Question
Solve: $23x + 29y = 98, 29x + 23y =110$
✓
Answer
The given equations are as follow:
$23x + 29y = 98 ...(i)$
$29x + 23y = 110 ....(ii)$
On adding $(i)$ and $(ii)$, we get,
$52x + 52y = 208$
$\Rightarrow x + y = 4 ....(iii)$
On subtracting $(i)$ from $(ii)$, we get:
$6x - 6y = 12$
$\Rightarrow x - y = 2 ....(iv)$
On additing $(iii)$ and $(iv)$, we get:
$2x = 6$
$\Rightarrow x = 3$
On subtracting $x = 3$ in $(iii)$, we get:
$3 + y = 4$
$\Rightarrow y = 4 - 3 = 1$
Hence, the required solution is $x = 3$ and $y = 1$.
$23x + 29y = 98 ...(i)$
$29x + 23y = 110 ....(ii)$
On adding $(i)$ and $(ii)$, we get,
$52x + 52y = 208$
$\Rightarrow x + y = 4 ....(iii)$
On subtracting $(i)$ from $(ii)$, we get:
$6x - 6y = 12$
$\Rightarrow x - y = 2 ....(iv)$
On additing $(iii)$ and $(iv)$, we get:
$2x = 6$
$\Rightarrow x = 3$
On subtracting $x = 3$ in $(iii)$, we get:
$3 + y = 4$
$\Rightarrow y = 4 - 3 = 1$
Hence, the required solution is $x = 3$ and $y = 1$.
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