Learning Objectives
After studying this chapter you should be able to:
- identify terms, variables, coefficients and constants in an algebraic expression;
- distinguish expressions in one variable from those in two variables;
- define a polynomial and state the degree of a given polynomial;
- recognise linear, quadratic, cubic and constant polynomials;
- compute the value of a polynomial for a given value of the variable;
- form a linear equation by setting a linear polynomial equal to a constant, and solve it;
- recognise a linear pattern and write its general term as a linear polynomial;
- distinguish linear growth from linear decay using the sign of the coefficient;
- find the linear relationship \(y = ax + b\) that fits two given points;
- draw the graph of a linear relationship and read its slope and y-intercept;
- use these ideas to solve simple real-life problems about cost, growth and decay.
1. Algebraic Expressions: The Language of Maths
An algebraic expression is built from numbers, letters and operation signs. The letters stand for numbers whose values may change, so we also call them variables. (They are sometimes called letter-numbers, because a letter simply takes the place of a number.)
- Terms: the pieces joined by \(+\) or \(-\). In \(4x + 5y + 3\) the terms are \(4x\), \(5y\) and \(3\).
- Variables: the letters used. In \(4x + 5y + 3\) the variables are \(x\) and \(y\).
- Coefficients: the numerical factors multiplying the variables. Here \(4\) is the coefficient of \(x\) and \(5\) is the coefficient of \(y\).
- Constant: the term with no variable. Here it is \(3\).
A shop sells notebooks at ₹35 each and pens at ₹12 each. Meera buys \(n\) notebooks and \(p\) pens and also pays ₹20 for delivery.
The cost of the notebooks is \(35n\), the cost of the pens is \(12p\), and the delivery is the constant. So the total cost is the expression
\(35n + 12p + 20\)
Terms: \(35n,\; 12p,\; 20\). Variables: \(n, p\). Coefficients: \(35\) and \(12\). Constant: \(20\).
An expression such as \(2x - 5\), \(4x + 7\) or \(6y - 1\) involves only one variable. An expression such as \(4x + 5y + 3\) involves two. This chapter deliberately restricts itself to expressions in one variable, because that is where the pattern of powers, degree and graphs becomes clean.
2. Polynomials and Their Degree
When an algebraic expression involves one variable together with its powers, it is called a one-variable polynomial (or univariate polynomial, meaning “having one variable”). When the context is clear we simply say polynomial.
The degree of a polynomial is the highest power of the variable appearing in it.
| Polynomial | Highest power | Degree | Type |
|---|---|---|---|
| \(5y^3 + y^2 + 2y - 1\) | 3 | 3 | cubic |
| \(x^2 + 5x + 1\) | 2 | 2 | quadratic |
| \(3z + 7\) | 1 | 1 | linear |
| \(8\) (that is, \(8x^0\)) | 0 | 0 | constant |
A polynomial of degree 1 is called a linear polynomial. Every linear polynomial can be written in the form
\(ax + b\), where \(a\) and \(b\) are real numbers and \(a \neq 0\).
Examples: \(2x + 3\), \(5 - 4y\), \(7x - 10\), \(\dfrac{1}{2}t + 6\).
Reading the parts of a polynomial
Take \(5y^3 + y^2 + 2y - 1\). In descending powers of \(y\):
- coefficient of \(y^3\) is \(5\);
- coefficient of \(y^2\) is \(1\) (the number 1 is never written);
- coefficient of \(y\) is \(2\);
- the constant term is \(-1\).
If a power is missing, its coefficient is understood to be \(0\). So in \(4z^3 + 5z^2 - 11\) the coefficient of \(z\) is \(0\), and the constant term is \(-11\).
The value of a polynomial
To find the value of a polynomial at a number, substitute that number everywhere the variable occurs and simplify.
\(P(4) = 3(4) - 5 = 12 - 5 = 7\)
\(P(0) = 3(0) - 5 = -5\)
\(P(-2) = 3(-2) - 5 = -6 - 5 = -11\)
Think of a linear polynomial as a machine. You feed in a number (the input), the machine works, and a single number comes out (the output). For \(2x + 3\): input \(4\) gives output \(11\); input \(-6\) gives output \(-9\).
3. Linear Equations
When a linear polynomial in one variable is equated to a constant, we get a linear equation.
Linear polynomial \(=\) constant → \(ax + b = c\)
Solving means finding the value(s) of \(x\) that make both sides equal.
The sum of the ages of a father and his son is 50 years, and the father is 5 years more than twice the son's age. Find both ages.
Let the son's age be \(s\) years. Then the father's age is \(2s + 5\). Using the sum:
\((2s + 5) + s = 50 \Rightarrow 3s + 5 = 50 \Rightarrow 3s = 45 \Rightarrow s = 15\)
Father's age \(= 2(15) + 5 = 35\) years.
Check: \(15 + 35 = 50\) and \(35 = 2 \times 15 + 5\). Both conditions hold.
Final answer: son 15 years, father 35 years.
4. Exploring Linear Patterns
A linear pattern is a sequence in which the difference between consecutive terms stays constant. Because the change per step never varies, a single linear polynomial describes the whole pattern, not just one term.
Figure 1: Stage \(n\) contains \(3n - 1\) squares, which is a linear polynomial in \(n\).
Finding the general term
Pattern: \(2,\; 5,\; 8,\; 11,\; \ldots\)
Step 1: terms are \(2, 5, 8, 11\). Step 2: differences are \(3, 3, 3\) — constant. Step 3: coefficient of \(n\) is \(3\). Step 4: at \(n = 1\), \(3(1) + b = 2\), so \(b = -1\).
General term: \(a_n = 3n - 1\). Test: \(n = 5\) gives \(3(5) - 1 = 14\), and indeed the next term after 11 is 14.
Final answer: \(a_n = 3n - 1\), a linear polynomial of degree 1.
5. Linear Growth and Linear Decay
Many real situations change by a fixed amount in each equal interval. We model such a situation with a linear polynomial written as
\(f(x) = ax + b\)
where \(b\) is the starting value and \(a\) is the amount of change per unit of \(x\).
- \(a > 0\) (positive coefficient) → the quantity grows as \(x\) increases. This is linear growth.
- \(a < 0\) (negative coefficient) → the quantity decreases as \(x\) increases. This is linear decay.
- \(a = 0\) is impossible here, because then the polynomial would not be linear; a constant is degree 0.
(i) A plant is 20 cm tall and grows 4 cm every month. After \(m\) months its height is \(h = 20 + 4m\). Here \(a = 4 > 0\), so this is linear growth. After 5 months: \(h = 20 + 20 = 40\) cm.
(ii) A tank holds 500 litres and leaks 10 litres every hour. After \(t\) hours the volume is \(v = 500 - 10t\). Here \(a = -10 < 0\), so this is linear decay. The tank is empty when \(500 - 10t = 0\), that is, after 50 hours.
6. Linear Relationships Between Two Quantities
Many situations link two quantities. Calling the independent one \(x\) and the dependent one \(y\), a linear relationship has the form
\(y = ax + b\)
If two points of the relationship are known, we can find \(a\) and \(b\).
A taxi charges a fixed fare of ₹50 plus ₹15 for each kilometre travelled. The fare is \(y\) rupees for a distance \(x\) km. Two trips are known: \(x = 4\) km gives \(y = 110\) rupees, and \(x = 7\) km gives \(y = 155\) rupees. Find the relationship.
Substituting: \(110 = 4a + b\) and \(155 = 7a + b\).
Subtracting: \(155 - 110 = 7a - 4a \Rightarrow 45 = 3a \Rightarrow a = 15\).
Then \(b = 110 - 4(15) = 110 - 60 = 50\).
Check: \(x = 7\) gives \(15(7) + 50 = 105 + 50 = 155\) rupees, as required.
Final answer: \(y = 15x + 50\), matching the fare rule.
7. Visualising Linear Relationships
Using the coordinate plane from Chapter 1, we can draw any relationship \(y = ax + b\).
- \(a\) is the slope of the line: it is the rise for every 1 unit run along the x-axis.
- \(b\) is the y-intercept: the line cuts the y-axis at the point \((0, b)\).
- Equations of the form \(y = ax\) pass through the origin \((0, 0)\).
- The steeper the slope, the faster the quantity changes: if \(a > 1\) the line is steeper than \(y = x\), and if \(a < 1\) it is less steep than \(y = x\).
- Lines with the same slope \(a\) but different y-intercepts are parallel.
A useful link back to patterns: the slope of the line is exactly the constant difference between consecutive terms of the sequence. For \(y = 2x - 1\) the terms are \(1, 3, 5, 7, \ldots\) and the slope is \(2\), which is the difference \(2\) between neighbouring terms.
How to draw the graph
Figure 2: The line \(y = 2x - 1\) cuts the y-axis at \((0, -1)\) and rises 2 units for every 1 unit to the right.
Figure 3: Two relationships with the same slope give parallel straight lines.
8. Solved Examples
A. Terms, coefficients and constants
Example 1
Question: In the expression \(9p - 4q^2 + 7r - 15\), name the terms, the variables, the coefficients of the variables and the constant term.
Concept: read the expression piece by piece.
Solution: Terms: \(9p,\; -4q^2,\; 7r,\; -15\). Variables: \(p, q, r\). Coefficients: \(9\) for \(p\), \(-4\) for \(q^2\), \(7\) for \(r\). Constant term: \(-15\).
Final answer: terms \(9p, -4q^2, 7r, -15\); coefficients \(9, -4, 7\); constant \(-15\).
B. Degree and type
Example 2
Question: State the degree and the type of each polynomial: (a) \(6t - 11\) (b) \(3w^2 - 4w + 9\) (c) \(2z^4 - z + 1\) (d) \(45\).
Concept: degree is the highest power present.
Solution:
- (a) highest power 1 → degree 1, linear
- (b) highest power 2 → degree 2, quadratic
- (c) highest power 4 → degree 4 (neither linear, quadratic nor cubic)
- (d) can be written \(45x^0\) → degree 0, constant
Final answer: (a) 1 linear, (b) 2 quadratic, (c) 4, (d) 0 constant.
Example 3
Question: What is the coefficient of \(x^2\) and the constant term in \(5x^3 - 8x^2 + 7\)?
Concept: a missing power has coefficient 0.
Solution: The \(x^2\) term is \(-8x^2\), so its coefficient is \(-8\). The term without a variable is \(7\), so the constant term is \(7\).
Final answer: coefficient of \(x^2\) is \(-8\); constant term is \(7\).
C. Values of a polynomial
Example 4
Question: If \(P(x) = 5x - 2\), find \(P(3)\), \(P(0)\) and \(P(-1)\).
Concept: substitute, then simplify carefully with negative values.
Solution:
- \(P(3) = 5(3) - 2 = 15 - 2 = 13\)
- \(P(0) = 5(0) - 2 = -2\)
- \(P(-1) = 5(-1) - 2 = -5 - 2 = -7\)
Final answer: \(P(3) = 13\), \(P(0) = -2\), \(P(-1) = -7\).
D. Linear equations
Example 5
Question: A pencil costs ₹8 and a notebook costs ₹45. A student buys \(c\) pencils and \(n\) notebooks and spends ₹310 in all. Form a linear equation and find the number of notebooks if 5 pencils are bought.
Concept: build the expression, then equate it to the total.
Solution: Total cost \(= 8c + 45n\), so \(8c + 45n = 310\). With \(c = 5\): \(8(5) + 45n = 310 \Rightarrow 40 + 45n = 310 \Rightarrow 45n = 270 \Rightarrow n = 6\).
Check: \(8(5) + 45(6) = 40 + 270 = 310\) rupees, as required.
Final answer: the equation is \(8c + 45n = 310\); with 5 pencils the student buys 6 notebooks.
Example 6
Question: Solve \(4x - 7 = 2x + 5\).
Concept: a linear polynomial set equal to a constant; move like terms together.
Solution: \(4x - 2x = 5 + 7 \Rightarrow 2x = 12 \Rightarrow x = 6\).
Check: LHS \(= 4(6) - 7 = 24 - 7 = 17\); RHS \(= 2(6) + 5 = 12 + 5 = 17\).
Final answer: \(x = 6\).
E. Linear patterns
Example 7
Question: A savings scheme gives ₹500 to start and adds ₹75 every month. Write an expression for the amount after \(m\) months, and find the amount after 8 months. Which linear polynomial is this?
Concept: starting value is \(b\), constant addition is \(a\).
Solution: \(A(m) = 500 + 75m\). After 8 months: \(A(8) = 500 + 75(8) = 500 + 600 = 1100\). This is the linear polynomial \(75m + 500\), of degree 1.
Final answer: \(A(m) = 75m + 500\); after 8 months the amount is ₹1100.
Example 8
Question: A queue of children is formed so that the first row has 4 children, the second 7, the third 10 and so on. How many children are in the 12th row?
Concept: constant difference 3, so the general term is linear.
Solution: Terms: \(4, 7, 10, \ldots\) with constant difference 3. Let \(a_n = 3n + b\). Using \(n = 1\): \(3 + b = 4 \Rightarrow b = 1\). So \(a_n = 3n + 1\). For \(n = 12\): \(a_{12} = 3(12) + 1 = 36 + 1 = 37\).
Check: \(a_2 = 3(2) + 1 = 7\), \(a_3 = 10\), matching the pattern.
Final answer: 37 children.
F. Growth and decay
Example 9
Question: A reservoir contains 600 litres of water and is drained at a constant 15 litres per hour. Write the expression for the water left after \(h\) hours, state whether it is growth or decay, and find when the reservoir is half empty.
Concept: negative coefficient means linear decay.
Solution: \(W(h) = 600 - 15h\). Since the coefficient of \(h\) is negative, this is linear decay. Half empty means \(W(h) = 300\): \(600 - 15h = 300 \Rightarrow 15h = 300 \Rightarrow h = 20\) hours.
Final answer: \(W(h) = 600 - 15h\), linear decay; half empty after 20 hours.
G. Linear relationships and graphs
Example 10
Question: The points \((2, 7)\) and \((6, 15)\) lie on a straight line. Find its equation in the form \(y = ax + b\), and state the slope and the y-intercept.
Concept: substitute both points and eliminate \(b\).
Solution: \(7 = 2a + b\) and \(15 = 6a + b\). Subtracting: \(15 - 7 = 6a - 2a \Rightarrow 8 = 4a \Rightarrow a = 2\). Then \(b = 7 - 2(2) = 7 - 4 = 3\). So \(y = 2x + 3\).
Check: at \(x = 6\), \(y = 2(6) + 3 = 15\), correct.
Final answer: \(y = 2x + 3\); slope \(2\), y-intercept \(3\) (the line cuts the y-axis at \((0, 3)\)).
Example 11
Question: Plot the graph of \(y = -3x + 6\) and state its slope, its y-intercept and the point where it cuts the x-axis.
Concept: start at the y-intercept, then step along the slope.
Solution: The y-intercept is \(b = 6\), so plot \((0, 6)\). The slope is \(-3\), so from \((0, 6)\) move 3 units down and 1 unit right to \((1, 3)\), then again to \((2, 0)\), and once more to \((3, -3)\). Join \((0,6)\), \((1,3)\), \((2,0)\), \((3,-3)\) with a ruler and extend the line. The x-axis crossing is where \(y = 0\): \(-3x + 6 = 0 \Rightarrow x = 2\), so the point is \((2, 0)\).
Final answer: slope \(-3\), y-intercept \(6\) at \((0, 6)\), x-intercept at \((2, 0)\).
Example 12
Question: Show that the lines \(y = 2x + 1\) and \(y = 2x - 4\) never meet.
Concept: equal slopes with different intercepts give parallel lines.
Solution: At any intersection the two expressions would be equal, so \(2x + 1 = 2x - 4 \Rightarrow 1 = -4\), which is impossible. Hence there is no solution and the lines are parallel. Both have slope 2, but the y-intercepts are \(1\) and \(-4\).
Final answer: the lines are parallel and never intersect.
H. Reasoning and application
Example 13
Question: The temperature of a freezer rises by \(2.5\) degrees Celsius every hour after it is switched off. At 12 noon it is \(-18\) degrees Celsius. Write the relationship and find the temperature at 4 p.m.
Concept: growth from a negative starting value; \(b\) may be negative.
Solution: \(T(t) = -18 + 2.5t\), where \(t\) is the number of hours after noon. The coefficient of \(t\) is positive, so the temperature is growing. At 4 p.m., \(t = 4\): \(T(4) = -18 + 2.5(4) = -18 + 10 = -8\) degrees Celsius.
Final answer: \(T(t) = -18 + 2.5t\); at 4 p.m. the freezer is at \(-8\) degrees Celsius.
Example 14
Question: A number is such that twice the number, increased by 7, equals the number increased by 19. Find the number.
Concept: turn the sentences into a linear equation in one variable.
Solution: Let the number be \(n\). "Twice the number, increased by 7" is \(2n + 7\). "The number increased by 19" is \(n + 19\). So \(2n + 7 = n + 19 \Rightarrow 2n - n = 19 - 7 \Rightarrow n = 12\).
Check: \(2(12) + 7 = 31\) and \(12 + 19 = 31\).
Final answer: the number is 12.
9. Common Mistakes
What went wrong: calling \(5\) the coefficient of \(y^2\) in \(5y^3 + y^2 + 2y - 1\).
Correction: the coefficient belongs to the power it is written with — \(5\) multiplies \(y^3\). Here the coefficient of \(y^2\) is \(1\).
What went wrong: saying \(4z^3 + 5z^2 - 11\) has "no constant term".
Correction: the constant term is \(-11\). Only a power that is missing has coefficient \(0\).
What went wrong: calling \(5y^3 + y^2 + 2y - 1\) a polynomial of degree 4 because it has four terms.
Correction: degree counts the highest power, not the terms. Here the degree is 3.
What went wrong: calling \(500 - 10t\) linear growth.
Correction: look at the sign of the coefficient of the variable. Negative means the quantity falls — linear decay.
What went wrong: saying the y-intercept of \(y = 2x - 1\) is \(2\).
Correction: the y-intercept is \(b\), found by putting \(x = 0\). Here it is \(-1\), so the line cuts the y-axis at \((0, -1)\).
What went wrong: joining plotted points with a smooth curve.
Correction: a linear relationship always gives a straight line. Use a ruler and extend it with arrowheads.
What went wrong: fitting a linear expression to squares, such as \(1, 4, 9, 16\).
Correction: the differences there are \(3, 5, 7\), which are not constant, so the pattern is not linear and no linear polynomial fits it.
10. Important Results at a Glance
Expression parts
Terms are joined by \(+\) or \(-\); coefficients multiply the variables; the constant has no variable.
Polynomial
An expression in one variable with its powers. Degree = highest power of the variable.
Linear polynomial
Degree 1, of the form \(ax + b\) with \(a \neq 0\).
Linear equation
A linear polynomial set equal to a constant: \(ax + b = c\).
Value of a polynomial
Substitute the given number for the variable and simplify.
Linear pattern
Consecutive differences are constant; the general term is a linear polynomial in \(n\).
Growth and decay
\(a > 0\) gives linear growth; \(a < 0\) gives linear decay.
Graph
\(y = ax + b\): slope \(a\), y-intercept \((0, b)\). Equal slopes with different intercepts are parallel.
11. Formulae and Quick Reference
| What you need | Formula or rule |
|---|---|
| Linear polynomial | \(ax + b\), where \(a \neq 0\) |
| Linear equation | \(ax + b = c\) |
| Degree | the highest power of the variable in the polynomial |
| Value of a polynomial | substitute: \(P(k) = ak + b\) for \(P(x) = ax + b\) |
| Slope | \(a = \dfrac{y_2 - y_1}{x_2 - x_1}\) |
| Slope-intercept form | \(y = ax + b\), y-intercept at \((0, b)\) |
| Lines of the form \(y = ax\) | \((0, 0)\) always lies on the line |
| x-intercept of \(y = ax + b\) | \(\left(-\dfrac{b}{a},\; 0\right)\), when \(b \neq 0\) |
12. Think and Understand
Q1.
Why must the coefficient of \(x\) be non-zero in a linear polynomial \(ax + b\)?
Q2.
A pattern gives \(1, 3, 6, 10, 15, \ldots\) Is it a linear pattern? Justify your answer.
Q3.
How does the sign of \(a\) tell you whether a relationship describes growth or decay? Give a real example of each.
Q4.
Two lines have the same slope but different y-intercepts. What can you say about where they meet?
Q5.
Why does the graph of every equation \(y = ax\) pass through the origin?
Q6.
If the y-intercept of a line is negative, what does that tell you about where the line meets the y-axis?
13. Exam Tips
- Always read the degree from the highest power, not from the number of terms.
- Remember that a missing power has coefficient \(0\), but the constant term is whatever number stands alone.
- Substitute values carefully with brackets, especially negative ones: \(3(-2) - 5 = -11\).
- Check that a pattern is linear by confirming the difference is constant before writing a general term.
- State clearly whether a situation is growth or decay, and say why (sign of the coefficient).
- When finding \(a\) and \(b\), show the two substituted equations and the subtraction step.
- Label both axes, mark a uniform scale and plot points accurately before joining them.
- Use a ruler: linear relationships give straight lines, never curves.
- Always give the slope and the y-intercept when a question asks about a graph.
- Verify your answer by substituting it back into the original condition.
14. Quick Revision
- Algebraic expression: numbers, variables and operation signs.
- Terms: \(4x,\; 5y,\; 3\) in \(4x + 5y + 3\).
- Coefficients: \(4\) and \(5\); constant: \(3\).
- Polynomial: expression in one variable with its powers; degree = highest power.
- Linear polynomial: degree 1, form \(ax + b\), \(a \neq 0\).
- Linear equation: \(ax + b = c\).
- Value: substitute and simplify.
- Linear pattern: constant difference; general term is linear, e.g. \(3n - 1\).
- Growth/decay: \(a > 0\) growth, \(a < 0\) decay.
- Relationship: \(y = ax + b\), found from two points.
- Graph: straight line; slope \(a\); y-intercept \((0, b)\).
- Parallel lines: same slope, different y-intercepts.
15. Chapter Checklist
- I can identify terms, variables, coefficients and constants.
- I can tell a one-variable expression from a two-variable expression.
- I can define a polynomial and find its degree.
- I know the names linear, quadratic, cubic and constant.
- I can compute the value of a polynomial at a given number.
- I can form a linear equation from a word problem and solve it.
- I can recognise a linear pattern and write its general term.
- I can distinguish linear growth from linear decay and explain why.
- I can find the equation of a line through two given points.
- I can plot a straight-line graph and label the slope and y-intercept.
- I can explain why equal slopes with different intercepts give parallel lines.
- I can use these ideas in real-life problems involving cost, growth and decay.