Exercise-4.2
Maths - Exercise
Infinite Solutions
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.
Finding Solutions
Write four solutions for each of the following equations:
- 2x + y = 7
- πx + y = 9
- x = 4y
Solution:
- 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: \(x,\,y\)=(\(0,7),(1,5),(2,3),(3,1)\)
- 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: \(x,\,y\)=(\(0,9),(\frac{1}{\pi},8),(1,9-\pi),(\frac{9}{\pi},0)\)
- 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: \(x,\,y\)=(\(-4,-1),(0,0),(4,1),(8,2)\)
Verify Solutions
Check which of the following are solutions of the equation \(x – 2y = 4\) and which are not:
- (0, 2)
- (2, 0)
- (4, 0)
- \(\sqrt{2},\,4\sqrt{2}\)
- (1.1)
Solutions:
- 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 \(x – 2y = 4\)
- 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 \(x – 2y = 4\)
- 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 \(x – 2y = 4\)
- 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 \(x – 2y = 4\)
- 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 \(x – 2y = 4\)
Value of Constant
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
\[
\begin{aligned}
2\cdot 2 + 3\cdot1&=k\\
4+3&=k\\
\implies k&=7
\end{aligned}
\]
