Question
Solve the following simultaneous equations:$41x + 53y = 135;53x + 41y = 147$
✓
Answer
The given equations are
$41x + 53y = 135 \dots....(i)$
$53x + 41y = 147 \dots....(ii)$
Multiplying eqn. $(i)$ by$ 53$ and eqn. $(ii)$ by $41,$ we get
$2173x + 2809y = 7155\dots ....(iii)$
$2173x + 1681y = 6027 \dots....(iv)$
Subtracting eqn. $(iv)$ from eq. $(iii)$, we get
$1128y = 1128$
$\Rightarrow y = 1$
Substituting the value of $y$ in eqn. $(i),$ we get
$41x + 53(1) = 135$
$\Rightarrow 41x + 53 = 135$
$\Rightarrow 41x = 135 - 53$
$\Rightarrow 41x = 82$
$\Rightarrow x = 2$
Thus, the solution set is $(2, 1).$
$41x + 53y = 135 \dots....(i)$
$53x + 41y = 147 \dots....(ii)$
Multiplying eqn. $(i)$ by$ 53$ and eqn. $(ii)$ by $41,$ we get
$2173x + 2809y = 7155\dots ....(iii)$
$2173x + 1681y = 6027 \dots....(iv)$
Subtracting eqn. $(iv)$ from eq. $(iii)$, we get
$1128y = 1128$
$\Rightarrow y = 1$
Substituting the value of $y$ in eqn. $(i),$ we get
$41x + 53(1) = 135$
$\Rightarrow 41x + 53 = 135$
$\Rightarrow 41x = 135 - 53$
$\Rightarrow 41x = 82$
$\Rightarrow x = 2$
Thus, the solution set is $(2, 1).$
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