Question

1. What is the solution to the system of equations?

Answer 1

Answer:

Step-by-step explanation:

(1) 3x + 4y = 12

(2) x + 2y = 10

(2)*3: 3x+6y=30

by elimination (2)*3 - (1)

6y-4y=30-12

2y=18

y=9

x+2(9)=10

x=-8

rearrange (2): x=10-2y

substitute in (1): 3(10-2y)+4y=12

30-6y+4y=12

2y=30-12

y=9

x=10-2(9)

=-8

same ans either way

Answer 2

Answer:

Step-by-step explanation:

(a) In order to use the elimination method, multiply the second equation by 2.

It becomes 2x + 4y = 20.

Subtract that from the first equation: 3x + 4y = 12

3x + 4y - 2x - 4y = 12 - 20

x = -8

Put that back in the second equation: -8 + 2y = 10

2y = 18

y = 9

(b) In order to use the substitution method, subtract 2y from the second equation.

It becomes x + 2y - 2y = 10 - 2y

x = 10 - 2y

Substitute that into the first equation:

3(10 - 2y) + 4y = 12

30 - 6y + 4y = 12

30 - 2y = 12

2y = 18

y = 9

Put back into the equation for x:

x = 10 - 2(9)

= 10 - 18

= -8