Question

level 2: use the following information to answer the next question.
three linear systems are given below.
i. $x + 2y = 0$ $2x + 4y = 0$
ii. $3x - y = 7$ $x + 4y = -3$
iii. $2x - 3y = 7$ $x + 5y = -3$

1. the values $x = 2$, and $y = -1$ are the solution for

a. i only b. i and ii c. iii only d. i and iii

Explanation:

Step1: Check System I

Substitute \(x = 2\), \(y=-1\) into \(x + 2y\): \(2+2\times(-1)=2 - 2 = 0\), which matches the first equation. Substitute into \(2x + 4y\): \(2\times2+4\times(-1)=4 - 4 = 0\), which matches the second equation. So \(x = 2\), \(y=-1\) is a solution for System I.

Step2: Check System II

Substitute \(x = 2\), \(y=-1\) into \(3x - y\): \(3\times2-(-1)=6 + 1 = 7\), which matches the first equation. Substitute into \(x + 4y\): \(2+4\times(-1)=2 - 4=-2
eq - 3\). So \(x = 2\), \(y=-1\) is not a solution for System II.

Step3: Check System III

Substitute \(x = 2\), \(y=-1\) into \(2x - 3y\): \(2\times2-3\times(-1)=4 + 3 = 7\), which matches the first equation. Substitute into \(x + 5y\): \(2+5\times(-1)=2 - 5=-3\), which matches the second equation. So \(x = 2\), \(y=-1\) is a solution for System III.

Answer:

D. I and III