Class 9 Simultaneous Linear Equations Mock Test (ICSE)

Topics in Simultaneous Linear Equations

• Substitution method
• Elimination method
• Cross-multiplication method
• Equations reducible to linear form
• Word problems on simultaneous equations

Simultaneous Linear Equations practice questions with answers

1. Q1Multiple choice·1 mark
   The solution of $\dfrac{2}{x + 1} = \dfrac{1}{x + 3}$ is

   1. (a)$x = -5$
   2. (b)$x = 5$
   3. (c)$x = -3$
   4. (d)$x = 3$
   Show answer and step marks

   1. Correct option: (a) $x = -5$+1

   Answer: (a)
2. Q2Multiple choice·1 mark
   To eliminate $y$ from the simultaneous equations $3x + 2y = 16$ and $5x - 2y = 0$ by the elimination method, which operation should be performed?

   1. (a)Subtract the second equation from the first
   2. (b)Multiply the second equation by $3$ and add it to the first
   3. (c)Add the two equations
   4. (d)Multiply the first equation by $5$ and subtract the second
   Show answer and step marks

   1. Correct option: (c) Add the two equations+1

   Answer: (c)
3. Q3Multiple choice·1 mark
   The simultaneous equations $5x - 3y = 7$ and $y = 2x - 1$ are to be solved by the substitution method. On substituting the value of $y$ from the second equation into the first, which equation in $x$ alone is obtained?

   1. (a)$5x - 6x - 3 = 7$
   2. (b)$5x - 2x + 1 = 7$
   3. (c)$5x - 6x + 3 = 7$
   4. (d)$5x + 6x - 3 = 7$
   Show answer and step marks

   1. Correct option: (c) $5x - 6x + 3 = 7$+1

   Answer: (c)
4. Q4Short answer·2 marks
   Solve the following system of linear equations using the substitution method: $x + 2y = 5$ $3x - y = 1$
   Show answer and step marks

   1. From the first equation, express $x$ in terms of $y$: $x = 5 - 2y$.+0.5
   2. Substitute $x$ into the second equation: $3(5 - 2y) - y = 1$, which gives $15 - 7y = 1$.+0.5
   3. Solve: $7y = 14 \implies y = 2$.+0.5
   4. Substitute back: $x = 5 - 2(2) = 1$.+0.5

   Answer: $x = 1, y = 2$
5. Q5Short answer·2 marks
   Solve the following system of linear equations using the substitution method: $x + 2y = 7$ $3x - y = 7$
   Show answer and step marks

   1. Express $x$ in terms of $y$ from the first equation: $x = 7 - 2y$.+0.5
   2. Substitute $x$ into the second equation: $3(7 - 2y) - y = 7$, which gives $21 - 7y = 7$.+0.5
   3. Solve: $7y = 14 \implies y = 2$.+0.5
   4. Substitute back: $x = 7 - 2(2) = 3$.+0.5

   Answer: $x = 3, y = 2$
6. Q6Long answer·4 marks
   Solve the following system of simultaneous linear equations using the elimination method: $4x - 3y = 15$ $2x + 5y = 1$
   Show answer and step marks

   1. Label the equations $4x - 3y = 15$ (1) and $2x + 5y = 1$ (2). Multiply equation (2) by 2 to make the coefficients of $x$ equal: $4x + 10y = 2$ --- (3)+1
   2. Subtract equation (3) from equation (1): $(4x - 3y) - (4x + 10y) = 15 - 2$ $-13y = 13$+1
   3. Solve: $y = -1$+1
   4. Substitute $y = -1$ into equation (2): $2x + 5(-1) = 1$ $2x = 6$ $x = 3$ Hence, $x = 3$ and $y = -1$.+1

   Answer: $x = 3, y = -1$

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