Solving Systems by Substitution

• Infinite number of solutions

• (0, 7)

• (-5, -7)

• (5, 7)

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

• No Solution

• (2, 2)

• (-1, 2)

• (-1, -2)

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

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

TEKS.MATH.A.5C

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

• (1,3)
• (-1,-3)
• (-1,3)
• (3,1)

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

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

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

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

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

6x + y = -1
y = -3x - 4

• (3, 1)
• (1, 7)
• (-3, 1)
• (1, -7)

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

y = -3x + 11
5x + y = 21

• (1, -5)
• (1, 5)
• (-4, 5)
• (5, -4)

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

3x + y = 1
y = 4x + 1

• (0, -1)
• No Solution
• (0, 8)
• (0, 1)

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6