NCERT Solutions · Class 9 Mathematics Introduction to Linear Polynomials

Build it term by term from the conditions.Two conditions must hold, and nothing else is fixed: The coefficients of \(\displaystyle x\) and of the constant term are not restricted at all, so we may choose them freely. Taking the \(\displaystyle x^3\) coefficient as $\displaystyle 1$, the \(\displaystyle x\) coefficient as $\displaystyle 2$ and the constant as $\displaystyle 5$ gives \[p(x) = x^3 - 7x^2 + 2x + 5 \]Check it against both conditions: the highest power is \(\displaystyle x^3\) with coefficient \(\displaystyle 1 \ne 0\), so the degree is $\displaystyle 3$; and the coefficient of \(\displaystyle x^2\) is \(\displaystyle -7\). Both conditions hold.This question has many correct answers. Any polynomial of the form \[p(x) = kx^3 - 7x^2 + mx + n, \qquad k \ne 0, \] with \(\displaystyle m\) and \(\displaystyle n\) any numbers, works. For instance \(\displaystyle 4x^3 - 7x^2\), \(\displaystyle -x^3 - 7x^2 + 9\) and \(\displaystyle x^3 - 7x^2 - \tfrac{1}{2}x + \sqrt{3}\) are all valid. Your answer need not match the one above; it only has to satisfy the two conditions.The one thing that would be wrong is taking \(\displaystyle k = 0\): \(\displaystyle -7x^2 + 2x + 5\) has the right \(\displaystyle x^2\) coefficient but degree $\displaystyle 2$, not 3.Answer: \(\displaystyle p(x) = x^{3} - 7x^{2} + 2x + 5\) is one such polynomial; any \(\displaystyle kx^{3} - 7x^{2} + mx + n\) with \(\displaystyle k \neq 0\) is equally correct.

Substitute the given value in place of the variable.The value of a polynomial at a number is found by replacing the variable by that number everywhere and simplifying.(i) \(\displaystyle 5x^{2} - 3x + 7\) at \(\displaystyle x = 1\).Put \(\displaystyle x = 1\) in each term: \[5(1)^{2} - 3(1) + 7 = 5 \times 1 - 3 + 7 = 5 - 3 + 7 = 9 \](A useful shortcut to check with: at \(\displaystyle x = 1\) the value of any polynomial is just the sum of its coefficients, since every power of $\displaystyle 1$ is 1. Here \(\displaystyle 5 + (-3) + 7 = 9\), which agrees.)(ii) \(\displaystyle 4t^{3} - t^{2} + 6\) at \(\displaystyle t = a\).Here the variable is being replaced by another letter, not by a number. That is allowed — substitution does not care whether the thing you put in is a number or a symbol. Replace every \(\displaystyle t\) by \(\displaystyle a\): \[4a^{3} - a^{2} + 6 \] This cannot be simplified further, because \(\displaystyle a\) stands for an unknown number, so \(\displaystyle 4a^3\) and \(\displaystyle a^2\) are not like terms.Answer: (i) \(\displaystyle 9\) (ii) \(\displaystyle 4a^{3} - a^{2} + 6\).

Translate the words into an equation, then undo the operations.Let the unknown number be \(\displaystyle x\). So the equation is \[\frac{5}{2}x + \frac{2}{3} = -\frac{7}{12} \]Undo the addition first. Subtract \(\displaystyle \dfrac{2}{3}\) from both sides: \[\frac{5}{2}x = -\frac{7}{12} - \frac{2}{3} \] The LCM of $\displaystyle 12$ and $\displaystyle 3$ is $\displaystyle 12$, and \(\displaystyle \dfrac{2}{3} = \dfrac{8}{12}\), so \[\frac{5}{2}x = -\frac{7}{12} - \frac{8}{12} = -\frac{15}{12} = -\frac{5}{4} \]Now undo the multiplication. Multiply both sides by \(\displaystyle \dfrac{2}{5}\) (the reciprocal of \(\displaystyle \dfrac{5}{2}\)): \[x = -\frac{5}{4} \times \frac{2}{5} = -\frac{10}{20} = -\frac{1}{2} \]Check by going forwards through the original words: \[\frac{5}{2} \times \left(-\frac{1}{2}\right) + \frac{2}{3} = -\frac{5}{4} + \frac{2}{3} = -\frac{15}{12} + \frac{8}{12} = -\frac{7}{12} \quad \checkmark \]Answer: the number is \(\displaystyle -\dfrac{1}{2}\).

Name the smaller number, express the other in terms of it, then test both readings.Let the smaller number be \(\displaystyle x\). Then the other number is \(\displaystyle 5x\), and since the numbers are positive, \(\displaystyle x > 0\).After $\displaystyle 21$ is added to each, the two new numbers are \[x + 21 \qquad \text{and} \qquad 5x + 21 \]The sentence says one of the new numbers is twice the other, without saying which. Since it does not say, we must try both readings and keep the one that gives positive numbers.Case $\displaystyle 1$: the bigger new number is twice the smaller new number. \[5x + 21 = 2(x + 21) \] \[5x + 21 = 2x + 42 \] Take \(\displaystyle 2x\) to the left and $\displaystyle 21$ to the right: \[3x = 21 \quad \Longrightarrow \quad x = 7 \] Then the numbers are \(\displaystyle x = 7\) and \(\displaystyle 5x = 35\), both positive — acceptable so far.Case $\displaystyle 2$: the smaller new number is twice the bigger new number. \[x + 21 = 2(5x + 21) \] \[x + 21 = 10x + 42 \quad \Longrightarrow \quad -21 = 9x \quad \Longrightarrow \quad x = -\frac{21}{9} = -\frac{7}{3} \] This is negative, so it is rejected: the question says the numbers are positive. (It is also unsurprising — a smaller number cannot be double a bigger one when both are positive.)Check Case $\displaystyle 1$ against the original statement. The numbers are $\displaystyle 7$ and $\displaystyle 35$, and indeed \(\displaystyle 35 = 5 \times 7\). Adding $\displaystyle 21$ to each gives \(\displaystyle 7 + 21 = 28\) and \(\displaystyle 35 + 21 = 56\), and \(\displaystyle 56 = 2 \times 28\). Both conditions hold.Answer: the numbers are $\displaystyle 7$ and 35.

A fixed start plus a constant monthly addition — a linear pattern.You begin with ₹$\displaystyle 800$ and add ₹$\displaystyle 250$ at the end of each month. After \(\displaystyle n\) months you have added \(\displaystyle 250n\), so the amount is \[A = 800 + 250n \](i) After $\displaystyle 6$ months.Here \(\displaystyle n = 6\): \[A = 800 + 250 \times 6 = 800 + 1500 = 2300 \] So you have ₹$\displaystyle 2,300$.(ii) After $\displaystyle 2$ years.First convert to months, because the saving happens monthly: \(\displaystyle 2 \text{ years} = 2 \times 12 = 24\) months. So \(\displaystyle n = 24\): \[A = 800 + 250 \times 24 = 800 + 6000 = 6800 \] So you have ₹$\displaystyle 6,800$.The linear pattern. \[A = 800 + 250n \]

\(\displaystyle n\) (months)	$\displaystyle 0$	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$	$\displaystyle 4$	$\displaystyle 5$	$\displaystyle 6$
\(\displaystyle A\) (in ₹)	$\displaystyle 800$	$\displaystyle 1050$	$\displaystyle 1300$	$\displaystyle 1550$	$\displaystyle 1800$	$\displaystyle 2050$	$\displaystyle 2300$

Each entry is $\displaystyle 250$ more than the one before, and the last column agrees with part (i).This is a linear polynomial in \(\displaystyle n\): the highest power of \(\displaystyle n\) is $\displaystyle 1$, the coefficient of \(\displaystyle n\) is the constant \(\displaystyle +250\) (equal amounts added in equal times, so the graph is a straight line rising steadily), and the constant term $\displaystyle 800$ is the amount at \(\displaystyle n = 0\), before any saving.Answer: (i) ₹$\displaystyle 2,300$ (ii) ₹$\displaystyle 6,800$; the pattern is \(\displaystyle A = 800 + 250n\), where \(\displaystyle n\) is the number of months.

Write a two-digit number in terms of its digits.Let the tens digit be \(\displaystyle a\) and the units digit be \(\displaystyle b\). Then the number itself is worth \[10a + b \] and the number formed by interchanging the digits is worth \[10b + a \]Use the "sum is $\displaystyle 143$" condition. \[(10a + b) + (10b + a) = 143 \] \[11a + 11b = 143 \quad \Longrightarrow \quad 11(a + b) = 143 \quad \Longrightarrow \quad a + b = \frac{143}{11} = 13 \]Notice what happened: reversing a two-digit number and adding always gives \(\displaystyle 11 \times (\text{sum of the digits})\), so this condition tells us the digit sum, nothing more.Use the "digits differ by $\displaystyle 3$" condition.So we need two digits with \[a + b = 13, \qquad |a - b| = 3 \] Adding \(\displaystyle a + b = 13\) and \(\displaystyle a - b = 3\) gives \(\displaystyle 2a = 16\), so \(\displaystyle a = 8\) and then \(\displaystyle b = 5\). Taking \(\displaystyle b - a = 3\) instead simply swaps the two, giving \(\displaystyle a = 5,\ b = 8\).So the digits are $\displaystyle 8$ and $\displaystyle 5$, and the two-digit numbers they form are $\displaystyle 85$ and $\displaystyle 58$. Either can be regarded as "the original" — the other is then the interchanged one.Check. The digits $\displaystyle 8$ and $\displaystyle 5$ differ by \(\displaystyle 8 - 5 = 3\) ✓. And \(\displaystyle 85 + 58 = 143\) ✓. Both are genuine two-digit numbers (neither has $\displaystyle 0$ in the tens place).Answer: the two numbers are $\displaystyle 85$ and 58.

Rewrite each equation as \(\displaystyle y = ax + b\), then read \(\displaystyle a\) and \(\displaystyle b\) straight off.Once an equation is in the form \(\displaystyle y = ax + b\), the slope is \(\displaystyle a\) and the y-intercept is \(\displaystyle b\); the line meets the y-axis at \(\displaystyle (0, b)\), because putting \(\displaystyle x = 0\) leaves \(\displaystyle y = b\).(i) \(\displaystyle y = -3x + 4\) is already in that form. Slope \(\displaystyle = -3\), y-intercept \(\displaystyle = 4\), cuts the y-axis at \(\displaystyle (0, 4)\). Plotting points: \(\displaystyle (0, 4),\ (1, 1),\ (2, -2)\).(ii) \(\displaystyle 2y = 4x + 7\). Divide every term by $\displaystyle 2$: \[y = 2x + \frac{7}{2} \] Slope \(\displaystyle = 2\), y-intercept \(\displaystyle = \dfrac{7}{2}\), cuts the y-axis at \(\displaystyle \left(0, \tfrac{7}{2}\right)\). Plotting points: \(\displaystyle (0, 3.5),\ (1, 5.5),\ (-1, 1.5)\).(iii) \(\displaystyle 5y = 6x - 10\). Divide by $\displaystyle 5$: \[y = \frac{6}{5}x - 2 \] Slope \(\displaystyle = \dfrac{6}{5}\), y-intercept \(\displaystyle = -2\), cuts the y-axis at \(\displaystyle (0, -2)\). Plotting points: \(\displaystyle (0, -2),\ (5, 4),\ (-5, -8)\) — multiples of $\displaystyle 5$ for \(\displaystyle x\) keep \(\displaystyle y\) a whole number.(iv) \(\displaystyle 3y = 6x - 11\). Divide by $\displaystyle 3$: \[y = 2x - \frac{11}{3} \] Slope \(\displaystyle = 2\), y-intercept \(\displaystyle = -\dfrac{11}{3}\), cuts the y-axis at \(\displaystyle \left(0, -\tfrac{11}{3}\right)\). Plotting points: \(\displaystyle \left(0, -\tfrac{11}{3}\right) \approx (0, -3.7)\) and \(\displaystyle \left(3, \tfrac{7}{3}\right) \approx (3, 2.3)\).Draw each line by plotting its two or three points on the same pair of axes and joining them with a ruler.Summary.

Equation	Slope	y-intercept	Point on the y-axis
\(\displaystyle y = -3x + 4\)	\(\displaystyle -3\)	\(\displaystyle 4\)	\(\displaystyle (0,\ 4)\)
\(\displaystyle 2y = 4x + 7\)	\(\displaystyle 2\)	\(\displaystyle \tfrac{7}{2}\)	\(\displaystyle \left(0,\ \tfrac{7}{2}\right)\)
\(\displaystyle 5y = 6x - 10\)	\(\displaystyle \tfrac{6}{5}\)	\(\displaystyle -2\)	\(\displaystyle (0,\ -2)\)
\(\displaystyle 3y = 6x - 11\)	\(\displaystyle 2\)	\(\displaystyle -\tfrac{11}{3}\)	\(\displaystyle \left(0,\ -\tfrac{11}{3}\right)\)

Are any of the lines parallel?Two lines are parallel exactly when they have the same slope but different y-intercepts (same slope and same intercept would make them the very same line).Looking down the slope column: \(\displaystyle -3\), \(\displaystyle 2\), \(\displaystyle \tfrac{6}{5}\), \(\displaystyle 2\). The value $\displaystyle 2$ occurs twice, for (ii) and (iv). Their intercepts \(\displaystyle \tfrac{7}{2}\) and \(\displaystyle -\tfrac{11}{3}\) are different, so they are two distinct lines with the same tilt.Answer: slopes and intercepts are \(\displaystyle -3\) and \(\displaystyle 4\); \(\displaystyle 2\) and \(\displaystyle \tfrac{7}{2}\); \(\displaystyle \tfrac{6}{5}\) and \(\displaystyle -2\); \(\displaystyle 2\) and \(\displaystyle -\tfrac{11}{3}\), so the lines cut the y-axis at \(\displaystyle (0,4)\), \(\displaystyle \left(0,\tfrac{7}{2}\right)\), \(\displaystyle (0,-2)\) and \(\displaystyle \left(0,-\tfrac{11}{3}\right)\). Yes — lines (ii) and (iv) are parallel, both having slope 2.

Substitute into the given rule for (i); reverse the steps for (ii).The rule is \[y = \frac{9}{5}(x - 273) + 32 \] with \(\displaystyle x\) in kelvin and \(\displaystyle y\) in degrees Fahrenheit.(i) Temperature in Fahrenheit when \(\displaystyle x = 313\) K.Substitute \(\displaystyle x = 313\), working inside the bracket first: \[y = \frac{9}{5}(313 - 273) + 32 = \frac{9}{5}(40) + 32 \] \[\frac{9}{5} \times 40 = 9 \times 8 = 72, \qquad \text{so } y = 72 + 32 = 104 \]So the temperature is $\displaystyle 104$ °F.(ii) Temperature in kelvin when \(\displaystyle y = 158\) °F.Now we know \(\displaystyle y\) and want \(\displaystyle x\), so we undo the operations in reverse order. Substitute \(\displaystyle y = 158\): \[158 = \frac{9}{5}(x - 273) + 32 \] Subtract $\displaystyle 32$ from both sides: \[126 = \frac{9}{5}(x - 273) \] Multiply both sides by \(\displaystyle \dfrac{5}{9}\): \[x - 273 = 126 \times \frac{5}{9} = 14 \times 5 = 70 \] Add $\displaystyle 273$ to both sides: \[x = 70 + 273 = 343 \]So the temperature is $\displaystyle 343$ K.Check part (ii) by putting $\displaystyle 343$ back into the original rule: \[\frac{9}{5}(343 - 273) + 32 = \frac{9}{5}(70) + 32 = 126 + 32 = 158 \quad \checkmark \]Answer: (i) $\displaystyle 104$ °F (ii) $\displaystyle 343$ K.

Turn the sentence into a formula, then fix the force to get a line.The statement "work done is the product of the constant force and the distance travelled in the direction of the force" says, in symbols, that for a constant force \(\displaystyle F\), \[w = F \times d \]Taking the constant force as $\displaystyle 3$ units, \(\displaystyle F = 3\), so the linear equation in the two variables \(\displaystyle w\) and \(\displaystyle d\) is \[w = 3d, \qquad \text{or equivalently} \qquad 3d - w = 0 \]This is a linear equation in two variables: each variable appears to the first power only, so its graph is a straight line. There is no constant term, so \(\displaystyle w = 0\) when \(\displaystyle d = 0\) — no distance, no work — and the line passes through the origin. Its slope is $\displaystyle 3$: every extra unit of distance costs $\displaystyle 3$ more units of work.Table of values for the graph.

\(\displaystyle d\) (units of distance)	$\displaystyle 0$	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$
\(\displaystyle w\) (units of work)	$\displaystyle 0$	$\displaystyle 3$	$\displaystyle 6$	$\displaystyle 9$

Drawing it. Take \(\displaystyle d\) along the horizontal axis and \(\displaystyle w\) along the vertical axis. Plot \(\displaystyle (0, 0)\), \(\displaystyle (1, 3)\), \(\displaystyle (2, 6)\) and \(\displaystyle (3, 9)\); they lie in a straight line, so join them with a ruler and extend. (Only \(\displaystyle d \ge 0\) is meaningful here, since a distance travelled cannot be negative.)Work done when the distance is $\displaystyle 2$ units. \[w = 3 \times 2 = 6 \]Verifying it on the graph. Start at \(\displaystyle d = 2\) on the horizontal axis, go straight up until you meet the line, then straight across to the vertical axis: you land on \(\displaystyle w = 6\). The point \(\displaystyle (2, 6)\) is one of the points already plotted from the table, so the graph and the calculation agree.Answer: \(\displaystyle w = 3d\); the work done over a distance of $\displaystyle 2$ units is $\displaystyle 6$ units, shown by the point \(\displaystyle (2, 6)\) on the line.

A linear polynomial has two unknowns; two points on the graph give two equations.A linear polynomial has the form \[p(x) = ax + b \]Saying the graph passes through a point means the coordinates of that point satisfy \(\displaystyle y = p(x)\).(i) Finding \(\displaystyle p(x)\).Through \(\displaystyle (1, 5)\): \(\displaystyle p(1) = 5\), so \[a(1) + b = 5 \quad \Longrightarrow \quad a + b = 5 \qquad \ldots (1) \] Through \(\displaystyle (3, 11)\): \(\displaystyle p(3) = 11\), so \[a(3) + b = 11 \quad \Longrightarrow \quad 3a + b = 11 \qquad \ldots (2) \]Subtract ($\displaystyle 1$) from ($\displaystyle 2$) to eliminate \(\displaystyle b\): \[(3a + b) - (a + b) = 11 - 5 \quad \Longrightarrow \quad 2a = 6 \quad \Longrightarrow \quad a = 3 \] Substitute in ($\displaystyle 1$): \(\displaystyle 3 + b = 5\), so \(\displaystyle b = 2\). Hence \[p(x) = 3x + 2 \]Check both given points: \(\displaystyle p(1) = 3 + 2 = 5\) ✓ and \(\displaystyle p(3) = 9 + 2 = 11\) ✓.(ii) Where the graph cuts the axes.The y-axis is the set of points with \(\displaystyle x = 0\). Put \(\displaystyle x = 0\): \[p(0) = 3(0) + 2 = 2 \] so the graph cuts the y-axis at \(\displaystyle (0,\ 2)\).The x-axis is the set of points with \(\displaystyle y = 0\). Put \(\displaystyle p(x) = 0\): \[3x + 2 = 0 \quad \Longrightarrow \quad 3x = -2 \quad \Longrightarrow \quad x = -\frac{2}{3} \] so the graph cuts the x-axis at \(\displaystyle \left(-\dfrac{2}{3},\ 0\right)\).(iii) Drawing the graph and verifying.

\(\displaystyle x\)	\(\displaystyle -1\)	$\displaystyle 0$	$\displaystyle 1$	$\displaystyle 3$
\(\displaystyle y = 3x + 2\)	\(\displaystyle -1\)	$\displaystyle 2$	$\displaystyle 5$	$\displaystyle 11$

Plot \(\displaystyle (-1, -1)\), \(\displaystyle (0, 2)\), \(\displaystyle (1, 5)\) and \(\displaystyle (3, 11)\) and join them with a ruler — they lie on one straight line, which confirms the polynomial is linear and that it does pass through the two given points.On that drawing, the line meets the vertical axis at height $\displaystyle 2$, matching \(\displaystyle (0, 2)\); and it crosses the horizontal axis a little to the left of the origin, between \(\displaystyle x = -1\) and \(\displaystyle x = 0\), at about \(\displaystyle -0.67\), which matches \(\displaystyle -\tfrac{2}{3}\).Answer: (i) \(\displaystyle p(x) = 3x + 2\). (ii) It cuts the y-axis at \(\displaystyle (0,\ 2)\) and the x-axis at \(\displaystyle \left(-\dfrac{2}{3},\ 0\right)\).