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Chapter 3: Pair of Linear Equations

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This page provides comprehensive Chapter 3: Pair of Linear Equations - Standard Worksheet - SJMaths. Standard level practice worksheet for Class 10 Pair of Linear Equations in Two Variables.

Standard Level Worksheet

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  1. Question 1: Solve the following pair of linear equations: $x + y = 14$ and $x - y = 4$.
    Solution:
    Step 1: Add the two equations: $(x + y) + (x - y) = 14 + 4 \Rightarrow 2x = 18$.
    Step 2: Solve for $x$: $x = 18/2 = 9$.
    Step 3: Substitute $x = 9$ in the first equation: $9 + y = 14 \Rightarrow y = 5$.
    Answer: $x = 9, y = 5$.
  2. Question 2: Solve $2x + 3y = 11$ and $2x - 4y = -24$ and hence find the value of 'm' for which $y = mx + 3$.
    Solution:
    Step 1: Subtract equation 2 from equation 1: $(2x + 3y) - (2x - 4y) = 11 - (-24) \Rightarrow 7y = 35$.
    Step 2: Solve for $y$: $y = 5$.
    Step 3: Substitute $y = 5$ in equation 1: $2x + 3(5) = 11 \Rightarrow 2x = 11 - 15 = -4 \Rightarrow x = -2$.
    Step 4: Substitute $x = -2, y = 5$ in $y = mx + 3$: $5 = m(-2) + 3 \Rightarrow 2 = -2m \Rightarrow m = -1$.
  3. Question 3: For what value of $k$ will the following pair of linear equations have infinitely many solutions?
    $kx + 3y - (k-3) = 0$
    $12x + ky - k = 0$
    Solution:
    Step 1: Condition for infinite solutions: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
    Step 2: Substitute values: $\frac{k}{12} = \frac{3}{k} = \frac{-(k-3)}{-k}$.
    Step 3: From first two ratios: $k^2 = 36 \Rightarrow k = \pm 6$.
    Step 4: Check $k=6$: $\frac{6}{12} = \frac{1}{2}$, $\frac{3}{6} = \frac{1}{2}$, $\frac{-(3)}{-6} = \frac{1}{2}$. Consistent.
    Step 5: Check $k=-6$: $\frac{-6}{12} = -\frac{1}{2}$, $\frac{3}{-6} = -\frac{1}{2}$, $\frac{-(-9)}{6} = \frac{3}{2}$. Inconsistent.
    Answer: $k = 6$.
  4. Question 4: Solve by elimination method: $3x - 5y - 4 = 0$ and $9x = 2y + 7$.
    Solution:
    Step 1: Rewrite equations: (1) $3x - 5y = 4$, (2) $9x - 2y = 7$.
    Step 2: Multiply Eq (1) by 3: $9x - 15y = 12$.
    Step 3: Subtract Eq (2) from new Eq (1): $(9x - 15y) - (9x - 2y) = 12 - 7 \Rightarrow -13y = 5 \Rightarrow y = -5/13$.
    Step 4: Substitute $y$ in Eq (1): $3x - 5(-5/13) = 4 \Rightarrow 3x + 25/13 = 4 \Rightarrow 3x = 27/13 \Rightarrow x = 9/13$.
  5. Question 5: The difference between two numbers is 26 and one number is three times the other. Find them.
    Solution:
    Step 1: Let numbers be $x$ and $y$ ($x > y$). Equations: $x - y = 26$ and $x = 3y$.
    Step 2: Substitute $x = 3y$ into the first equation: $3y - y = 26 \Rightarrow 2y = 26$.
    Step 3: Solve for $y$: $y = 13$.
    Step 4: Find $x$: $x = 3(13) = 39$.
    Answer: The numbers are 39 and 13.
  6. Question 6: Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son. What are their present ages?
    Solution:
    Step 1: Let Jacob's age $= x$, Son's age $= y$.
    Step 2: Five years hence: $x+5 = 3(y+5) \Rightarrow x - 3y = 10$ (Eq 1).
    Step 3: Five years ago: $x-5 = 7(y-5) \Rightarrow x - 7y = -30$ (Eq 2).
    Step 4: Subtract Eq 2 from Eq 1: $4y = 40 \Rightarrow y = 10$.
    Step 5: Find $x$: $x - 3(10) = 10 \Rightarrow x = 40$.
    Answer: Jacob is 40, Son is 10.
  7. Question 7: Solve for $x$ and $y$: $\frac{x}{2} + \frac{2y}{3} = -1$ and $x - \frac{y}{3} = 3$.
    Solution:
    Step 1: Simplify Eq 1 (multiply by 6): $3x + 4y = -6$.
    Step 2: Simplify Eq 2 (multiply by 3): $3x - y = 9$.
    Step 3: Subtract Eq 2 from Eq 1: $5y = -15 \Rightarrow y = -3$.
    Step 4: Substitute $y = -3$ in Eq 2: $3x - (-3) = 9 \Rightarrow 3x = 6 \Rightarrow x = 2$.
  8. Question 8: A fraction becomes 9/11 if 2 is added to both the numerator and the denominator. If 3 is added to both, it becomes 5/6. Find the fraction.
    Solution:
    Step 1: Let fraction be $x/y$.
    Step 2: Case 1: $\frac{x+2}{y+2} = \frac{9}{11} \Rightarrow 11x + 22 = 9y + 18 \Rightarrow 11x - 9y = -4$ (Eq 1).
    Step 3: Case 2: $\frac{x+3}{y+3} = \frac{5}{6} \Rightarrow 6x + 18 = 5y + 15 \Rightarrow 6x - 5y = -3$ (Eq 2).
    Step 4: Solving the system (e.g., substitution or elimination) yields $x = 7, y = 9$.
    Answer: Fraction is $7/9$.
  9. Question 9: For which value of $p$ does the pair of equations given below has unique solution?
    $4x + py + 8 = 0$
    $2x + 2y + 2 = 0$
    Solution:
    Step 1: Condition for unique solution: $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
    Step 2: Substitute coefficients: $\frac{4}{2} \neq \frac{p}{2}$.
    Step 3: Simplify: $2 \neq \frac{p}{2} \Rightarrow p \neq 4$.
    Answer: All real values of $p$ except 4.
  10. Question 10: The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?
    Solution:
    Step 1: Let number be $10x + y$. Reversed: $10y + x$.
    Step 2: Sum condition: $(10x+y) + (10y+x) = 66 \Rightarrow 11(x+y) = 66 \Rightarrow x+y = 6$.
    Step 3: Difference condition: $x-y=2$ or $y-x=2$.
    Step 4: Case 1 ($x-y=2$): Solving with $x+y=6$ gives $x=4, y=2$. Number is 42.
    Step 5: Case 2 ($y-x=2$): Solving with $x+y=6$ gives $x=2, y=4$. Number is 24.
    Answer: 42 and 24.
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