Solving Two Variable System of Equations

Solving Two Variable System of Equations

9th Grade

18 Qs

quiz-placeholder

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Solving Two Variable System of Equations

Solving Two Variable System of Equations

Assessment

Quiz

Mathematics

9th Grade

Hard

CCSS
8.EE.C.8B, HSA.REI.C.9, 8.EE.C.8A

+1

Standards-aligned

Created by

Anthony Clark

FREE Resource

18 questions

Show all answers

1.

DRAG AND DROP QUESTION

1 min • 2 pts

In this system

x+3y=9

4x-2y=-6

it'll be easiest to start by solving the first equation for x.

The result of doing so is x=​9-3y.

Fill in the blanks to plug that expression in and solve for y:

4(​9-3y)-2​y =-6

36-​12y-2y=​-6

36-​14y =-6

-14y=​ (a)  

y=​ (b)  

-42

3

30

-3

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

2.

REORDER QUESTION

1 min • 5 pts

Put the steps for solving by ELIMINATION in order:

Subtract the equations to eliminate one variable

Solve for the remaining variable

Multiply one or both equations by a number to get common but opposite coefficients

Make sure the equations are lined up

Plug that number into either original equation and solve for the other variable

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

3.

DRAG AND DROP QUESTION

1 min • 5 pts

In this system

x+3y=9

4x-2y=-6

we have figured out that x=9-3y, and y=3.

Complete the steps to solve for x:

x=9-3(​ (a)   )

x=9-​ (b)  

x=​ (c)  

The solution to the system is (​ (d)   ,​ (e)   ).

3

9

0

y

x

y=3

3y

-9

18

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

4.

DRAG AND DROP QUESTION

1 min • 3 pts

Consider the system

4x+3y=-1

5x+4y=1

Let's say I want to eliminate the y's. What is the least common multiple of 3 and 4?

LCM = ​ (a)  

Multiply each equation by a number so that the coefficient of y will be 12, with one positive and one negative.

​ (b)   (4x+3y=-1)

​ (c)   (5x+4y=1)

12

4

-3

3

5

1

-1

-12

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

5.

DRAG AND DROP QUESTION

1 min • 5 pts

Consider the system

4x+3y=-1

5x+4y=1

After we do this multiplication, what will the new equations be?

​ 4(4x+3y=-1) ---> ​ (a)   x +​ (b)   y=​ (c)  

​-3(5x+4y=1)​ ---> ​ (d)   x -​ (e)   y= -3

16

12

-4

-15

3

5

1

-1

-12

-3

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

6.

DRAG AND DROP QUESTION

1 min • 4 pts

Consider the system:

4x+3y=-1

5x+4y=1

Which we converted to:

16 x +​ 12y=​-4

-15x -​ 12y= -3

Combining the equations straight down gives us:

​ (a)   x+​ (b)   y=​ (c)  

Solving for x, we get x=​ (d)  

1

0

-7

3

5

-1

-12

-3

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

7.

DRAG AND DROP QUESTION

1 min • 5 pts

Consider the system:

4x+3y=-1

5x+4y=1

Since we now know that x=-7, solve for y:

4(​ (a)   )+3y=-1

​ (b)   +3y=-1

​ (c)   =​ (d)  

y=​ (e)  

-7

-28

3y

27

9

5

7

-12

3

y

Tags

CCSS.HSA.REI.C.9

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