NCERT Solutions: Class 9 Maths Chapter 2 - Introduction to Linear Polynomials

The degree of a polynomial is the highest power of the variable with non-zero coefficient. Thus $2x^2-5x+3$ has highest power $2$, $y^3+2y-1$ has highest power $3$, the non-zero constant $-9$ has degree $0$, and $4z-3$ has highest power $1$.

Each example has a non-zero term with the required highest power: $x+2$ has highest power $1$, $x^2+3x+1$ has highest power $2$, and $x^3-x+5$ has highest power $3$.

There is no $z$ term in $4z^3+5z^2-11$. We can write it as $4z^3+5z^2+0z-11$, so the coefficient of $z$ is $0$.

Substitute each value of $x$ in $5x-3$: for $x=0$, $5(0)-3=-3$; for $x=-1$, $5(-1)-3=-8$; for $x=2$, $5(2)-3=7$.

For $s=0$, $7(0)^2-4(0)+6=6$. For $s=-3$, $7(9)-4(-3)+6=63+12+6=81$. For $s=4$, $7(16)-4(4)+6=112-16+6=102$.

Let Salil's present age be $x$ years. His mother's present age is $3x$ years. After 5 years, their ages will be $x+5$ and $3x+5$. So $(x+5)+(3x+5)=70$, hence $4x+10=70$, $4x=60$, and $x=15$. Therefore the mother's age is $3x=45$ years.

Let the integers be $2x$ and $5x$. Their difference is $5x-2x=3x=63$, so $x=21$. Hence the integers are $2x=42$ and $5x=105$.

Let the number of five-rupee coins be $x$. Then the number of two-rupee coins is $3x$. The total value is $5x+2(3x)=88$, so $11x=88$ and $x=8$. Thus she has $8$ five-rupee coins and $3x=24$ two-rupee coins.

Let the shorter piece be $x$ feet. Then the longer piece is $4x$ feet. Since the total length is $300$ feet, $x+4x=300$, so $5x=300$ and $x=60$. The longer piece is $4x=240$ feet.

Let the width be $w$ cm. Then the length is $2w+3$ cm. The perimeter is $2(l+w)=24$, so $2((2w+3)+w)=24$. Thus $6w+6=24$, $6w=18$, and $w=3$. The length is $2(3)+3=9$ cm.

She starts with `500 and adds `150 every month. After $n$ months, the amount is $500+150n$. For $n=2$, this is $500+300=800$; for $n=3$, it is $950$; for $n=4$, it is $1100$.

The rally loses 9 members each hour. So after $n$ hours, $9n$ members have dropped out and $120-9n$ remain. Substituting $n=1,2,3$ gives $111,102,93$.

Area of a rectangle is length times breadth. Since the length is $13$ cm, $A=13b$. For $b=12$, $A=156$; for $b=10$, $A=130$; for $b=8$, $A=104$.

Volume of a rectangular box is length times breadth times height. Here $V=7\times 11\times h=77h$. For $h=5,9,13$, the volumes are $385$, $693$ and $1001\text{ cm}^3$.

In $d$ days Sarita reads $20d$ pages. Pages left $=500-20d$. For $d=15$, pages left $=500-20(15)=500-300=200$.

The starting height is $1.75$ feet. After $t$ months, the increase is $0.5t$ feet, so $h=1.75+0.5t$. For $t=7$, $h=1.75+3.5=5.25$ feet. The table follows by substituting $t=0$ to $10$.

The initial value is `10000 and the yearly decrease is `800. After $t$ years, $v=10000-800t$. For $t=3$, $v=10000-2400=7600$. Substituting $t=0$ to $8$ gives the table.

The population starts at $750$ and grows by $50$ every year. Hence $P=750+50t$. For $t=6$, $P=750+300=1050$. The table is obtained by adding $50$ for each next year.

After $x$ days, the total reduction is `$(15x)$, so $b(x)=600-15x$. The balance runs out when $600-15x=0$, so $15x=600$ and $x=40$. The table is made by subtracting `15 each day from the previous balance.

The two observations give $400=10a+b$ and $500=14a+b$. Subtracting the first equation from the second gives $100=4a$, so $a=25$. Then $400=10(25)+b$, so $b=150$.

The data gives $800=10a+b$ and $1100=15a+b$. Subtracting gives $300=5a$, so $a=60$. Then $800=10(60)+b$, so $b=200$.

Using $°C=a\,°F+b$, the melting point gives $0=32a+b$ and the boiling point gives $100=212a+b$. Subtracting the first equation from the second gives $100=180a$, so $a=\dfrac{5}{9}$. Then $0=32\cdot\dfrac{5}{9}+b$, so $b=-\dfrac{160}{9}$. Hence $°C=\dfrac{5}{9}°F-\dfrac{160}{9}$.

For a line $y=ax+b$, $a$ controls the slope and $b$ controls the y-intercept. Points for graphing include: $y=4x$: $(0,0),(1,4)$; $y=2x$: $(0,0),(1,2)$; $y=x$: $(0,0),(1,1)$. For $y=-6x,-3x,-x$, use $(0,0)$ and $(1,-6),(1,-3),(1,-1)$ respectively. For $y=5x$ and $y=-5x$, use $(0,0),(1,5)$ and $(0,0),(1,-5)$. For $y=3x-1,3x,3x+1$, use y-intercepts $(0,-1),(0,0),(0,1)$ and one more point on each line; equal slopes make them parallel. For $y=-2x-3$ and $y=-2x$, equal slopes make them parallel; $y=2x+3$ has opposite positive slope.

The highest power is $3$, so the polynomial has degree $3$, and the coefficient of the $x^2$ term is $-7$.

For (i), substitute $x=1$: $5(1)^2-3(1)+7=5-3+7=9$. For (ii), substitute $t=a$: $4a^3-a^2+6$.

Let the number be $x$. Then $\dfrac{5}{2}x+\dfrac{2}{3}=-\dfrac{7}{12}$. So $\dfrac{5}{2}x=-\dfrac{7}{12}-\dfrac{2}{3}=-\dfrac{7}{12}-\dfrac{8}{12}=-\dfrac{15}{12}=-\dfrac{5}{4}$. Multiplying by $\dfrac{2}{5}$ gives $x=-\dfrac{1}{2}$.

Let the smaller positive number be $x$. Then the other number is $5x$. After adding $21$, the numbers become $x+21$ and $5x+21$. Since the larger new number is twice the smaller new number, $5x+21=2(x+21)$. Thus $5x+21=2x+42$, so $3x=21$ and $x=7$. The numbers are $7$ and $35$.

After $n$ months, the saved amount added is `$(250n)$, so $A=800+250n$. For 6 months, $A=800+250(6)=2300$. Two years is $24$ months, so $A=800+250(24)=6800$.

Let the tens digit be $a$ and the ones digit be $b$. The original number is $10a+b$ and the reversed number is $10b+a$. Their sum is $11a+11b=11(a+b)=143$, so $a+b=13$. The digits differ by $3$, so the digits are $5$ and $8$. Therefore the possible numbers are $58$ and $85$.

Write each equation in the form $y=ax+b$. Then $a$ is the slope and $b$ is the y-intercept. The y-axis is cut where $x=0$, so the point is $(0,b)$. Thus the four equations give the listed slopes and y-axis intersection points.