Question
Solve the given pair of linear equation by the elimination method and the substitution method: $3x – 5y – 4 = 0$ and $9x = 2y + 7$
✓
Answer
- By Elimination method,
The given system of equations is :
$3 x - 5 y - 4 = 0............(1)$
$9 x = 2 y + 7$
$9 x - 2 y - 7 = 0.............(2)$
Multiplying equation $(1)$ by $3,$ we get
$9 x - 15 y - 12 = 0.............(3)$
Subtracting equation $(3)$ from equation $(2) ,$ we get
$13 y + 5 = 0$
$\Rightarrow \quad 13 y = - 5 \Rightarrow y = \frac { - 5 } { 13 }$
Substituting this value of $y$ in equation $(1),$ we get
$3 x - 5 \left( \frac { - 5 } { 13 } \right) - 4 = 0$
$\Rightarrow \quad 3 x + \frac { 25 } { 13 } - 4 = 0 \Rightarrow 3 x - \frac { 27 } { 13 } = 0$
$\Rightarrow \quad 3 x = \frac { 27 } { 13 } \Rightarrow x = \frac { 9 } { 13 }$
So, the solution of the given system of equation is
$x = \frac { 9 } { 13 } , y = \frac { - 5 } { 13 }$ - By Substitution method:
The given system of equation is:
$3 x - 5 y - 4 = 0.............(1)$
$9 x = 2 y + 7...................(2)$
From equation $(2),$
$x = \frac { 2 y + 7 } { 9 }..................(3)$
Substituting this value of x in equation(1), we get
$3 \left( \frac { 2 y + 7 } { 9 } \right) - 5 y - 4 = 0$
$\Rightarrow \quad \frac { 2 y + 7 } { 3 } - 5 y - 4 = 0$
$\Rightarrow \quad 2 y + 7 - 15 y - 12 = 0$
$\Rightarrow -13y - 5 = 0$
$\Rightarrow 13y = -5$
$\Rightarrow \quad y = \frac { - 5 } { 13 }$
Substituting this value of $y$ in equation$(3),$ we get
$x = \frac { 2 \left( \frac { - 5 } { 13 } \right) + 7 } { 9 } = \frac { - \frac { 10 } { 13 } + 7 } { 9 } = \frac { - 10 + 91 } { 117 } = \frac { 81 } { 117 } = \frac { 9 } { 13 }$
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