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Simultaneous (Linear) Equations — ICSE Class 9 Maths Important Questions

13 hand-picked ICSE Class 9 Maths important questions for Simultaneous (Linear) Equations, each with a full model answer — the formats and topics most likely to appear in your board exam.

13
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6
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32
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₹0
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Quick answer

High-yield ICSE Simultaneous Linear Equations questions ask you to solve a pair like 2x+3y=12, 3x−2y=52x+3y=12,\ 3x-2y=52x+3y=12, 3x-2y=5 by substitution, elimination, and cross-multiplication, and to form and solve two equations from word problems on numbers, ages, fractions, and money. Method-specific solving and word problems appear almost every year.

Syllabus reference (2026–27): CISCE (Council for the Indian School Certificate Examinations)

About Simultaneous (Linear) Equations

In the ICSE Class 9 Maths chapter Simultaneous Linear Equations you solve a pair of linear equations in two variables xxx and yyy using substitution, elimination, and cross-multiplication, and you translate word problems into two equations before solving them. A solution is the pair (x,y)(x,y)(x,y) that satisfies both equations at once.

Solving by substitutionSolving by eliminationSolving by cross-multiplicationEquations reducible to linear formWord problems

Key concepts & formulas

Elimination method

Make the coefficient of one variable equal in both equations, then add or subtract to eliminate it. For a1x+b1y=c1a_1x+b_1y=c_1a_1x+b_1y=c_1 and a2x+b2y=c2a_2x+b_2y=c_2a_2x+b_2y=c_2, multiply to match coefficients before combining.

Cross-multiplication

For a1x+b1y+c1=0a_1x+b_1y+c_1=0a_1x+b_1y+c_1=0 and a2x+b2y+c2=0a_2x+b_2y+c_2=0a_2x+b_2y+c_2=0, xb1c2−b2c1=yc1a2−c2a1=1a1b2−a2b1\dfrac{x}{b_1c_2-b_2c_1}=\dfrac{y}{c_1a_2-c_2a_1}=\dfrac{1}{a_1b_2-a_2b_1}x/b_1c_2-b_2c_1=y/c_1a_2-c_2a_1=1/a_1b_2-a_2b_1.

Reducible equations

Equations in 1x\dfrac1x1x and 1y\dfrac1y1y become linear by substituting u=1x, v=1yu=\dfrac1x,\ v=\dfrac1yu=1x, v=1y; solve for u,vu,vu,v then back-substitute.

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Important questions with answers

Try each on paper first, then reveal the model answer to check your method.

Question typeCountMarks
MCQ41
Assertion–Reason11
Very Short22
Short Answer33
Long Answer25
Case-based14

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The solution of x+y=7x+y=7x+y=7 and x−y=3x-y=3x-y=3 is:

  1. (a)

    x=5, y=2x=5,\ y=2x=5, y=2

  2. (b)

    x=2, y=5x=2,\ y=5x=2, y=5

  3. (c)

    x=4, y=3x=4,\ y=3x=4, y=3

  4. (d)

    x=3, y=4x=3,\ y=4x=3, y=4

Show model answer

Answer: (a) x=5, y=2x=5,\ y=2x=5, y=2.

Adding the two equations: 2x=10⇒x=52x=10\Rightarrow x=52x=10 x=5. Then y=7−5=2y=7-5=2y=7-5=2.

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Q2MCQEasy1 mark

For the equations a1x+b1y+c1=0a_1x+b_1y+c_1=0a_1x+b_1y+c_1=0 and a2x+b2y+c2=0a_2x+b_2y+c_2=0a_2x+b_2y+c_2=0, the cross-multiplication denominator of 1 \dfrac{1}{\ }1/ is:

  1. (a)

    a1b2−a2b1a_1b_2-a_2b_1a_1b_2-a_2b_1

  2. (b)

    b1c2−b2c1b_1c_2-b_2c_1b_1c_2-b_2c_1

  3. (c)

    c1a2−c2a1c_1a_2-c_2a_1c_1a_2-c_2a_1

  4. (d)

    a1a2−b1b2a_1a_2-b_1b_2a_1a_2-b_1b_2

Show model answer

Answer: (a) a1b2−a2b1a_1b_2-a_2b_1a_1b_2-a_2b_1.

The standard formula is xb1c2−b2c1=yc1a2−c2a1=1a1b2−a2b1\dfrac{x}{b_1c_2-b_2c_1}=\dfrac{y}{c_1a_2-c_2a_1}=\dfrac{1}{a_1b_2-a_2b_1}x/b_1c_2-b_2c_1=y/c_1a_2-c_2a_1=1/a_1b_2-a_2b_1, so the last denominator is a1b2−a2b1a_1b_2-a_2b_1a_1b_2-a_2b_1.

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Q3MCQModerate1 mark

If 2x+3y=2\dfrac{2}{x}+\dfrac{3}{y}=22/x+3/y=2 and 5x−3y=5\dfrac{5}{x}-\dfrac{3}{y}=55/x-3/y=5, then xxx equals:

  1. (a)

    111

  2. (b)

    222

  3. (c)

    12\dfrac1212

  4. (d)

    333

Show model answer

Answer: (a) 111.

Add the equations: 7x=7⇒x=1\dfrac{7}{x}=7\Rightarrow x=17/x=7 x=1.

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Q4MCQHOTS1 mark

The pair 3x−2y=43x-2y=43x-2y=4 and 9x−6y=k9x-6y=k9x-6y=k has infinitely many solutions when kkk equals:

  1. (a)

    121212

  2. (b)

    444

  3. (c)

    666

  4. (d)

    999

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Answer: (a) 121212.

For infinitely many solutions a1a2=b1b2=c1c2\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}a_1/a_2=b_1/b_2=c_1/c_2. Here 39=−2−6=13\dfrac{3}{9}=\dfrac{-2}{-6}=\dfrac133/9=-2/-6=13, so 4k=13⇒k=12\dfrac{4}{k}=\dfrac13\Rightarrow k=124/k=13 k=12.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): The equations x+2y=5x+2y=5x+2y=5 and 2x+4y=102x+4y=102x+4y=10 have infinitely many solutions.

Reason (R): Two linear equations have infinitely many solutions when they represent the same straight line.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

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Answer: (a) The second equation is exactly twice the first, so both represent the same line and there are infinitely many solutions; R correctly explains A.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Solve by substitution: y=2x−3y=2x-3y=2x-3 and x+y=9x+y=9x+y=9.

Show model answer

Substitute y=2x−3y=2x-3y=2x-3 into x+y=9x+y=9x+y=9:

x+(2x−3)=9⇒3x=12⇒x=4.x+(2x-3)=9\Rightarrow 3x=12\Rightarrow x=4.x+(2x-3)=9 3x=12 x=4.

Then y=2(4)−3=5y=2(4)-3=5y=2(4)-3=5.

Solution: x=4, y=5x=4,\ y=5x=4, y=5.

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Q7Very ShortModerate2 marks

Solve by elimination: 3x+4y=103x+4y=103x+4y=10 and 2x−2y=22x-2y=22x-2y=2.

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Multiply the second equation by 222: 4x−4y=44x-4y=44x-4y=4.

Add to 3x+4y=103x+4y=103x+4y=10:

7x=14⇒x=2.7x=14\Rightarrow x=2.7x=14 x=2.

From 2x−2y=22x-2y=22x-2y=2: 4−2y=2⇒y=14-2y=2\Rightarrow y=14-2y=2 y=1.

Solution: x=2, y=1x=2,\ y=1x=2, y=1.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

Solve by cross-multiplication: 2x+3y=172x+3y=172x+3y=17 and 3x−2y=63x-2y=63x-2y=6.

Show model answer

Write in the form ax+by+c=0ax+by+c=0ax+by+c=0:

2x+3y−17=0,3x−2y−6=0.2x+3y-17=0,\qquad 3x-2y-6=0.2x+3y-17=0, 3x-2y-6=0.

Apply cross-multiplication:

x(3)(−6)−(−2)(−17)=y(−17)(3)−(−6)(2)=1(2)(−2)−(3)(3).\frac{x}{(3)(-6)-(-2)(-17)}=\frac{y}{(-17)(3)-(-6)(2)}=\frac{1}{(2)(-2)-(3)(3)}.x/(3)(-6)-(-2)(-17)=y/(-17)(3)-(-6)(2)=1/(2)(-2)-(3)(3).

x−18−34=y−51+12=1−4−9=1−13.\frac{x}{-18-34}=\frac{y}{-51+12}=\frac{1}{-4-9}=\frac{1}{-13}.x/-18-34=y/-51+12=1/-4-9=1/-13.

x−52=1−13⇒x=4,y−39=1−13⇒y=3.\frac{x}{-52}=\frac{1}{-13}\Rightarrow x=4,\qquad \frac{y}{-39}=\frac{1}{-13}\Rightarrow y=3.x/-52=1/-13 x=4, y/-39=1/-13 y=3.

Solution: x=4, y=3x=4,\ y=3x=4, y=3.

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Q9Short AnswerModerate3 marks

Solve: 2x+3y=13\dfrac{2}{x}+\dfrac{3}{y}=132/x+3/y=13 and 5x−4y=−2\dfrac{5}{x}-\dfrac{4}{y}=-25/x-4/y=-2.

Show model answer

Let u=1xu=\dfrac1xu=1x and v=1yv=\dfrac1yv=1y. The equations become:

2u+3v=13,5u−4v=−2.2u+3v=13,\qquad 5u-4v=-2.2u+3v=13, 5u-4v=-2.

Multiply the first by 444 and the second by 333:

8u+12v=52,15u−12v=−6.8u+12v=52,\qquad 15u-12v=-6.8u+12v=52, 15u-12v=-6.

Add: 23u=46⇒u=223u=46\Rightarrow u=223u=46 u=2. Then 2(2)+3v=13⇒v=32(2)+3v=13\Rightarrow v=32(2)+3v=13 v=3.

So 1x=2⇒x=12\dfrac1x=2\Rightarrow x=\dfrac121x=2 x=12 and 1y=3⇒y=13\dfrac1y=3\Rightarrow y=\dfrac131y=3 y=13.

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Q10Short AnswerEasy3 marks

The sum of two numbers is 353535 and their difference is 131313. Find the numbers.

Show model answer

Let the numbers be xxx and yyy with x>yx>yx>y.

x+y=35,x−y=13.x+y=35,\qquad x-y=13.x+y=35, x-y=13.

Add: 2x=48⇒x=242x=48\Rightarrow x=242x=48 x=24.

Subtract: 2y=22⇒y=112y=22\Rightarrow y=112y=22 y=11.

The numbers are 242424 and 111111.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

A fraction becomes 12\dfrac1212 when 111 is subtracted from the numerator, and it becomes 13\dfrac1313 when 888 is added to the denominator. Find the fraction.

Show model answer

Let the fraction be xy\dfrac{x}{y}x/y.

Condition 1: x−1y=12⇒2(x−1)=y⇒2x−y=2.\dfrac{x-1}{y}=\dfrac12\Rightarrow 2(x-1)=y\Rightarrow 2x-y=2.x-1/y=12 2(x-1)=y 2x-y=2.

Condition 2: xy+8=13⇒3x=y+8⇒3x−y=8.\dfrac{x}{y+8}=\dfrac13\Rightarrow 3x=y+8\Rightarrow 3x-y=8.x/y+8=13 3x=y+8 3x-y=8.

Subtract the first from the second:

(3x−y)−(2x−y)=8−2⇒x=6.(3x-y)-(2x-y)=8-2\Rightarrow x=6.(3x-y)-(2x-y)=8-2 x=6.

From 2x−y=22x-y=22x-y=2: 12−y=2⇒y=1012-y=2\Rightarrow y=1012-y=2 y=10.

The fraction is 610\dfrac{6}{10}6/10, i.e. 35\dfrac{3}{5}3/5.

Check: 6−110=510=12\dfrac{6-1}{10}=\dfrac{5}{10}=\dfrac126-1/10=5/10=12 and 610+8=618=13\dfrac{6}{10+8}=\dfrac{6}{18}=\dfrac136/10+8=6/18=13. Correct.

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Q12Long AnswerHOTS5 marks

Five years ago a man was seven times as old as his son. Five years hence the father will be three times as old as his son. Find their present ages.

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Let the present ages be xxx years (father) and yyy years (son).

Five years ago: father =x−5=x-5=x-5, son =y−5=y-5=y-5.

x−5=7(y−5)⇒x−5=7y−35⇒x−7y=−30.(1)x-5=7(y-5)\Rightarrow x-5=7y-35\Rightarrow x-7y=-30.\quad(1)x-5=7(y-5) x-5=7y-35 x-7y=-30.(1)

Five years hence: father =x+5=x+5=x+5, son =y+5=y+5=y+5.

x+5=3(y+5)⇒x+5=3y+15⇒x−3y=10.(2)x+5=3(y+5)\Rightarrow x+5=3y+15\Rightarrow x-3y=10.\quad(2)x+5=3(y+5) x+5=3y+15 x-3y=10.(2)

Subtract (1)(1)(1) from (2)(2)(2):

(x−3y)−(x−7y)=10−(−30)⇒4y=40⇒y=10.(x-3y)-(x-7y)=10-(-30)\Rightarrow 4y=40\Rightarrow y=10.(x-3y)-(x-7y)=10-(-30) 4y=40 y=10.

From (2)(2)(2): x−30=10⇒x=40x-30=10\Rightarrow x=40x-30=10 x=40.

The father is 404040 years old and the son is 101010 years old.

Check: Five years ago 35=7×535=7\times535=7×5; five years hence 45=3×1545=3\times1545=3×15. Correct.

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

At a stationery shop, Ravi buys 222 pens and 333 notebooks for Rs 80\textrm{Rs }80Rs 80, while Meena buys 333 pens and 222 notebooks for Rs 70\textrm{Rs }70Rs 70. Taking the cost of a pen as Rs x\textrm{Rs }xRs x and a notebook as Rs y\textrm{Rs }yRs y:

(i) Form the two equations.

(ii) Find the cost of one pen.

(iii) Find the cost of one notebook.

(iv) Find the cost of 555 pens and 444 notebooks.

Show model answer

(i) From Ravi: 2x+3y=802x+3y=802x+3y=80. From Meena: 3x+2y=703x+2y=703x+2y=70.

(ii) Multiply the first by 333 and the second by 222:

6x+9y=240,6x+4y=140.6x+9y=240,\qquad 6x+4y=140.6x+9y=240, 6x+4y=140.

Subtract: 5y=100⇒y=205y=100\Rightarrow y=205y=100 y=20. Then 2x+3(20)=80⇒2x=20⇒x=102x+3(20)=80\Rightarrow 2x=20\Rightarrow x=102x+3(20)=80 2x=20 x=10.

Cost of one pen =Rs 10=\textrm{Rs }10=Rs 10.

(iii) Cost of one notebook =Rs 20=\textrm{Rs }20=Rs 20.

(iv) 5x+4y=5(10)+4(20)=50+80=Rs 1305x+4y=5(10)+4(20)=50+80=\textrm{Rs }1305x+4y=5(10)+4(20)=50+80=Rs 130.

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All ICSE Class 9 Maths Chapters

Rational and Irrational NumbersCompound Interest (Without Using Formula)Compound Interest (Using Formula)ExpansionsFactorisationSimultaneous (Linear) EquationsIndices (Exponents)LogarithmsTriangles (Congruency in Triangles)Isosceles TrianglesInequalitiesMid-point and Its Converse (Including Intercept Theorem)Pythagoras TheoremRectilinear Figures (Quadrilaterals)Construction of PolygonsArea TheoremsCircleStatistics (Classification of Data, Tabulation)Mean and MedianArea and Perimeter of Plane FiguresSolids (Surface Area and Volume of 3-D Solids)Trigonometrical RatiosTrigonometrical Ratios of Standard AnglesSolution of Right TrianglesComplementary AnglesCo-ordinate GeometryGraphical Solution (Simultaneous Linear Equations, Graphically)Distance Formula

Frequently asked questions

  • Are these Simultaneous (Linear) Equations important questions free?
    Yes. All 13 ICSE Class 9 Maths important questions for Simultaneous (Linear) Equations are free, with full model answers and no login required.
  • Do these Simultaneous (Linear) Equations questions follow the latest ICSE syllabus?
    Yes — they are aligned to the CISCE latest syllabus syllabus for ICSE Class 9 Maths, so nothing here is outside the current course.
  • How should I practise the Simultaneous (Linear) Equations important questions?
    Attempt each question on paper first, then reveal the model answer to check your method — not just the final result. Re-do anything you got wrong the same day.
  • What types of questions are covered for Simultaneous (Linear) Equations?
    A full mix — multiple-choice questions, assertion–reason questions, very short answer questions, short answer questions, long answer questions, case-based questions — so every format in the ICSE paper is covered.

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