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

13 ICSE Class 9 Maths practice questions on Simultaneous (Linear) Equations, each with a full model answer, covering 5 topics from the chapter in the question formats used in the exam.

By The Classmate AI Editorial Team

Reviewed by Classmate AI Team · 18 October 2026

Last updated 25 October 2026

13
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Key concepts
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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 elimination, substitution and cross-multiplication, and to frame and solve two equations from word problems on numbers, ages, fractions and money.

Syllabus reference (2026–27): ICSE 2028 Mathematics syllabus (PDF) · CISCE (Council for the Indian School Certificate Examinations)

About Simultaneous (Linear) Equations

Within 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 in $\frac1x$ and $\frac1y$Word 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_1a₁x+b₁y=c₁ and a2x+b2y=c2a_2x+b_2y=c_2a₂x+b₂y=c₂, multiply to match coefficients before combining.

Cross-multiplication

For a1x+b1y+c1=0a_1x+b_1y+c_1=0a₁x+b₁y+c₁=0 and a2x+b2y+c2=0a_2x+b_2y+c_2=0a₂x+b₂y+c₂=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/(b₁c₂-b₂c₁)=y/(c₁a₂-c₂a₁)=1/(a₁b₂-a₂b₁).

Reducible equations

Equations in 1x\dfrac1x1/x and 1y\dfrac1y1/y become linear by substituting u=1x, v=1yu=\dfrac1x,\ v=\dfrac1yu=1/x, v=1/y; 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.

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

  1. (a)

    a1b2−a2b1a_1b_2-a_2b_1a₁b₂-a₂b₁

  2. (b)

    b1c2−b2c1b_1c_2-b_2c_1b₁c₂-b₂c₁

  3. (c)

    c1a2−c2a1c_1a_2-c_2a_1c₁a₂-c₂a₁

  4. (d)

    a1a2−b1b2a_1a_2-b_1b_2a₁a₂-b₁b₂

Show model answer

Answer: (a) a1b2−a2b1a_1b_2-a_2b_1a₁b₂-a₂b₁.

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/(b₁c₂-b₂c₁)=y/(c₁a₂-c₂a₁)=1/(a₁b₂-a₂b₁), so the last denominator is a1b2−a2b1a_1b_2-a_2b_1a₁b₂-a₂b₁.

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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\dfrac121/2

  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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Q4MCQEasy1 mark

The solution of 3x+2y=123x+2y=123x+2y=12 and x−y=−1x-y=-1x-y=-1 is:

  1. (a)

    x=2, y=3x=2,\ y=3x=2, y=3

  2. (b)

    x=3, y=2x=3,\ y=2x=3, y=2

  3. (c)

    x=4, y=0x=4,\ y=0x=4, y=0

  4. (d)

    x=1, y=2x=1,\ y=2x=1, y=2

Show model answer

Answer: (a) x=2, y=3x=2,\ y=3x=2, y=3. From x−y=−1x-y=-1x-y=-1, x=y−1x=y-1x=y-1. Then 3(y−1)+2y=12⇒5y=15⇒y=33(y-1)+2y=12\Rightarrow5y=15\Rightarrow y=33(y-1)+2y=12⇒5y=15⇒ y=3, x=2x=2x=2. Check: 6+6=126+6=126+6=12 and 2−3=−12-3=-12-3=-1.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonEasy1 mark

Assertion (A): In the substitution method, the value of one variable taken from one equation is put into the other equation.

Reason (R): This turns the pair into a single equation in one variable, which can be solved directly.

  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

Show model answer

Answer: (a) Both are true and R explains A: substituting removes one variable, leaving one linear equation in one variable.

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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.

Show model answer

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=1/x and v=1yv=\dfrac1yv=1/y. 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=\dfrac121/x=2⇒ x=1/2 and 1y=3⇒y=13\dfrac1y=3\Rightarrow y=\dfrac131/y=3⇒ y=1/3.

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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\dfrac121/2 when 111 is subtracted from the numerator, and it becomes 13\dfrac131/3 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=1/2⇒ 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)=1/3⇒ 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}=\dfrac12(6-1)/10=5/10=1/2 and 610+8=618=13\dfrac{6}{10+8}=\dfrac{6}{18}=\dfrac136/(10+8)=6/18=1/3. 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.

Show model answer

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

  • What are the methods to solve simultaneous linear equations in ICSE Class 9?
    Substitution, elimination (equal coefficients) and cross-multiplication, plus reducible equations in 1x\tfrac1x1/x and 1y\tfrac1y1/y. This page has 13 questions using all of them with answers, including age, fraction and shop-purchase word problems.
  • How do you solve 2x + 3y = 17 and 3x − 2y = 6 by elimination?
    Make the yyy coefficients equal: multiply the first by 222 and the second by 333 to get 4x+6y=344x+6y=344x+6y=34 and 9x−6y=189x-6y=189x-6y=18. Adding gives 13x=5213x=5213x=52, so x=4x=4x=4, and then y=3y=3y=3. Check in both original equations.
  • How do you solve equations with 1/x and 1/y?
    Substitute u=1xu=\tfrac1xu=1/x and v=1yv=\tfrac1yv=1/y to make them linear. For 2x+3y=13\tfrac2x+\tfrac3y=132/x+3/y=13 and 5x−4y=−2\tfrac5x-\tfrac4y=-25/x-4/y=-2: 2u+3v=132u+3v=132u+3v=13 and 5u−4v=−25u-4v=-25u-4v=-2 give u=2u=2u=2, v=3v=3v=3, so x=12x=\tfrac12x=1/2 and y=13y=\tfrac13y=1/3. Remember to convert back at the end; leaving the answer as uuu and vvv loses the final mark.
  • How do you solve age problems with simultaneous equations?
    Let the present ages be fff and sss and write one equation for each condition. Five years ago the father was seven times the son: f−5=7(s−5)f-5=7(s-5)f-5=7(s-5); in five years he will be three times: f+5=3(s+5)f+5=3(s+5)f+5=3(s+5). Solving gives a son of 101010 and a father of 404040.
  • Is there a graphical method chapter for simultaneous equations in ICSE Class 9?
    Yes — Graphical Solution (Chapter 27) solves the same pairs by drawing lines. Simultaneous (Linear) Equations is Chapter 6, after Factorisation (Chapter 5); both chapters have important-questions pages on Classmate AI.
  • Which method should I use for simultaneous equations in the exam?
    Use the method the question names. If none is named, elimination is usually fastest; substitution suits equations where one variable is already alone, like y=2x−3y=2x-3y=2x-3. Cross-multiplication avoids fractions when coefficients are awkward.
  • Can I download Simultaneous Linear Equations ICSE questions?
    Yes. Enter your email in the free PDF box on this page to open all 13 simultaneous-equation questions with worked answers as a printable worksheet.

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