Respuesta :
Answer:
C is the only one equivalent
Step-by-step explanation:
A. 5x − 3y = 6
6x − 2y = 8
Resolving the orginial system and this system we have
5x − 3y = 6 → 5x − 3y = 6 → -8y = - 4 → y=2 → x=0
(-5)(x + y = 2) -5x -5y = -10
A. 5x − 3y = 6 → 5x − 3y = 6 → -4x = -6 → x=3 → y=3
( -3/2)(6x − 2y = 8) - 9x + 3y = -12
A is not an equivalent system because the 2nd eq of each one have a different solution
B. 5x − 3y = 6
2x + 2y = 2
Resolving the orginial system and this system we have
5x − 3y = 6 → 5x − 3y = 6 → -8y = - 4 → y=2 → x=0
(-5)(x + y = 2) -5x -5y = -10
B 5x − 3y = 6 → 5x − 3y = 6 → 8x = 9 → x=9/8→ y= -1/8
(3/2)(2x + 2y = 2) 3x + 3y = 3
B is not an equivalent system because the 2nd eq of each one have a different solution
C. −5x − 3y = 6
x + y = 2
C is the same system of the original
D. 5x − 3y = 6
6x − 2y = 2
Resolving the orginial system and this system we have
5x − 3y = 6 → 5x − 3y = 6 → -8y = - 4 → y=2 → x=0
(-5)(x + y = 2) -5x -5y = -10
D 5x − 3y = 6 → 5x − 3y = 6 → 2x = 2 → x=1→ y= - 1/3
(-3/2)(6x − 2y = 2) - 3x + 3y = -3
D is not an equivalent system because the 2nd eq of each one have a different solution
Answer:
A.
Step-by-step explanation:
To find the equivalent system of equations, we first need to find the solutions of the given system, and then, see which option has the same solutions.
So, we solve the system of equations given by elimination, to do so, we multiply the second equation by 3, and then sum both of them:
[tex]\left \{ {{5x-3y=6} \atop {x+y=2}} \right.\\\left \{ {{5x-3y=6} \atop {3x+3y=6}} \right.\\8x=12\\x=\frac{12}{8}=\frac{3}{2}[/tex]
Now, we replace this value in one of the equation to find the other solution:
[tex]x+y=2\\\frac{3}{2}+y=2\\y=2-\frac{3}{2}\\y=\frac{4-3}{2}=\frac{1}{2}[/tex]
Therefore, the solution is [tex](\frac{3}{2};\frac{1}{2})[/tex]
Now, we test these values in each option to find the answer that matches.
Option A.
[tex]\left \{ {{5x-3y=6} \atop {6x-2y=8}} \right.\\\left \{ {{5\frac{3}{2} -3\frac{1}{2} =6} \atop {6\frac{3}{2} -2\frac{1}{2} =8}} \right.\\\left \{ {{\frac{15}{2}-\frac{3}{2}=6} \atop {9-1=8}} \right.\\\\\\\left \{ {{\frac{12}{2} =6} \atop {8=8}} \right.[/tex]
As you can see, the first option matches perfectly, the solutions of the given system are also the solutions of the Option A. Therefore they are equivalent systems of equations.