Solve by Substitution

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

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
-3y = -6x -15

• (-1.5,-4)
• (1.5,4)
• No solution (parallel lines)
• Infinitely many solutions

Tags

CCSS.8.EE.C.8B

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

y = -2x + 18
y = 8

• (5, 8)
• (-5, 8)
• (2, 8)
• (-2, 8)

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

3x + 2y = 8
-x + y = -11

• (-5,6)
• (6,-11)
• (-6,-5)
• (6,-5)

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

Solve for x and y
3x + 2y = 16
7x + y = 19

• (-2,5)
• (-2,-5)
• (2,-5)
• (2,5)

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

Solve for x and y.
3x+y=-5
6x+2y=10

• No Solution
• Infinite Number of Solutions
• (2,3)
• (6,-2)

Tags

CCSS.8.EE.C.8B

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

• (1.4, -0.3)
• (3.1, 5)
• (-2.5, 3.5)
• (6.9, 1.02)

Tags

CCSS.HSA.REI.C.9

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