Gyaan

Linear Equations

beginner linear equations simultaneous equations algebra word problems

Linear equations are the bread and butter of algebra. If we can solve these fast, we unlock speed in almost every other aptitude topic because so many word problems boil down to “set up an equation, solve it.”

A linear equation is simply an equation where the variable has power 1 — no squares, no cubes, just plain old x. Think of it like a balance scale: whatever we do to one side, we do to the other, and the scale stays balanced.

Single Variable Equations

The goal is simple: get x alone on one side.

Key Rules
1. Add/subtract the same value on both sides
2. Multiply/divide the same value on both sides
3. Cross-multiply when we have fractions: a/b = c/d → ad = bc
4. Always simplify before solving — combine like terms first

Quick trick: When we see an equation like 3x + 7 = 22, just think: “move 7 to the other side, flip its sign.” So 3x = 22 - 7 = 15, then x = 5. We don’t need to formally “subtract 7 from both sides” every time — just move and flip.

Example 1: Basic single variable

Solve: 5x - 3 = 2x + 12

Move all x terms to the left, constants to the right:

  • 5x - 2x = 12 + 3
  • 3x = 15
  • x = 5

Example 2: Fractions involved

Solve: (2x + 1)/3 = (x - 2)/2

Cross-multiply:

  • 2(2x + 1) = 3(x - 2)
  • 4x + 2 = 3x - 6
  • 4x - 3x = -6 - 2
  • x = -8

Simultaneous Linear Equations (Two Variables)

When we have two unknowns, we need two equations. There are two main methods: substitution and elimination. Both get us to the same answer — pick whichever feels faster for the problem.

Key Formulas — Simultaneous Equations
Substitution: Solve one equation for one variable, plug into the other
Elimination: Make coefficients of one variable equal, then add/subtract equations
Cross-multiplication shortcut for a₁x + b₁y = c₁ and a₂x + b₂y = c₂:
x = (b₂c₁ - b₁c₂) / (a₁b₂ - a₂b₁)
y = (a₁c₂ - a₂c₁) / (a₁b₂ - a₂b₁)

Method 1: Substitution

Best when one equation already has a variable with coefficient 1 (like y = 3x + 2).

Method 2: Elimination

Best when coefficients are easy to match up.

Example 3: Elimination method

Solve: 3x + 2y = 16 and 5x - 2y = 24

Notice the y coefficients are already +2 and -2. Just add the equations:

  • (3x + 2y) + (5x - 2y) = 16 + 24
  • 8x = 40
  • x = 5

Plug back: 3(5) + 2y = 16 → 2y = 1 → y = 0.5

Answer: x = 5, y = 0.5

Example 4: Substitution method

Solve: x + 3y = 10 and 2x - y = 3

From equation 1: x = 10 - 3y

Plug into equation 2: 2(10 - 3y) - y = 3

  • 20 - 6y - y = 3
  • -7y = -17
  • y = 17/7
  • x = 10 - 3(17/7) = 10 - 51/7 = 19/7

Answer: x = 19/7, y = 17/7

Word Problem Translation — The Real Skill

This is where most people get stuck. The math is easy; the translation from English to equation is the hard part. Here are the magic translations:

English → Math Dictionary
"the sum of two numbers"
→ x + y
"one number is 14 more than the other"
→ x = y + 14
"twice a number"
→ 2x
"a number decreased by 5"
→ x - 5
"the product of two numbers is 48"
→ xy = 48
"ratio of A to B is 3:5"
→ A/B = 3/5 or A = 3k, B = 5k
"is" or "was" or "will be"
→ = (equals)
"of" (as in "30% of")
→ × (multiply)

Example 5: Classic word problem

“The sum of two numbers is 50. One number is 14 more than the other. Find the numbers.”

Let the smaller number = x. Then the larger = x + 14.

  • x + (x + 14) = 50
  • 2x + 14 = 50
  • 2x = 36
  • x = 18

So the numbers are 18 and 32.

Quick check: 18 + 32 = 50 and 32 - 18 = 14. Done!

Example 6: Age problem

“A father is 3 times as old as his son. 12 years later, he will be twice as old as his son. Find their present ages.”

Let son’s age = x. Father’s age = 3x.

After 12 years: son = x + 12, father = 3x + 12.

Given: 3x + 12 = 2(x + 12)

  • 3x + 12 = 2x + 24
  • x = 12

Son is 12, father is 36.

Quick check: After 12 years — son is 24, father is 48. Is 48 = 2 × 24? Yes!

Shortcut: The “Assumed Mean” Trick for Symmetric Equations

When we see something like: “The sum of two numbers is 100, and the difference is 20.”

Instead of setting up full equations, just think:

  • Average = 100/2 = 50
  • Half the difference = 20/2 = 10
  • Numbers are 50 + 10 = 60 and 50 - 10 = 40

This works every time and saves a ton of time. In simple language, we’re just splitting the total evenly and then adjusting by half the gap.

Special Cases to Watch For

  1. No solution: When we end up with something like 0 = 5 — the equations are parallel lines that never meet (inconsistent system).

  2. Infinite solutions: When we get 0 = 0 — both equations are actually the same line (dependent system).

  3. Number of solutions test: For a₁x + b₁y = c₁ and a₂x + b₂y = c₂:

    • Unique solution: a₁/a₂ ≠ b₁/b₂
    • No solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂
    • Infinite solutions: a₁/a₂ = b₁/b₂ = c₁/c₂

Common Exam Variations

  • Digit problems: “A two-digit number…” → number = 10a + b (tens digit × 10 + units digit). If digits are reversed: 10b + a.
  • Fraction problems: “If 1 is added to numerator and 2 to denominator, fraction becomes 1/2” → (x+1)/(y+2) = 1/2.
  • Speed/distance word problems that reduce to linear equations.
  • Ages — present, past, future relationships.

Practice Problems

Problem 1: The sum of two numbers is 84. One is 3 times the other. Find them.

Problem 2: A two-digit number is 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.

Problem 3: Solve: 2x + 3y = 13 and 5x - y = 7.

Answers

Problem 1: Let smaller = x. Larger = 3x. So x + 3x = 84 → 4x = 84 → x = 21. Numbers are 21 and 63.

Problem 2: Let tens digit = a, units digit = b. Number = 10a + b.

  • 10a + b = 4(a + b) → 10a + b = 4a + 4b → 6a = 3b → b = 2a
  • 10a + b + 18 = 10b + a → 9a - 9b = -18 → a - b = -2
  • Substituting b = 2a: a - 2a = -2 → a = 2, b = 4. Number is 24.

Problem 3: From equation 2: y = 5x - 7. Plug into equation 1: 2x + 3(5x - 7) = 13 → 2x + 15x - 21 = 13 → 17x = 34 → x = 2, y = 3. Answer: x = 2, y = 3.

References