Answer:
A
Step-by-step explanation:
Given :
2x + 4y = 14 ---------- eq 1
4x + y = 20 ---------- eq 2
if you multiply eq 2 by -4 on both sides, you get
-4 (4x + y = 20) = -4 (20)
-16x -4y = -80 --------- eq3
we can see that eq. 1 and eq 2 together forms the system of equations presented in option A, Hence A is equvalent to the orginal system of equations given in the question.
Answer:
Step-by-step explanation:
[tex]\left\{\begin{array}{ccc}2x+4y=14&(1)\\4x+y=20&(2)\end{array}\right\\\\\left\{\begin{array}{ccc}2x+4y=14&(1)\\4x+y=20&\text{multiply both sides by (-4)}\end{array}\right\\\left\{\begin{array}{ccc}2x+4y=14&(1)\\-16x-4y=-80&(2)\end{array}\right\to \boxed{A.}[/tex]
B.
[tex]\left\{\begin{array}{ccc}2x+4y=14&(1)\\4x+y=20&\text{change the signs}\end{array}\right\\\\\left\{\begin{array}{ccc}2x+4y=14&(1)\\-4x-y=-20&\text{it's different to (2)}\end{array}\right[/tex]
C.
[tex]\left\{\begin{array}{ccc}2x+4y=14&\text{multiply both sides by 2}\\4x+y=20&(2)\end{array}\right\\\left\{\begin{array}{ccc}4x+8y=28&\text{different to (1)}\\4x+y=20&(2)\end{array}\right[/tex]
D.
[tex]\left\{\begin{array}{ccc}2x+4y=14&\text{change the signs}\\4x+y=20&(2)\end{array}\right\\\left\{\begin{array}{ccc}-2x-4y=-14&\text{different to (1)}\\4x+y=20&(2)\end{array}\right\\\\A.[/tex]
