Systems by Substitution Part 1

Solve this system of equations.

y = 2x + 1

y = 4x - 1

What is the solution? (Please type your answer as (x value, y value). For example if you solve and get x = 8 and y = 6 then you should put (8, 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

y = -2x + 18
y = 8

• (5, 8)
• (-5, 8)
• (2, 8)
• (-2, 8)

Tags

CCSS.8.EE.C.8B

CCSS.HSA.REI.C.6

y = -5x + 6

y = x - 6

• (2, - 4)

• (- 4 , 6)

• ( - 4 , 2)

• (6, 4)

y = - 3x - 5

y = - 4x - 5

• (4 - 5)

• ( - 5 , 1)

• (0, - 5)

• ( - 5 , 4)

y = - 2x - 1

Y = 4x + 17

• ( 3, - 3)

• ( - 3 , - 5 )

• ( - 3 , 5)

• ( 3 , 5)

y = - 7

y = 2x - 13

• (- 3 , - 7)

• (3 , - 7)

• ( - 3 , 7)

• ( 6 , 3)

y = x - 4

y = 2x - 11

What is the solution? (Please type your answer as (x value, y value). For example if you solve and get x = 8 and y = 6 then you should put (8, 6)

y = - 1

y = 4x - 1

• (0, - 7)

• (0, 7)

• (0, 8)

• (0, - 1)

y = 3x - 9

y = 3x - 4

• (6, - 5)

• infinitely many

• no solution

• (5, 6)