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Pair of Linear Equations in Two variables — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsPair of Linear Equations in Two variables5 Marks
Question

Solve the pairs of linear equation by the elimination method and the substitution method:$\frac{x}{2} + \frac{{2y}}{3} = - 1\,and\,x - \frac{y}{3} = 3$

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Answer

  1. By Elimination method
    The given system of equation is
    $\frac { x } { 2 } + \frac { 2 y } { 3 } = - 1$...............(1)
    $x - \frac { y } { 3 } = 3$....................(2)
    Multiplying equation (2) by 2, we get
    $2 x - \frac { 2 y } { 3 } = 6$......................(3)
    Adding equation(1) and equation (2), we get
    $\frac { 5 } { 2 } x = 5 \Rightarrow x = \frac { 5 \times 2 } { 5 } \Rightarrow \quad x = 2$
    Substituting this value of x in equation(2), we get
    $2 - \frac { y } { 3 } = 3 \Rightarrow \frac { y } { 3 } = 2 - 3 = - 1 \Rightarrow \quad y = - 3$
    So, the solution of the given system of equation is
    x = 2, y = -3
  2. By substitution method
    The given system of equation is
    $\frac { x } { 2 } + \frac { 2 y } { 3 } = - 1$...............(1)
    $x - \frac { y } { 3 } = 3$....................(2)
    From equation (2),
    $x = \frac { y } { 3 } + 3$....................(3)
    Substituting this value of x in (1),
    $\frac { 1 } { 2 } \left( \frac { y } { 3 } + 3 \right) + \frac { 2 y } { 3 } = - 1$
    $\Rightarrow \quad \frac { y } { 6 } + \frac { 3 } { 2 } + \frac { 2 y } { 3 } = - 1 \Rightarrow \quad \frac { 5 y } { 6 } = - 1 - \frac { 3 } { 2 }$
    $\Rightarrow \quad \frac { 5 y } { 6 } - - \frac { 5 } { 2 } \Rightarrow y = - 3$
    Substituting this value of y in equation (3), we get
    $x = - \frac { 3 } { 3 } + 3 = - 1 + 3 - 2$
    So, the solution of the given system of equations is x = 2, y = -3
    Verification: Substituting x = 2, y = -3, we find that both the equation (1) and (2) are satisfied as shown below:
    $\frac { x } { 2 } + \frac { 2 y } { 3 } = \frac { 2 } { 2 } + \frac { 2 ( - 3 ) } { 3 } = 1 - 2 = - 1$

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