Solving Systems by Substitution

What is the first step in solving a system by Substitution?

• Make sure both equations are in standard form.
• Make sure at least one equation is solved for one variable.
• Make sure both equations are in slope-intercept form.
• Make sure both equations can be solved.

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

Solve this system of equations. 
y = 2x + 1
y = 4x - 1

• (1,3)
• (-1,-3)
• (-1,3)
• (3,1)

Tags

CCSS.HSA.REI.C.6

CCSS.8.EE.C.8B

y = -2x + 5
y = 2x - 11

• No Solution
• (0, 0)
• (3, -1)
• (4, -3)

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

Solve the system by substitution.
 
5x + 4y= −14
y =  −7x  −  15 

• (-2, -1)
• (1, -2)
• (-2, 1)
• (-1, -2)

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

Solve:

y = 2x -11

6x - 3y = -15

• (2, -4)

• (2, 4)

• No solution

• Infinite solutions

Tags

CCSS.HSA.REI.C.6

x + 2y = 2
x = -4y + 2

• (-3, 0)
• (0, 2)
• (-3, -2)
• (2, 0)

Tags

CCSS.HSA.REI.C.6

CCSS.8.EE.C.8B

x - 3y = -13
4x + 2y = 4

• (1, 4)
• (-1, -4)
• (-1, 4)
• (1, -4)

Tags

CCSS.HSA.REI.C.6

CCSS.8.EE.C.8B

y = -6x + 5
-2x + y = 5

• (-3, -6)
• (-6, 3)
• (0, 5)
• (-3, 5)

Tags

CCSS.HSA.REI.C.6

CCSS.8.EE.C.8B

y = -3x + 11
y = -6x - 13

• (-8, 35)
• (8, -13)
• (1, 8)
• (2, 13)

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

2x + y = 5

y = 2x - 11

• No Solution

• (0, 0)

• (3, -1)

• (4, -3)

Tags

CCSS.HSA.REI.C.6

CCSS.HSA.REI.D.10

CCSS.HSA.REI.D.11