12 Linear Equations
12.1 Introduction
Linear equations form the backbone of algebra and quantitative aptitude.
They represent relationships between variables with degree one (highest power of variable = 1).
In exams, linear equations appear in forms of simple solving, word problems, equations in two or more variables, and applications in ages, mixtures, percentages, and time–work–distance.
12.2 1) Basics of Linear Equations
General form (one variable):
\[ax+b=0\] where \(a\neq0\).
Solution: \(x=-\frac{b}{a}\).General form (two variables):
\[ax+by+c=0\]
Represents a straight line in the coordinate plane.Linear equation in \(n\) variables:
\[a_1x_1+a_2x_2+\dots+a_nx_n+b=0\]
12.3 2) Types of Solutions (Two Variables)
System of two equations:
\[a_1x+b_1y=c_1\]
\[a_2x+b_2y=c_2\]
Unique solution (intersecting lines):
\(\frac{a_1}{a_2}\neq \frac{b_1}{b_2}\)No solution (parallel lines):
\(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\)Infinite solutions (coincident lines):
\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
12.4 3) Methods of Solving
12.4.1 3.1 Substitution Method
- Express one variable in terms of another and substitute.
- Example: Solve \(2x+y=10\), \(x-y=1\).
From second: \(x=y+1\). Sub into first: \(2(y+1)+y=10 \Rightarrow 3y+2=10 \Rightarrow y=8/3\), \(x=11/3\).
12.4.2 3.2 Elimination Method
- Multiply equations if needed and eliminate one variable.
- Example: Solve \(3x+2y=16\), \(2x+3y=13\).
Multiply first by 3: \(9x+6y=48\).
Multiply second by 2: \(4x+6y=26\).
Subtract: \(5x=22 \Rightarrow x=22/5\), then \(y=(16-3x)/2=14/10=7/5\).
12.4.3 3.3 Cross Multiplication (for two variables)
For system \(a_1x+b_1y+c_1=0\), \(a_2x+b_2y+c_2=0\):
\[ \frac{x}{b_1c_2-b_2c_1} = \frac{y}{c_1a_2-c_2a_1} = \frac{1}{a_1b_2-a_2b_1} \]
12.5 4) Applications in Word Problems
12.5.1 4.1 Age Problems
Father is 3 times son’s age. In 10 years, difference = 20. Find ages.
Let son=x, father=3x. After 10 yrs: (3x+10)−(x+10)=2x=20 ⇒ x=10, father=30.
12.5.2 4.2 Mixture Problems
A fruit seller mixes 10 kg apples at 80/kg with 15 kg apples at 100/kg. Selling price = 96/kg. Find profit/loss.
CP = (10×80+15×100)/(25)=92. SP=96 ⇒ Profit=4 on 92=4.35%.
12.5.3 4.3 Time–Work
A does work in 6 days, B in 8 days. Together in 1 day?
Rate: 1/6+1/8=7/24 ⇒ total time=24/7≈3.43 days.
12.5.4 4.4 Percent/Ratio Problems
Twice a number plus thrice another = 36. Their difference=6. Find numbers.
Let x,y. 2x+3y=36, x−y=6 ⇒ Solve.
12.6 5) Linear Equations in Three Variables
System:
\[a_1x+b_1y+c_1z=d_1\]
\[a_2x+b_2y+c_2z=d_2\]
\[a_3x+b_3y+c_3z=d_3\]
- Solve using elimination/substitution or determinants (Cramer’s rule).
- Exam questions often simple, solvable by stepwise substitution.
12.7 6) Special Forms
- Equations reducible to linear: e.g., \(\frac{1}{x}+\frac{1}{y}=3\), \(\frac{1}{x}-\frac{1}{y}=1\) → substitute \(p=1/x\), \(q=1/y\).
- Word equations: “If you double the number and add 5, result is 21” → \(2x+5=21\).
12.8 7) Solved Examples
Ex 1 Solve: \(x+2y=8\), \(3x-y=5\).
From first: \(x=8-2y\). Sub: \(3(8-2y)-y=5 \Rightarrow 24-6y-y=5 \Rightarrow y=19/7\), \(x=30/7\).
Ex 2 A and B together earn 1,200. A earns 200 more than B. Find amounts.
Let B=x ⇒ A=x+200. Then x+x+200=1200 ⇒ 2x=1000 ⇒ x=500, A=700.
Ex 3 Solve by cross multiplication: 2x+3y=8, 4x−y=2.
Cross method ⇒ x=1, y=2.
Ex 4 In 3 variables: x+y+z=6, x+2y+3z=14, 2x+y+z=10.
From first: x=6−y−z. Sub into others → solve → (x,y,z)=(3,1,2).
12.9 8) Practice Set – Level 1
- Solve: 5x+2y=20, 3x−y=1.
- A number plus its double equals 21. Find number.
- Two numbers differ by 4. Their sum=20. Find numbers.
- A pen and pencil cost 18, two pens and three pencils cost 48. Find cost of each.
- Solve: x+y=10, x−y=4.
12.10 Practice Set – Level 2
- 3x+4y=25, 5x+6y=41. Solve.
- A father is twice his son’s age. In 10 yrs sum of ages=80. Find ages.
- Three numbers add to 60. Their ratio 2:3:5. Find numbers.
- Solve: 2x+y+z=8, x−y+2z=3, 3x+2y−z=7.
- The cost of 2 apples and 3 bananas=40, 4 apples and 9 bananas=100. Find apple and banana cost.
12.11 9) Common Mistakes
- Forgetting to check solution type (unique/no/infinite).
- Mixing CP, SP context with linear algebra.
- Arithmetic slips in substitution/elimination.
- Misreading word problems (e.g., “more than,” “less than”).
12.12 Summary
- Linear equations = degree 1 relations.
- Solving methods: substitution, elimination, cross multiplication.
- Word problems → translate carefully into equations.
- In 3 variables, use substitution/determinants.
- Key exam skill: speed in setting up + eliminating without mistakes.