Exercise 3.1 - Chapter 3 Algebra 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Ex $3.1$ : Chapter 3 - Algebra - 10th Maths Guide Samacheer Kalvi Solutions
Question $1 .$
Solve the following system of linear equations in three variables
(i) $x+y+z=5 ; 2 x-y+z=9 ; x-2 y+3 z=16$
(ii) $\frac{1}{x}-\frac{2}{y}+4=0 ; \frac{1}{y}-\frac{1}{z}+1=0 ; \frac{2}{z}+\frac{3}{x}=14$
(iii) $\mathrm{x}+20=\frac{3 y}{2}+10=2 \mathrm{z}+5=110-(\mathrm{y}+\mathrm{z})$
Solutions:
(i) $x+y+z=5$ ... (1)
$2 x-y+z=9$ ....(2)
$2 x-y+z=9$ $x-2 y+3 z=16$ ...(3)

Substitute $z=4$ in (4)
$\begin{aligned}
&3 x+2(4)=14 \\
&3 x+8=14 \\
&3 x=6 \\
&x=2
\end{aligned}$
Substitute $x=2, z=4$ in (1)
$\begin{aligned}
&2+y+4=5 \Rightarrow y=-1 \\
&x=2, y=-1, z=4
\end{aligned}$

$\begin{aligned}
&\text { (ii) } \frac{1}{x}-\frac{2}{y}+4=0 \\
&\frac{1}{y}-\frac{1}{z}+1=0 \\
&\frac{2}{z}+\frac{3}{x}=14 \\
&\text { Put } \frac{1}{x}=a \\
&\frac{1}{y}=\mathrm{b} \quad \frac{1}{z}=\mathrm{c} \text { in }(1),(2) \&(3) \\
&\mathrm{a}-2 \mathrm{~b}+4=0 \Rightarrow \mathrm{a}-2 \mathrm{~b}=-4 \ldots \ldots \ldots \ldots . \text { (1) } \\
&\mathrm{b}-\mathrm{c}+1=0 \Rightarrow \mathrm{b}-\mathrm{c}=-1 \ldots \ldots \ldots \ldots \text { (2) }
\end{aligned}$

2c + 3a = 14 ⇒ 2c + 3a = 14 …………. (3)

Substitute $a=2$ in (1), we get
$\begin{aligned}
2-2 b &=-4 \\
-2 b &=-6 \\
b &=3
\end{aligned}$
Substitute $b=3$ in (2), we get
$\begin{aligned}
&3-c=-1 \\
&\begin{aligned}
-c &=-4 \\
\Rightarrow &=4 \\
a=\frac{1}{x}=2 \Rightarrow x=\frac{1}{2}
\end{aligned} \\
&b=\frac{1}{y}=3 \Rightarrow y=\frac{1}{3} \\
&c=\frac{1}{z}=4 \Rightarrow z=\frac{1}{4} \\
&\text { (iii) } x+20=\frac{3 y}{2}+10=2 \mathrm{z}+5=110-(\mathrm{y}+\mathrm{z}) \\
&\mathrm{x}=\frac{3 y}{2}-10 \ldots \ldots \ldots \ldots(\mathrm{l}) \\
&2 \mathrm{z}+5=110-(\mathrm{y}+\mathrm{z}) \\
&2 \mathrm{z}=105-\mathrm{y}-\mathrm{z} \\
&\mathrm{y}=105-3 \mathrm{z} \ldots \ldots \ldots \ldots(\mathrm{a}) \\
&\text { Substitute }(2) \text { in }(1), \mathrm{x}=\frac{315}{2}-\frac{9 z}{2}-10 \\
&=2 \mathrm{z}+5-20 \\
&\therefore 315-9 \mathrm{z}-20=4 \mathrm{z}-30 \\
&x=\frac{1}{2}, y=\frac{1}{3}, z=\frac{1}{4}
\end{aligned}$

$13 \mathrm{z}=315-20+30$ $=325$ $z=\frac{325}{13}=25$ $x+20=2 z+5$ $x+20=50+5$ $x=35$ Substitute $z=25$ in $(2)$ $y=105-3 z=105-75=30$ $y=30$ $x=35, y=30, z=25$
The system has unique solutions.

Question 2.
Discuss the nature of solutions of the following system of equations
(i) $x+2 y-z=6 ;-3 x-2 y+5 z=-12 ; x-2 z=3$
(ii) $2 \mathrm{y}+\mathrm{z}=3(-\mathrm{x}+1) ;-\mathrm{x}+3 \mathrm{y}-\mathrm{z}=-43 \mathrm{x}+2 \mathrm{y}+\mathrm{z}=-\frac{1}{2}$
(iii) $\frac{y+z}{4}=\frac{z+x}{3}=\frac{x+y}{2} ; \mathrm{x}+\mathrm{y}+\mathrm{z}=27$
Solution:
(i) $x+2 y-z=6$ (1)
$-3 x-2 y+5 z=-12$ (2) 
$x-2 z=3$ (3)

We see that the system has an infinite number of solutions.
(ii) $2 y+z=3(-x+1)$;
$-x+3 y-z=-4$ $3 x+2 y+z=-\frac{1}{2}$ $2 y+z+3 x=3 \Rightarrow 3 x+2 y+z=3$ $-x+3 y-z=-4 \ldots \ldots \ldots \ldots .(2)$ $3 x+2 y+z=-\frac{1}{2} \ldots \ldots \ldots \ldots \ldots \ldots .$ (3)

This is a contradiction. This means the system is inconsistent and has no solutions

Sub. $x=3$ in $(4) \Rightarrow 5(3)-z=0$
$15-z=0$ $-z=-15$ $z=15$ Sub, $x=3, z=15$ in $(3)$ $x+y+z=27$ $3+y+15=27$ $y=27-18=9$ $x=3, y=9, z=15$
$\therefore$ The system has unique solutions.
 

Question $3 .$
Vani, her father and her grand father have an average age of 53 . One-half of her grand father's age plus one-third of her father's age plus one fourth of Vani's age is 65 . Four years ago if Vani's
grandfather was four times as old as Vani then how old are they all now?
Solution:
Let Vani's age be $x$
Let Vani's father's age be $y$
Let Vani's grand father's age be $z$.
$\begin{aligned}
&\frac{x+y+z}{3}=53 \Rightarrow x+y+z=159 \\
&\frac{x}{4}+\frac{y}{3}+\frac{z}{2}=65 \Rightarrow 3 x+4 y+6 z=780
\end{aligned}$

Sub, $z=84$ in (3), we get
$4 x-84=12$ $4 x=96$ $x=24$ Sub, $x=24, z=84$ in $(1)$ we get $24+y+84=159$ $y=159-108$ $=51$ $\therefore$ Vani's age $=24$ years Her father's age $=51$ years Her grand father's age $=84$ years.
$\therefore$ Vani's age $=24$ years
Her father's age $=51$ years
Her grand father's age $=84$ years.

Question $4 .$
The sum of the digits of a three-digit number is 11 . If the digits are reversed, the new number is 46 more than five times the former number. If the hundreds digit plus twice the tens digit is equal to
the units digit, then find the original three digit number?
Solution:
Let the number be $100 \mathrm{x}+10 \mathrm{y}+\mathrm{z}$.
Reversed number be $100 z+10 y+x$.
$\begin{aligned}
&x+y+z=11 \ldots \ldots \ldots \ldots .(1) \\
&100 z+10 y+x=5(100 x+10 y+z)+46 \\
&100 z+10 y+x=500 x+50 y+5 z+46 \\
&499 x+40 y-95 z-46 \ldots \ldots \ldots \ldots .(2) \\
&x+2 y=z \\
&x+2 y-z=0
\end{aligned}$

Sub. $x=1$ in (4)
$\begin{aligned}
2 \times 1+3 y &=11 \\
3 y &=9 \\
y &=3
\end{aligned}$
Sub, $x=1, y=3$ in (1)
$\begin{aligned}
1+3+z &=11 \\
z &=7
\end{aligned}$
$\therefore$ The number is $x y z=137$
 

Question $5 .$
There are 12 pieces of five, ten and twenty rupee currencies whose total value is $\square 105$. When first 2 sorts are interchanged in their numbers its value will be increased by $\square 20$. Find the number of
currencies in each sort.
Solution:
Let $x, y$ and $z$ be number of currency pieces of $5,10,20$ rupees
$\begin{aligned}
&x+y+z=12 \ldots \ldots \ldots(1) \\
&5 x+10 y+20 z=105 \ldots \ldots \\
&10 x+5 y+20 z=125 \ldots \ldots .
\end{aligned}$

Sub, $z=2$ in (5), we get $15 y+20 \times 2=85$
$15 y=45$
$y=3$
Sub; $y=3, z=2$ in (1)
$x+y+z=12$
$\mathrm{x}=7$
$\therefore$ The solutions are
the number of $\square 5$ are 7
the number of $\square 10$ are 3
the number of $\square 20$ are 2