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State Board · Class 10 · Mathematics Part 1

Lesson 1: Linear Equations in Two Variables

5 concepts · 40 practice questions with answers · Free to read

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Linear equations and simultaneous equations

From daily life

A shopkeeper says: "2 notebooks and 3 pens cost ₹ 70." That alone does not tell the price of either item. With one more fact, such as "3 notebooks and 2 pens cost ₹ 80", both prices can be found. Two facts, two unknowns.

Simple definition

A linear equation in two variables (दोन चलांतील रेषीय समीकरण) has the form ax + by + c = 0, where a, b, c are real numbers and a, b are not both zero.

Rule 1: Linear or not? — The degree of each term with a variable must be 1. 2x + 3y = 7 becomes 2x + 3y − 7 = 0, so it is linear. But xy = 6 and x2 + y = 3 are not linear, because xy and x2 have degree 2.

  • Easy : Easy: 5x − 2y = 9 is linear.
  • Medium : Medium: 3x = 4y − 12 becomes 3x − 4y + 12 = 0.
  • Exam-type : Exam: Is 2xy + 3 = 0 linear? No, the term xy has degree 2.

Rule 2: Simultaneous equations — Two linear equations taken together are simultaneous equations (एकसामायिक समीकरणे). Solve by elimination (remove one variable) or substitution. x + y = 10 and x − y = 2: adding gives 2x = 12, so x = 6 and y = 4. 3x + 2y = 16 and x − y = 2: put x = y + 2, so 5y + 6 = 16, y = 2, x = 4.

  • Easy : Easy: x + y = 7 and x − y = 1 give x = 4, y = 3.
  • Medium : Medium: 2x + y = 11 and x − y = 1: add to get 3x = 12, x = 4, y = 3.
  • Exam-type : Exam: 4x + 3y = 17 and 3x + 4y = 18 (coefficients swapped): adding gives x + y = 5; subtracting gives x − y = −1; so x = 2 and y = 3.
Remember

General form ax + by + c = 0. Always check the answer in both equations.

Common mistake

Checking the answer in only one equation. It must satisfy both.

Shortcut

Memory clue: "दोन चल = दोन समीकरणे".

Spot the difference

MethodIdea
EliminationAdd or subtract to remove one variable
SubstitutionPut one variable in terms of the other

Practice questions

Easy
  1. 1What is the general form of a linear equation in two variables?
  2. 2Is x2 + y = 5 linear?
  3. 3Write 4x = 3y + 8 in general form.
Medium
  1. 4Solve x + y = 9, x − y = 3.
  2. 5Solve 2x + y = 10, x = 3.
  3. 6Solve 3x + y = 14, x − y = 2.
Challenging
  1. 7Solve 5x + 2y = 26, 3x + 4y = 24.
  2. 8Why must the solution satisfy both equations?
Show answers (try them yourself first)
  1. 1ax + by + c = 0.
  2. 2No.
  3. 34x − 3y − 8 = 0.
  4. 4x = 6, y = 3.
  5. 5y = 4.
  6. 64x = 16, x = 4, y = 2.
  7. 7Double the first: 10x + 4y = 52; subtract: 7x = 28, x = 4, y = 3.
  8. 8Simultaneous means both are true at the same time.

Graphical method

From daily life

Draw the path of a cyclist who goes 1 km further every minute: you get a straight line. Every linear equation in two variables makes a straight line on a graph. Two such lines either cross, run side by side, or lie on top of each other.

Simple definition

The graph of a linear equation in two variables is a straight line. The solution of simultaneous equations is the point of intersection of their lines.

Rule 1: Drawing a line and solving graphically — Find at least 3 or 4 pairs (put x = 0, then y = 0, to get easy points) and check that they lie on one line. For 2x + y = 6: (0, 6), (3, 0), (1, 4), (2, 2). The line y = 3 is parallel to the X-axis and x = −2 is parallel to the Y-axis. To solve x + y = 5 and x − y = 1, draw both lines; they meet at (3, 2), so x = 3, y = 2.

Two lines x plus y equals 6 and x minus y equals 2 cross at the point (4, 2)−112345678−2−11234567xySolving x + y = 6 and x − y = 2 by graph(4, 2)x + y = 6x − y = 2

In the figure : In the figure the solution of simultaneous equations is the point where their lines meet.

  • Easy : Easy: (1, 4) is on 2x + y = 6 since 2 + 4 = 6.
  • Medium : Medium: x + y = 5 passes through (0, 5) and (5, 0).
  • Exam-type : Exam: The lines x + y = 5 and x − y = 1 meet at (3, 2): 3 + 2 = 5 and 3 − 2 = 1.

Rule 2: Three possibilities — For a1x + b1y = c1 and a2x + b2y = c2: if the lines cross, there is exactly one solution; if they are parallel, there is no solution; if they coincide, there are infinitely many. x + y = 3 and 2x + 2y = 6 are the same line (coincide). 3x + 2y = 6 and 6x + 4y = 18 are parallel, since 36 = 24 = 12 but 618 = 13.

Three cases of two linear equations: crossing lines, parallel lines and coinciding linesOne solutionlines crossa₁/a₂ ≠ b₁/b₂No solutionparallel linesa₁/a₂ = b₁/b₂ ≠ c₁/c₂Infinite solutionssame linea₁/a₂ = b₁/b₂ = c₁/c₂

In the figure : In the figure crossing, parallel and coinciding lines give one, no or infinitely many solutions.

Live figure

Graphical method

In the app you can move this figure with your finger.

Open in the app
  • Easy : Easy: Crossing lines give one solution.
  • Medium : Medium: Parallel lines give no solution.
  • Exam-type : Exam: x + 2y = 4 and 3x + 6y = 12: 13 = 26 = 412, so the lines coincide.
Remember

Cross: 1 solution. Parallel: none. Same line: infinitely many.

Common mistake

Reading the point off carelessly. Always check the coordinates in both equations.

Shortcut

Memory clue: "छेद = एक; समांतर = शून्य; एकरूप = अनंत".

Spot the difference

LinesSolutions
IntersectingOne
ParallelNone
CoincidingInfinitely many

Practice questions

Easy
  1. 1What is the graph of a linear equation in two variables?
  2. 2How do we solve simultaneous equations graphically?
  3. 3Find the x- and y-intercepts of x + y = 6.
Medium
  1. 4Is (2, 3) on 3x − y = 3?
  2. 5Number of solutions if the lines are parallel?
  3. 6Number of solutions if the lines coincide?
Challenging
  1. 7Do x + y = 2 and 2x + 2y = 7 have a solution?
  2. 8Solve graphically: x + y = 6, x − y = 2.
Show answers (try them yourself first)
  1. 1A straight line.
  2. 2Find the point where the lines meet.
  3. 3(6, 0) and (0, 6).
  4. 46 − 3 = 3, yes.
  5. 5None.
  6. 6Infinitely many.
  7. 7No, they are parallel.
  8. 8They meet at (4, 2).

Determinants and Cramer's method

From daily life

Think of a small machine that takes the numbers from two equations and gives out the answer by a fixed recipe: multiply crosswise, subtract. That recipe is a determinant, and using it to solve equations is Cramer's method.

Simple definition

A determinant (सारणिक) of rows (a, b) and (c, d) has the value ad − bc. Cramer's method uses D, Dx, Dy: x = DxD, y = DyD.

Rule 1: Determinant value — Rows (2, 5) and (1, 3): value = 2 × 3 − 5 × 1 = 1. Rows (4, 3) and (2, 1): 4 × 1 − 3 × 2 = −2. Rows (−3, 2) and (5, 4): −12 − 10 = −22.

  • Easy : Easy: Rows (3, 1) and (2, 4): 12 − 2 = 10.
  • Medium : Medium: Rows (5, −2) and (1, 3): 15 + 2 = 17.
  • Exam-type : Exam: Rows (6, 3) and (4, 2): 12 − 12 = 0.

Rule 2: Cramer's method — For a1x + b1y = c1 and a2x + b2y = c2: D = a1b2 − a2b1 (must not be 0); Dx replaces the x-column by the constants; Dy replaces the y-column by the constants. Example 3x + 2y = 7 and x + 4y = 9: D = 12 − 2 = 10; Dx = 7 × 4 − 9 × 2 = 10; Dy = 3 × 9 − 1 × 7 = 20. So x = 1 and y = 2. Check: 3 + 4 = 7 and 1 + 8 = 9.

Three determinants D, Dx and Dy for 2x + y = 7 and x + 3y = 11 giving x = 2 and y = 32x + y = 7 and x + 3y = 11D21136 − 1 = 5Dx7111321 − 11 = 10Dy2711122 − 7 = 15x = Dx ÷ D = 10 ÷ 5 = 2y = Dy ÷ D = 15 ÷ 5 = 3(the orange column is replaced by the constants 7 and 11)

In the figure : In the figure Cramer's method: D keeps the coefficients; in Dx and Dy one column is replaced by the constants.

  • Easy : Easy: x + y = 5 and x − y = 1: D = −2, Dx = −6, Dy = −4, so x = 3, y = 2.
  • Medium : Medium: 2x − y = 3 and x + 3y = 5: D = 7, Dx = 14, Dy = 7, so x = 2, y = 1.
  • Exam-type : Exam: 2x + 4y = 6 and x + 2y = 5 have D = 4 − 4 = 0, so Cramer's method cannot give a unique solution.
Remember

D must not be zero. x = Dx ÷ D, y = Dy ÷ D.

Common mistake

Writing the equations without bringing them to the form ax + by = c. Arrange them first.

Shortcut

Memory clue: "क्रॉस गुणा - वजा; D शून्य नको".

Spot the difference

DeterminantReplace
DNothing
Dxx-column by constants
Dyy-column by constants

Practice questions

Easy
  1. 1Value of the determinant with rows (a, b) and (c, d)?
  2. 2After whom is Cramer's method named?
  3. 3Rows (7, 2) and (3, 1): value?
Medium
  1. 4Formula for x?
  2. 5What if D = 0?
  3. 6Solve x + 2y = 8, 3x − y = 3 by Cramer.
Challenging
  1. 7Solve 4x + y = 14, 2x + 3y = 12 by Cramer.
  2. 8Find the number of solutions of 2x + 3y = 6 and 4x + 6y = 12.
Show answers (try them yourself first)
  1. 1ad − bc.
  2. 2Gabriel Cramer, a Swiss mathematician.
  3. 37 − 6 = 1.
  4. 4Dx ÷ D.
  5. 5No unique solution.
  6. 6D = −7, Dx = −14, Dy = −21; x = 2, y = 3.
  7. 7D = 10, Dx = 30, Dy = 20; x = 3, y = 2.
  8. 8D = 0 and the lines coincide: infinitely many.

Equations reducible to linear equations

From daily life

Some equations look hard because the unknown is in the denominator. But if you give a nickname to "1 over x", the equation turns into an ordinary linear one.

Simple definition

Some non-linear equations become linear when suitable expressions are replaced by new variables. The denominator of a fraction can never be zero.

Rule 1: Fractions with x and y — 2/x + 3/y = 12 and 1/x + 1/y = 5. Put m = 1x, n = 1y. Then 2m + 3n = 12 and m + n = 5. From the second m = 5 − n, so 10 − 2n + 3n = 12, n = 2, m = 3. Hence x = 13, y = 12. Check: 2 ÷ 13 = 6, 3 ÷ 12 = 6, sum 12; and 3 + 2 = 5.

Four steps: replace 1/x and 1/y by m and n, solve, then go back to x and yMaking it linear with new variables2/x + 3/y = 81/x + 1/y = 3Let 1/x = m, 1/y = n2m + 3n = 8, m + n = 3Solve:m = 1, n = 2x = 1/m = 1y = 1/n = 1/2

In the figure : In the figure new variables turn the fractional equations into linear ones.

  • Easy : Easy: If 1x = 2, then x = 12.
  • Medium : Medium: If m = 5 and n = 4 then x = 15 and y = 14.
  • Exam-type : Exam: x = 0 is not allowed, since 1/x would be undefined.

Rule 2: Fractions with x + y and x − y — Solve 10/(x + y) + 2/(x − y) = 4 and 15/(x + y) − 4/(x − y) = −1. Put a = 1(x + y), b = 1(x − y). Then 10a + 2b = 4 and 15a − 4b = −1. Double the first: 20a + 4b = 8; add: 35a = 7, a = 15, and b = 1. So x + y = 5, x − y = 1, giving x = 3, y = 2.

  • Easy : Easy: a = 15 means x + y = 5.
  • Medium : Medium: b = 1 means x − y = 1.
  • Exam-type : Exam: Check: 10 ÷ 5 + 2 ÷ 1 = 4 and 15 ÷ 5 − 4 ÷ 1 = −1.
Remember

Replace, solve for the new variables, then go back to x and y.

Common mistake

Stopping at m and n. The question asks for x and y.

Shortcut

Memory clue: "बदला → सोडवा → परत या".

Spot the difference

OriginalNew variable
1xm
1yn
1(x + y)a

Practice questions

Easy
  1. 1Why do we use new variables?
  2. 2If m = 1x and m = 4, find x.
  3. 3Can the denominator be zero?
Medium
  1. 4Write 6/x + 2/y = 8 in terms of m and n.
  2. 5Solve 1/x + 1/y = 5, 1/x − 1/y = 1.
  3. 6Solve 2/x + 1/y = 7, 1/x + 1/y = 4.
Challenging
  1. 7If x + y = 5 and x − y = 1, find 1/(x + y) and 1/(x − y).
  2. 8Why do we substitute back at the end?
Show answers (try them yourself first)
  1. 1To make the equations linear.
  2. 214.
  3. 3No.
  4. 46m + 2n = 8.
  5. 5m = 3, n = 2; x = 13, y = 12.
  6. 6Subtract: m = 3, n = 1; x = 13, y = 1.
  7. 715 and 1.
  8. 8The answer is needed for x and y, not for the new variables.

Applications of simultaneous equations

From daily life

A word problem is a story with two hidden numbers. Give the hidden numbers names x and y, turn each sentence of the story into an equation, and you have two simultaneous equations.

Simple definition

To solve a word problem: assume two variables, form two equations from the two conditions, solve them, and check the answer in the story.

Rule 1: Shapes and prices — A rectangle has perimeter 50 cm and its length is 5 cm more than its breadth. l + b = 25 and l − b = 5, so l = 15 cm and b = 10 cm. Two notebooks and three pens cost ₹ 70; three notebooks and two pens cost ₹ 80. 2x + 3y = 70 and 3x + 2y = 80: adding gives x + y = 30; subtracting gives x − y = 10, so a notebook costs ₹ 20 and a pen ₹ 10.

Steps for a word problem: choose variables, write two equations, solve and checkFrom a story to two equationsPen = x, book = y2x + y = 25x + 2y = 35Solve:x = 5, y = 15Check:10 + 15 = 25 ✓5 + 30 = 35 ✓

In the figure : In the figure a word problem becomes two equations, which are solved and checked.

  • Easy : Easy: Two numbers have sum 12 and difference 2: they are 7 and 5.
  • Medium : Medium: A rectangle has l + b = 25 and l − b = 5, so l = 15 and b = 10.
  • Exam-type : Exam: Check the prices: 2 × 20 + 3 × 10 = 70 and 3 × 20 + 2 × 10 = 80.

Rule 2: Digits and boats — A two-digit number has digit sum 9 and the number formed by reversing the digits is 27 more. With tens digit x and units digit y: x + y = 9 and 9(y − x) = 27, so y − x = 3. Then y = 6, x = 3, and the number is 36 (63 − 36 = 27). A boat of speed x km/h in a stream of speed y km/h: upstream speed x − y, downstream speed x + y. If it covers 15 km upstream and 27 km downstream in 6 hours, and 25 km upstream and 18 km downstream in 7 hours, put a = 1(x − y), b = 1(x + y): 15a + 27b = 6 and 25a + 18b = 7 give a = 15, b = 19. So x − y = 5, x + y = 9, and the boat's speed is 7 km/h and the stream's 2 km/h.

  • Easy : Easy: Digits 3 and 6 add to 9.
  • Medium : Medium: Upstream speed = 7 − 2 = 5 km/h; downstream = 9 km/h.
  • Exam-type : Exam: Check the boat: 15 ÷ 5 + 27 ÷ 9 = 6 and 25 ÷ 5 + 18 ÷ 9 = 7.
Remember

Name the unknowns, make two equations, solve, check in the story.

Common mistake

Giving an answer without units or without checking. Always state the units and check.

Shortcut

Memory clue: "नाव द्या → समीकरण → सोडवा → तपासा".

Spot the difference

TypeTypical equations
Sum and differencex + y = s, x − y = d
Pricesax + by = p, cx + dy = q
Boat and streamspeeds x − y and x + y

Practice questions

Easy
  1. 1What is the first step in a word problem?
  2. 2How many equations do we need for two variables?
  3. 3Two numbers add to 20 and differ by 4. Find them.
Medium
  1. 4A pen and a pencil cost ₹ 15; the pen costs ₹ 5 more. Find both prices.
  2. 5A two-digit number has digit sum 7 and the reversed number is 9 more. Find it.
  3. 6A boat's upstream speed is 6 km/h and downstream 10 km/h. Find the boat's and stream's speeds.
Challenging
  1. 7The perimeter of a rectangle is 36 cm; length is double the breadth. Find the sides.
  2. 8Why must we check the answer in the story?
Show answers (try them yourself first)
  1. 1Assume variables for the unknowns.
  2. 2Two.
  3. 312 and 8.
  4. 4Pen ₹ 10, pencil ₹ 5.
  5. 5y − x = 1, so 34.
  6. 6Boat 8 km/h, stream 2 km/h.
  7. 7l + b = 18, l = 2b: b = 6, l = 12.
  8. 8To make sure it fits the conditions and makes sense.

Quick revision

  • Graphical method: the solution is the point of intersection; Cramer's rule: x = Dx ÷ D and y = Dy ÷ D with D not equal to zero.
  • Replace 1/x and 1/y, or 1/(x − y) and 1/(x + y), by new variables to get linear equations.
  • In word problems, form two equations from the two conditions and solve them.