Question I need to solve these two equations using either substitution or elimination 1/4x – 1/2 = 5 3x + 2/3y = -40

Answer: y = -15 x = -10 Step-by-step explanation: 1/4x – 1/2 = 5 (1) 3x + 2/3y = -40 (2) Multiply (1) by 4 1/4x – 1/2 = 5 1/4x * 4 – 1/2y * 4 = 5 * 4 4/4x – 4/2y = 20 x – 2y = 20 (3) Multiply (2) by 3 3x + 2/3y = -40 (2) 3x * 3 + 2/3y * 3 = -40 * 3 9x + 6/3y = -120 9x + 2y = -120 (4) x – 2y = 20 (3) 9x + 2y = -120 (4) Using elimination method, Add (3) and (4) 9x + x = -120 + 20 10x = -100 x = -100/10 x = -10 Substitute x = -10 into (3) x – 2y = 20 (3) -10 – 2y = 20 -2y = 20 + 10 -2y = 30 y = 30/-2 y = -15 y = -15 x = -10 Log in to Reply

Answer:

y = -15

x = -10

Step-by-step explanation:

1/4x – 1/2 = 5 (1)

3x + 2/3y = -40 (2)

Multiply (1) by 4

1/4x – 1/2 = 5

1/4x * 4 – 1/2y * 4 = 5 * 4

4/4x – 4/2y = 20

x – 2y = 20 (3)

Multiply (2) by 3

3x + 2/3y = -40 (2)

3x * 3 + 2/3y * 3 = -40 * 3

9x + 6/3y = -120

9x + 2y = -120 (4)

x – 2y = 20 (3)

9x + 2y = -120 (4)

Using elimination method, Add (3) and (4)

9x + x = -120 + 20

10x = -100

x = -100/10

x = -10

Substitute x = -10 into (3)

x – 2y = 20 (3)

-10 – 2y = 20

-2y = 20 + 10

-2y = 30

y = 30/-2

y = -15

y = -15

x = -10