Exercise 3.10 - Chapter 3 Algebra 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.10$
Question $1 .$
Draw the graph for the following
(i) $\mathrm{y}=2 \mathrm{x}$
(ii) $\mathrm{y}=4 \mathrm{x}-1$
(iii) $\mathrm{y}=\left(\frac{3}{2}\right) \mathrm{x}+3$
(iv) $3 x+2 y=14$
Solution:
(i) Put $x=-1, y=2 \times-1=-2$
When $x=0, y=2 \times 0=0$
When $x=1, y=2 \times 1=2$

The points $(x, y)$ to be plotted: $(-1,-2),(0,0),(1,2)$
(ii) When $\mathrm{x}=-1 \Rightarrow \mathrm{y}=4(-1)-1$
$\begin{aligned}
&y=-4-1=-5 \\
&x=0 \Rightarrow y=4 \times 0-1=-1 \\
&x=1 \Rightarrow y=4 \times 1-1=3
\end{aligned}$

The points (x, y) to be plotted: (-1, -5), (0, -1), (1, 3)

(iii) $x$
$\begin{aligned}
&x=-2 \Rightarrow y=\left(\frac{3}{2}\right)(-2)+3=-3+3=0 \\
&x=0 \Rightarrow y=\left(\frac{3}{2}\right)(0)+3=0+3=3 \\
&x=2 \Rightarrow y=\left(\frac{3}{2}\right)(2)+3=3+3=6
\end{aligned}$

The points to be plotted: (-2, 0), (0, 3), (2, 6)

(iv)
$\begin{aligned}
&\begin{aligned}
2 y &=-3 x+14 \\
y &=\frac{-3 x+14}{2} \\
\Rightarrow \quad \quad y &=\left(\frac{-3}{2}\right) x+7
\end{aligned} \\
&\text { When } x=-2, \quad y=\left(\frac{-3}{2}\right)(-2)+7=3+7=10 \\
&\text { When } x=0, \quad y=\left(\frac{-3}{2}\right)(0)+7=7 \\
&\text { When } x=2, \quad y=-3+7=4
\end{aligned}$

The points to be plotted: $(-2,10),(0,7),(2,4)$
 

Question 2.

Solve graphically
(i) $x+y=7 ; x-y=3$
(ii) $3 x+2 y=4 ; 9 x+6 y-12=0$
(iii) $\frac{x}{2}+\frac{y}{4}=1 ; \frac{x}{2}+\frac{y}{4}=2$
(iv) $x-y=0 ; y+3=0$
(v) $y=2 x+1 ; y+3 x-6=0$
(vi) $x=-3, y=3$
Solution:
(i) We can find $x$ and $y$ intericepts and thus of the two points on the lines (1), (2) $x+y=7 \ldots \ldots . .(1), x-y=3 \ldots \ldots \ldots . .$ (2)
To draw the graph of (1)
Put $x=0$ in (1)
$0+\mathrm{y}=7 \Rightarrow \mathrm{y}=7$
Thus $\mathrm{A}(0,7)$ is a point on the line
Put $\mathrm{y}=0$ in (1)
$x+0=7 \Rightarrow x=7$
Thus $B(7,0)$ is another point on the line Plot $\mathrm{A}$ and $\mathrm{B}$. Join them to produce the line (1). To draw the graph of $(2)$, we can adopt the same procedure.

When $x=0,(2) \Rightarrow x-y=3$
$0-y=3 \Rightarrow y=-3$
$P(0,-3)$ is a point on the line.
Put $y=0$ in $(2) ; x-0=3$
$\mathrm{x}=3$
$\therefore \mathrm{Q}(3,0)$ is another point on the line (2)
Plot P, Q
1 The point of intersection $(5,2)$ of lines (1), (2) is a solution
(ii) $3 x+2 y=4 \ldots \ldots$ (1)
$9 x+6 y=12 \ldots \ldots . . .(2)$
To draw the graph of (1)
Put $x=0$ in $(1) \Rightarrow 3(0)+2 y=4$
$2 y=4$
$y=2$
$\therefore \mathrm{A}(0,2)$ is a point on the line (1)
Put $y=0$ in $(1) \Rightarrow 3 x+2(0)=4$
$3 x=4$
$\mathrm{x}=\frac{4}{3}=1.3$
$\therefore \mathrm{B}(1.3,0)$ is another point on the line (1)
Plot the points A, B. Join them to produce the line (1)
To draw the graph of $(2)$
Put $x=0$ in $(2) \Rightarrow 9(0)+6 y=12$

$\begin{aligned}
&6 y=12 \\
&y=2 \\
&\therefore P(0,2) \text { is a point on the line }(2) \\
&\text { Put } y=0 \text { in }(2) \Rightarrow 9 x+6(0)=12 \\
&9 x=12 \\
&x=\frac{12}{9}=\frac{4}{3} \\
&x=1.3
\end{aligned}$

$Q(1.3,0)$ is another point on the line (2)
Plot P, Q. Join them to produce the line (2).
The point of intersection of the lines (1), (2) is a solution. $[\mathrm{A}, \mathrm{B}],[\mathrm{P}, \mathrm{Q}]$ represent the same line.
$\therefore$ All the points on one line are also on the other.
This means we have an infinite number of solutions.

(iii) $\frac{x}{2}+\frac{y}{4}=1$
$\frac{x}{2}+\frac{y}{4}=2$.
$\begin{aligned}(1) \Rightarrow \frac{x}{2}+& \frac{y}{4}=1 \Rightarrow \frac{2 x+y}{4}=1 \\ & \Rightarrow 2 x+y=4 \end{aligned}$
$\begin{aligned} \Rightarrow 2 x+y &=4 \\ y &=-2 x+4 \end{aligned}$
$\therefore \quad(1) \Rightarrow y=-2 x+4$
(2) $\Rightarrow \frac{x}{2}+\frac{y}{4}=2$
$\frac{2 x+y}{4}=2 \Rightarrow 2 x+y=8$
(2) $\Rightarrow y=-2 x+8$
$\therefore$ Comparing (1), (2) we can conclude their slopes are equal
$\therefore$ The lines are parallel and will not meet at any point and hence no solution exists.
Let us draw the graphs of (1) \& (2)
(1) $\Rightarrow 2 x+y=4$, Put $x=0$ in (1) $\Rightarrow y=4$
$\therefore \mathrm{A}(0,4)$ is a point on (1)
Put $y=0$ in (1) $\Rightarrow 2 \mathrm{x}=4$
$x=2$
$\therefore \mathrm{B}(2,0)$ is another point on (1)
Plot A, B ; Join them to produce the line (1)
(2) $\Rightarrow 2 x+y=8$, Put $x=0$ in (2)
$2(0)+y=8$
$\mathrm{y}=8, \mathrm{P}(0,8)$ is a point
Put $y=0$ in $(2) \Rightarrow 2 x+0=8$

$2(0)+y=8$
$y=8, P(0,8)$ is a point
Put $y=0$ in $(2) \Rightarrow 2 x+0=8$
$2 x=8$
$x=4, Q(4,0)$ is another point on the line (2)

Plot $P, Q:$ Join them to produce the line (2)
(iv) $x-y=0$(1) $\Rightarrow-y=-x \Rightarrow y=x$

Put $x=0$ in (1), $0-y=0$
$-\mathrm{y}=0 \Rightarrow \mathrm{y}=0$
A $(0,0)$ is a point on the line (1)
Put $\mathrm{y}=0$ in $(1) \Rightarrow \mathrm{x}-0=0 \Rightarrow \mathrm{x}=0$
$B(-3,-3)$ is also the same point as $A$
(2) $\Rightarrow y+3=0$
$y=-3$
$\therefore$ from (1) $\mathrm{y}=-3=\mathrm{x}$
$\therefore \mathrm{B}(-3,-3)$ is the solution

(v) y = 2x + 1 …………… (1)

The points to be plotted $(-1,9),(0,6),(1,3)$
The points to be plotted $(-1,-1),(0,1),(1,3)$

The points of intersection $(1,3)$ of the lines $(1)$ and $(2)$ is a solution. The solution is the point that is common to both the lines.
$\therefore$ The solution is as $\mathrm{x}=1, \mathrm{y}=3$.
(vi) The point of intersection $(-3,3)$ is a solution.

$x+y=7 \ldots \ldots . .(1), x-y=3$
To draw the graph of (1)
Put $x=0$ in (1) $0+y=7 \Rightarrow y=7$
Thus $\mathrm{A}(0,7)$ is a point on the line
Put $y=0$ in (1)
$x+0=7 \Rightarrow x=7$
Thus $B(7,0)$ is another point on the line
Plot $\mathrm{A}$ and $\mathrm{B}$. Join them to produce the line (1).
To draw the graph of $(2)$, we can adopt the same procedure.
When $x=0, \ldots \ldots . .(2) \Rightarrow x-y=3$
$0-\mathrm{y}=3 \Rightarrow \mathrm{y}=-3$
$P(0,-3)$ is a point on the line.
Put $y=0$ in $(2) ; x-0=3$
$x=3$
$\therefore \mathrm{Q}(3,0)$ is another point on the line (2) Plot $\mathrm{P}, \mathrm{Q}$

Question $3 .$
Two cars are 100 miles apart. If they drive towards each other they will meet in 1 hour. If they drive in the same direction they will meet in 2 hours. Find their speed by using graphical method.
Solution:
Let $x, y$ be the speed of the two cars. If the two cars travel towards each other they will meet in $1 \mathrm{hr}$. The distance between them $\mathrm{d}=100 ; \frac{d}{s}=\mathrm{t}$
i.e., $\frac{100}{x+y}=1 \Rightarrow \mathrm{x}+\mathrm{y}=100 \ldots \ldots \ldots$ (1)
If the two cars travel in the same direction they will meet in $2 \mathrm{hrs}$. $x+y=\frac{100}{2} \Rightarrow x-y=50 \ldots . . . \ldots . . .$ (2)
(1) $\Rightarrow x+y=100$

The points to be plotted $(0,100),(100,0)$
(2) $\Rightarrow x-y=50 \Rightarrow-y=-x+50 \Rightarrow y=x-50$

The points to be plotted
$(-100,-50),(0,-50)$

$x+y=10 \ldots \ldots \ldots \ldots \ldots(1)$
Put $\mathrm{x}=0$ in (1), then $0+\mathrm{y}=100 \Rightarrow \mathrm{y}=100$
$\mathrm{A}(0,100)$ is a point on (1)
Put $y=0$ in (1), then $x+0=100 \Rightarrow x=100$
$B(100,0)$ is another point on (1)
Plot A \& B. Join them to produce the line (1)
Similarly by $x-y=50$
Put $x=0$ in (2), then $0-y=50 \Rightarrow y=-50$
$P(0,-50)$ is a point on (2)
Put $y=0$ in (2), then $x-0=50 \Rightarrow x=50$
$Q(50,0)$ is another point on (2)

Plot P \& Q. Join them to produce the line (2)
The point of intersection $(75,25)$ of the two lines $(1) \&(2)$ is the solution. $\therefore$ The solution i.e., the speed of the two cars $x$ and $y$ is given by $x=75 \mathrm{~km}$ and $y=25$ $\mathrm{km}$