In Text Questions (Text Book Page No. 95,99,100,102,104,109,111,114,117) - Chapter 3 - Algebra - 8th Maths Guide Samacheer Kalvi Solutions

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Try These (Text Book Page No. 95)
Question $1 .$
$3 y+6$
Answer:
$3 y+6$
$3 y+6=3 \times y+2 \times 3$
Taking out the common factor 3 from each term we get $3(y+2)$
$\therefore 3 \mathrm{y}+6=3(\mathrm{y}+2)$
 

Question 2 .
$10 x^{2}+15 y^{2}$
Answer:
$10 x^{2}+15 y^{2}$
$10 x^{2}+15 y^{2}=(2 \times 5 \times x \times x)+(3 \times 5 \times y \times y)$
Taking out the common factor 5 we have
$10 x^{2}+15 y^{2}=5\left(2 x^{2}+3 y^{2}\right)$
 

Question $3 .$
$7 m(m-5)+1(5-m)$
Answer:
$7 m(m-5)+1(5-m)$
$7 \mathrm{~m}(\mathrm{~m}-5)+1(5-\mathrm{m})=7 \mathrm{~m}(\mathrm{~m}-5)+(-1)(-5+\mathrm{m})$
$=7 m(m-5)-1(m-5)$
Taking out the common binomial factor $(m-5)=(m-5)(7 m-1)$

Question $4 .$
$64-x^{2}$
Answer:
$64-x^{2}$
$64-x^{2}=82-x^{2}$
This is of the form $a^{2}-b^{2}$
Comparing with $a^{2}-b^{2}$ we have $a=8, b=x$
$a^{2}-b^{2}=(a+b)(a-b)$
$64-x^{2}=(8+x)(8-x)$

Question $5 .$
$x^{2}-3 x+2$
Answer:

$\begin{aligned}
&x^{2}-3 x+2=x^{2}-2 x-x+2 \\
&=x(x-2)-(x-2) \\
&=(x-2)(x-1)
\end{aligned}$

Question $6 .$
$y^{2}-4 y-32$
Answer:

$\begin{aligned}
&y^{2}-4 y-32=y 2-8 y+4 y-32 \\
&=y(y-8)+4(y-8) \\
&=(y-8)(y+4)
\end{aligned}$

Question $7 .$
$\mathrm{p}^{2}+2 \mathrm{p}-15$
Answer:

$\begin{aligned}
&\mathrm{p}^{2}+2 \mathrm{p}-15=\mathrm{p}^{2}+5 \mathrm{p}-3 \mathrm{p}-15 \\
&=\mathrm{p}(\mathrm{p}+5)-3(\mathrm{p}+5) \\
&=(\mathrm{p}+5)(\mathrm{p}-3)
\end{aligned}$

Question 8 .
$m^{2}+14 m+48
$Answer:
$\begin{aligned}
&m^{2}+14 m+48=m^{2}+8 m+6 m+48 \\
&=m(m+8)+6(m+8) \\
&=(m+6)(m+8)
\end{aligned}$

Question $9 .$
$x^{2}-x-90$
Answer:
$\begin{aligned}
&x^{2}-x-90=x^{2}-10 x+9 x-90 \\
&=x(x-10)+9(x-10) \\
&=(x+9)(x-10)
\end{aligned}$

Question $10 .$

$\begin{aligned}
&9 x^{2}-6 x-8 \\
&\text { Answer: } \\
&9 x^{2}-6 x-8=9 x^{2}-12 x+6 x-8 \\
&=3 x(3 x-4)+2(3 x-4) \\
&=(3 x+2)(3 x-4)
\end{aligned}$

Try These (Text Book Page No. 99)

Identify which among the following are linear equations.
(i) $2+x=19$
Answer:
Linear as degree of the variable $x$ is 1
(ii) $7 \mathrm{x}^{2}-5=3$
Answer:
not linear as highest degree of $x$ is 2
(iii) $4 \mathrm{p}^{3}=12$
Answer:
not linear as highest degree of $\mathrm{p}$ is 3
(iv) $6 m+2$
Answer:
Linear, but not an equation
(v) $n=10$
Answer:
Linear equation as degree of $n$ is 1
(vi) $7 \mathrm{k}-12=0$
Answer:
Linear equation as degree of $k$ is 1
(vii) $\frac{6 x}{8}+y=1$
Answer:
Linear equation as degree of $x$ \& $y$ is 1
(vii) $5+y=3 x$
Answer:
Linear equation as degree of $\mathrm{y}$ \& $\mathrm{x}$ is 1
(ix) $10 p+2 q=3$
Answer:
Linear equation as degree of $p$ \& $q$ is 1
(x) $x^{2}-2 x-4$
Answer:
not linear equation as highest degree of $x$ is 2

Think (Text Book Page No. 99)
(i) Is $t(t-5)=10$ a linear equation? Why?
Answer:
$\begin{aligned}
&t(t-5)=10 \\
&=t \times t-5 \times t=10
\end{aligned}$
$=t^{2}-5 t=10$
This is not a linear equations as the highest degree of the variable ' $t$ ' is 2
(ii) Is $x^{2}=2 x$, a linear equation? Why?
Answer:
$\begin{aligned}
&x^{2}=2 x \\
&=x^{2}-2 x=0
\end{aligned}$
This is not a linear equations as the highest degree of the variable ' $x$ ' is 2
 

Try These (Text Book Page No. 100)
Convert the following statements into linear equations:
Question $1 .$
On subtracting 8 from the product of 5 and a number, I get 32 .
Answer:
Convert to linear equations:
Given that on subtracting 8 from product of 5 and $a$, we get 32
$\begin{aligned}
&\therefore 5 \times x-8=32 \\
&\therefore 5 x-8=32
\end{aligned}$

Question $2 .$
The sum of three consecutive integers is 78 .
Answer:
Sum of 3 consecutive integers is 78
Let integer be bx
$\begin{aligned}
&\therefore \mathrm{x}+(\mathrm{x}+1)+(\mathrm{x}+2)=78 \\
&\therefore \mathrm{x}+\mathrm{x}+1+\mathrm{x}+2=78 \\
&\therefore 3 \mathrm{x}+3=78
\end{aligned}$
 

Question $3 .$
Peter had a Two hundred rupee note. After buying 7 copies of a book he was left with 60 .
Answer:
Let cost of one book be ' $x$ '
$\therefore$ Given that $200-7 \times x=60$
$\therefore 200-7 \mathrm{x}=60$

Question $4 .$
The base angles of an isosceles triangle are equal and the vertex angle measures $80^{\circ}$.
Answer:
Let base angles each be equal to $x$ \& vertex bottom angle is $80^{\circ}$. Applying triangle property, sum of all angles is $180^{\circ}$
$\begin{aligned}
&\therefore \mathrm{x}+\mathrm{x}+80=180^{\circ} \\
&\therefore 2 \mathrm{x}+80=180^{\circ}
\end{aligned}$

Question $5 .$
In a triangle $\mathrm{ABC}, \angle \mathrm{A}$ is 100 more than $\angle \mathrm{B}$. Also $\angle \mathrm{C}$ is three times $\angle \mathrm{A}$. Express the equation in terms of angle $\mathrm{B}$.

In a triangle $\mathrm{ABC}, \angle \mathrm{A}$ is 100 more than $\angle \mathrm{B}$. Also $\angle \mathrm{C}$ is three times $\angle \mathrm{A}$. Express the equation in terms of angle $B$.
Answer:
Let $\angle B=b$
Given $\angle \mathrm{A}=10^{\circ}+\angle \mathrm{B}=10+\mathrm{b}$
Also given that $\angle \mathrm{C}=3 \times \angle \mathrm{A}=3 \times(10+\mathrm{b})=30+3 \mathrm{~b}$
Sum of the angles $=180^{\circ}$
$\begin{aligned}
&\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ} \\
&10+\mathrm{b}+\mathrm{b}+30+3 \mathrm{~b}=180^{\circ} \\
&\therefore 5 \mathrm{~b}+40=180^{\circ}
\end{aligned}$

Think (Text Book Page No. 101)
Can you get more than one solution for a linear equation?
Answer:
Yes, we can get. Consider the below line or equation.
$x+y=5$
here, when $x=1, y=4$
$\begin{aligned}
&\text { when } x=2, y=3 \\
&x=3, y=2 \\
&x=4, y=1
\end{aligned}$
Hence, we get multiple solutions for the saine linear equation.

Try These (Text Book Page No. 101)
Identify which among the following are linear equations.
(i) $2+x=10$
Answer:
$\begin{aligned}
&2+x=10 \\
&\Rightarrow x=\frac{10}{2}=5
\end{aligned}$
(ii) $3+x=5$
Answer:
$\begin{aligned}
&3+x \Rightarrow 5 \\
&x=5-3=2
\end{aligned}$
(iii) $x-6=10$
Answer:
$\begin{aligned}
&x-6=10 \\
&x=10+6=16
\end{aligned}$

(iv) $3 x+5=2$
Answer:
$\begin{aligned}
&\Rightarrow 3 x+5=2 \\
&3 x=2-5=-3
\end{aligned}$
(v) $\frac{2 x}{7}=3$
Answer:
$\begin{aligned}
&\Rightarrow 2 \mathrm{x}=3 \times 7=21 \\
&\mathrm{x}=\frac{21}{2}
\end{aligned}$
(vi) $-2=4 m-6$
Answer:
$\begin{aligned}
&\Rightarrow-2 x=4 m-6 \\
&-2+6=4 m \\
&4=4 m \\
&m=\frac{4}{4}=1
\end{aligned}$
(vii) $4(3 \mathrm{x}-1)=80$
Answer:
$\begin{aligned}
&\Rightarrow 4(3 x-1)=80 \\
&12 x-4=80 \\
&12 x=80+4=84 \\
&x=\frac{84}{12}=7
\end{aligned}$

(viii) $3 x-8=7-2 x$
Answer:
$\begin{aligned}
&\Rightarrow 3 x-8=7-2 x \\
&3 x+2 x=7+8=15 \\
&5 x=15 \\
&x=\frac{15}{5}=3
\end{aligned}$
(ix) $7-y=3(5-y)$
Answer:
$\begin{aligned}
&\Rightarrow 7-y=3(5-y) \\
&7-y=15-3 y \\
&3 y-y=15-7 \\
&2 y=8 \\
&y=\frac{8}{2}=4
\end{aligned}$
(x) $4(1-2 y)-2(3-y)=0$
Answer:
$\begin{aligned}
&\Rightarrow 4(1-2 y)-2(3-y)=0 \\
&4-8 y-\dot{0}-2 y=0 \\
&-2-6 y=0 \\
&6 y=-2 \\
&y=\frac{-2}{6}=\frac{-1}{3}
\end{aligned}$

Think (Text Book Page No. 102)
Question $1 .$
"An equation is multiplied or divided by a non zero number on either side:" Will there be any change in the solution?
Answer:
Not be any change in the solution

Question $2 .$
"An equation is multiplied or divided by two different numbers on either side. What will happen to the equation?
Answer:
When an equation is multiplied or divided by 2 different numbers on either side, there will be a change in the equation \& accordingly, solution will also change.

Think (Text Book Page No. 104)
Suppose we take the second piece to be $\mathrm{x}$ and the first piece to be $(200-\mathrm{x})$, how will the steps vary ?
Will the answer be different?
Answer:
Let $2^{\text {nd }}$ piece be ' $x$ ' \& $1^{\text {st }}$ piece is $200-x$
Given that lst piece is $40 \mathrm{~cm}$ smaller than hence the other piece
$\begin{aligned}
&\therefore 200-\mathrm{x}=2 \times \mathrm{x}-40 \\
&200-x=2 x-40 \\
&\therefore 200+40=2 \mathrm{x}+\mathrm{x} \\
&240=3 \mathrm{x} \\
&\therefore \mathrm{x}=\frac{240}{3}=80 \\
&\therefore 1^{\text {st }} \text { piece }=200-\mathrm{x}=200-80=120 \mathrm{~cm} \\
&2^{\text {nd }} \text { piece }=\mathrm{x}=80 \mathrm{~cm}
\end{aligned}$
The answer will not change
 

Think (Text Book Page No. 109)
If instead of $(4,3)$, we write $(3,4)$ and tn to mark it, will it represent ' $M$ ' again?
Answer:
Let 3,4 be M. when we mark, we find that it is a different point and not ' $M$ '
 

Try These (Text Book Page No. 111)
Question $1 .$
Complete the table given below.

Answer:

Question $2 .$
Write the coordinates of the points marked in the following figure

Answer:
$\begin{aligned}
&\mathrm{A}-(-3,2) \\
&\mathrm{B}-(5,2) \\
&\mathrm{C}-(5,-3) \\
&\mathrm{D}-(-3,3) \\
&\mathrm{E}-(-1,4) \\
&\mathrm{F}-(1,2) \\
&\mathrm{G}-(7,4) \\
&\mathrm{H}-(0,2) \\
&\mathrm{I}-(0,3) \\
&\mathrm{J}-(-3,0) \\
&\mathrm{K}-(5,0) \\
&\mathrm{L}-(-1,0) \\
&\mathrm{M}-(-2,0) \\
&\mathrm{N}-(-2,-1) \\
&\mathrm{O}-(0,0) \\
&\mathrm{P}-(-1,-1) \\
&\mathrm{Q}-(1,-1) \\
&\mathrm{R}-(2,-1) \\
&\mathrm{S}-(0,-3) \\
&\mathrm{T}-(7,0) \\
&\mathrm{U}-(7,-2)
\end{aligned}$

Think (Text Book Page No. 114)
Which of the points $(5,-10)(0,5)(5,20)$ lie on the straight line $x=5$ ?
Answer:
All points on the line $\mathrm{X}=5$ will have $\mathrm{X}$-coordinate as 5 . Therefore, any point with $\mathrm{X}$-coordinate as 5 will lie on $X=5$ line. Hence the points $(5,-10) \&(5,20)$ will lie on $X=5$

Try These (Text Book Page No. 117)
Identify and correct the errors

Answer:

(ii)

Answer:

Answer: