Solving System of Equations Using Substitution

Solve the following systems of equations using substitution:

y = 3 - x

3y + x = 5

• No solution

• (1, 2)

• (3, 0)

• (2, 1)

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

Which answer has a correct first step for this system:

• The variable is already isolated, so we've already solved the system.

• The variable is already isolated for y, so we substitute it in the first equation and solve for x

  3x + 3(3x + 5) = 3

• Free sha vaca doo

• The variable is already isolated for y, so we substitute it in the second equation and solve for x

  y = 3(3x + 3y) = 3

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

Solve this system using substitution

y = 3x - 5

2x + 4y = -6

• (1, -2)

• (2, -2)

• (1, 2)

• Why have you done this?

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

Solve with substitution

x = -3y - 8

x - 2y = -3

• (5, -1)

• (-5, 1)

• (-5,-1)

• (1,-5)

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

Solve the system of equations.

• (-34, -17)

• (8, 4)

• (10, 20)

• No solution

• Infinite solutions

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

Solve the system of equations.

• (-6, 1)

• (-12, -12)

• (1, -6)

• No solution

• Infinite solutions

Tags

CCSS.8.EE.C.8B

What is the solution to this system?
2x + 3y = 4
y= 5x - 27

• (5,-2)

• (5, 2)

• (2,3)

• (3,2)

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

6x = y

3y - 4x = 42

• (3, 18)

• (8, 2)

• (2, 8)

• (1, 6)

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

x = 3y

2x - y =10

• (4, 5)

• (1, 1)

• (5, 8)

• (6, 2)

Tags

CCSS.HSA.REI.C.7

y = x + 1

2x - 3 = y

• (3, 5)

• (5, 4)

• (4, 3)

• (4, 5)