Which one of the following options is true, and why?
y = 3x + 5 has
(i) a unique solution,
(ii) only two solutions,
(iii) infinitely many solutions

Solution:
The correct option is (iii) infinitely many solutions.
The given equation \(y=3x+5\) is a linear equation in two variables \(x\) and \(y\) Such an equation represents a straight line when plotted on a Cartesian plane. The line passes through an infinite number of points, meaning:
For each real value of \(x\), there is exactly one corresponding value of \(y\) calculated as \(y=3x+5\).
Therefore, infinitely many ordered pairs\((x,y)\) satisfy the equation.

Write four solutions for each of the following equations:

1. 2x + y = 7
2. πx + y = 9
3. x = 4y

Solution:

1. To find the solutions, choose convenient values of \(x\) and compute corresponding values of \(y\) using \[ 2x + y = 7\\\\ \begin{array}{r|l} x&y\\\hline 0&7\\\hline 1&5\\\hline 2&3\\\hline 3&1 \end{array} \] Solutions: \(x,\,y\)=(\(0,7),(1,5),(2,3),(3,1)\)
2. ii. To find the solutions, choose convenient values of \(x\) and compute corresponding values of \(y\) using \[ πx + y = 9\\\\ \begin{array}{r|l} x&y\\\hline 0&9\\\hline \frac{1}{\pi}&8\\\hline 1&9-\pi\\\hline \frac{9}{\pi}&0\\\hline \end{array} \] Solutions: \(x,\,y\)=(\(0,9),(\frac{1}{\pi},8),(1,9-\pi),(\frac{9}{\pi},0)\)
3. iii. To find the solutions, choose convenient values of \(x\) and compute corresponding values of \(y\) using \[ x = 4y\\\\ \begin{array}{r|l} x&y\\\hline -4&-1\\\hline 0&0\\\hline 4&1\\\hline 8&2\\\hline \end{array} \] Solutions: \(x,\,y\)=(\(-4,-1),(0,0),(4,1),(8,2)\)

Check which of the following are solutions of the equation \(x – 2y = 4\) and which are not:

1. (0, 2)
2. (2, 0)
3. (4, 0)
4. \(\sqrt{2},\,4\sqrt{2}\)
5. (1.1)

Solutions:

1. Substitute values of \(x =0\) and \(y=2\) in \(x – 2y = 4\) \[\begin{aligned} x – 2y &= 4\\ 0-2\times 2&=4\\ 0-4&=4\\ -4&=4\implies \text{ not true} \end{aligned}\] hence, (0,2) is not a solution of \(x – 2y = 4\)
2. Substitute values of \(x =2\) and \(y=0\) in \(x – 2y = 4\) \[\begin{aligned} x – 2y &= 4\\ 2-2\times 0&=4\\ 2-0&=4\\ 2&=4\implies \text{ not true} \end{aligned}\] hence, (2,0) is not a solution of \(x – 2y = 4\)
3. Substitute values of \(x =4\) and \(y=0\) in \(x – 2y = 4\) \[\begin{aligned} x – 2y &= 4\\ 4-2\times 0&=4\\ 4-0&=4\\ 4&=4\implies \text{ true} \end{aligned}\] hence, (4,0) is a solution of \(x – 2y = 4\)
4. Substitute values of \(x =\sqrt{2}\) and \(y=4\sqrt{2}\) in \(x – 2y = 4\) \[\begin{aligned} x – 2y &= 4\\ \sqrt{2}-2\times 4\sqrt{2}&=4\\ \sqrt{2}-8\sqrt{2}&=4\\ \sqrt{2}(1-8)&=4\\ -7\sqrt{2}&=4\implies \text{ not true} \end{aligned}\] hence, \((\sqrt{2},4\sqrt{2})\) is not a solution of \(x – 2y = 4\)
5. Substitute values of \(x =1\) and \(y=1\) in \(x – 2y = 4\) \[\begin{aligned} x – 2y &= 4\\ 1-2\times 1&=4\\ 1-2&=4\\ -1&=4\implies \text{ not true} \end{aligned}\] hence, (1,1) is not a solution of \(x – 2y = 4\)

Find the value of \(k\), if \(x = 2,\; y = 1\) is a solution of the equation \(2x + 3y = k\).

Solution:
Find the value of \(k\), if \(x = 2, y = 1\) is a solution of the equation 2x + 3y = k. Given that \(x=2\) and \(y=1\) is a soultion of equation \(2x + 3y = k\) hence it should satitfy the eqauation \[ 2x + 3y = k\] Substituting the values of \(x\) and \(y\) in the equation \[ \begin{aligned} 2\cdot 2 + 3\cdot1&=k\\ 4+3&=k\\ \implies k&=7 \end{aligned} \]