ , $ax+by=-8$ , $4x+3y=26$ , and $2ax-3by=44$ all meet at one point. Find the values of the constants $a$ and $b$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Four lines pass through one common point. Two of the lines contain the unknown constants $a$ and $b$ ; the other two are fully known. We need $a$ and $b$ .

Givens

The lines $3x+4y=30$ , $ax+by=-8$ , $4x+3y=26$ , $2ax-3by=44$ .
All four lines meet at one point.

Unknowns

The constants $a$ and $b$ .

Constraints

The common point lies on every one of the four lines.

2 · Planchoose the strategy

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

Change focus from the unknown constants to the common point. The two lines with no constants already determine that point. Once it is known, plugging it into the other two lines gives a system whose unknowns are $a$ and $b$ .

3 · Execute4 carry out the plan

1Find the point from the two known lines

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

Solve

3x+4y=30

and

4x+3y=26

. Adding them gives

7x+7y=56

, so

x+y=8

; subtracting gives

-x+y=4

, so

x-y=-4

.

x = 2, \quad y = 6

Two different lines meet in exactly one point, so this is the shared point of all four.

2Put the point into the lines with constants

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

The point

(2, 6)

is on

ax+by=-8

and on

2ax-3by=44

.

2a + 6b = -8, \quad 4a - 18b = 44

Now

a

and

b

are the unknowns of a new linear system.

3Simplify both equations

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

Divide the first by

2

and the second by

2

.

a + 3b = -4, \quad 2a - 9b = 22

Smaller coefficients mean fewer arithmetic slips.

4Solve for $a$ and $b$

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

Substitute

a = -4 - 3b

into the second equation.

2(-4 - 3b) - 9b = 22 \;\Rightarrow\; -8 - 15b = 22 \;\Rightarrow\; b = -2, \quad a = -4 - 3 \times (-2) = 2

One substitution finishes the system.

Answer:

a=2

,

b=-2

4 · Reviewdoes it hold up?

With $a=2$ , $b=-2$ the lines are $2x-2y=-8$ and $4x+6y=44$ . At $(2, 6)$ : $4-12=-8$ and $8+36=44$ . Both pass through $(2, 6)$ , as do $3x+4y=30$ and $4x+3y=26$ .

Another way: Eliminate instead: doubling $a+3b=-4$ gives $2a+6b=-8$ ; subtracting $2a-9b=22$ gives $15b=-30$ , so $b=-2$ and $a=2$ .

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Finding the intersection point of two lines and solving the system it produces for $a$ and $b$.

💡Takeaway. When several lines share a point, let the fully known lines find the point first.

Find the common point from the known lines · 12 practice problems

Generated variants — 12

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