Question

30 points. Help please

Consider this system of equations:

2x + 1/4y = 3 ( equation A )
2/3x - y = 6 ( equation B )

The expressions that give the value of y are

a) A - 3B
b) A - (3B/2)
c) 2A - 3B
d) (A/3) + B)

and

a) A + 3B
b) A+ (3B/2)
c) (A/3) - B
d) (A/3) - 2B

The solution for the given system is

a) (27/13, 60/13)
b) (-27/13, 60/13)
c) (27/13, -60/13)
d) (-27/13, -60/13)

Answer 1

A - 3B and (A/3) - B

solving :-

2x + 1/4 y = 3

2/3x - y = 6 Multiply this by 3

2x - 3y = 18 subtract this from first equation

0 + 13/4y = -15

y = -60/13

x = (3 - 1/4*-60/13) / 2 = 27/13

Explanation:

Answer 2

A - 3B and (A/3) - B

solving :-
2x + 1/4 y = 3
2/3x - y = 6 Multiply this by 3
2x - 3y = 18 subtract this from first equation
0 + 13/4y = -15
y = -60/13

x = (3 - 1/4*-60/13) / 2 = 27/13

Thats C