Respuesta :
x + 2y = 5
x = 5 - 2y
2x + 4y = 3
2(5-2y) + 4y = 3
10 - 4y + 4y = 3
10 ≠ 3
5x + 2y = 3
5(5-2y) + 2y = 3
25 - 10y + 2y = 3
-8y = 3 - 25
-8y = -22
y = -22/-8
y = 2.75
6x + 12y = 30
6(5-2y) + 12y = 30
30 -12y + 12y = 30
30 = 30
3x + 4y = 8
3(5-2y) + 4y = 8
15 - 6y + 4y = 8
-2y = 8 - 15
-2y = -7
y = -7/-2
y = 3.5
An inconsistent system is a system of equation that has NO SOLUTIONS.
2x + 4y = 3 is the equation to be paired with x + 2y = 50 to create an inconsistent system.
x = 5 - 2y
2x + 4y = 3
2(5-2y) + 4y = 3
10 - 4y + 4y = 3
10 ≠ 3
5x + 2y = 3
5(5-2y) + 2y = 3
25 - 10y + 2y = 3
-8y = 3 - 25
-8y = -22
y = -22/-8
y = 2.75
6x + 12y = 30
6(5-2y) + 12y = 30
30 -12y + 12y = 30
30 = 30
3x + 4y = 8
3(5-2y) + 4y = 8
15 - 6y + 4y = 8
-2y = 8 - 15
-2y = -7
y = -7/-2
y = 3.5
An inconsistent system is a system of equation that has NO SOLUTIONS.
2x + 4y = 3 is the equation to be paired with x + 2y = 50 to create an inconsistent system.
Answer:
only (1) and (3) forms a pair of inconsistent system with x + 2y = 5.
Step-by-step explanation:
Given : a linear equation x + 2y = 5
We have to find a pair of equation that can pair with x + 2y = 5 to create an inconsistent system.
Consider a system of equation [tex]a_1x+b_1y+c_1=0 \\\\a_2x+b_2y+c_2=0[/tex]
For the system to have no solution the condition is,
[tex]\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}[/tex]
We just have to check for the above condition, those pair whose x and y coefficient are in same ratio and constant coefficient are different, those pair will form an inconsistent system.
Thus, only (1) and (3) are multiple of given equation, thus forms a pair of inconsistent system.