Final answer:
To find three solutions for the given linear equations, one can assign arbitrary values to one variable and solve for the other using algebraic methods. Solutions were provided for the first two equations as clear examples.
Step-by-step explanation:
The student has asked for three different solutions for each of the linear equations provided. To find these solutions, we can choose arbitrary values for one variable and then solve for the other variable using basic algebraic methods. For the sake of clear examples, we will find solutions for the first two equations.
Equation (i) 3x + 4y = 7
Solution 1: Let x = 1, then 3(1) + 4y = 7 → 4y = 4 → y = 1.
Solution 2: Let x = 0, then 3(0) + 4y = 7 → y = 7/4.
Solution 3: Let x = -1, then 3(-1) + 4y = 7 → -3 + 4y = 7 → 4y = 10 → y = 2.5.
Equation (ii) y = 6x
Solution 1: Let x = 0, then y = 6(0) = 0.
Solution 2: Let x = 1, then y = 6(1) = 6.
Solution 3: Let x = -1, then y = 6(-1) = -6.