Question

$x \3 + y \4 = 1 \\answer 4x + 3y - 12 = 0$
explain how the answer came
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Answer 1

Question :-

Solve :-

➩ 3x + 4y = 1

➩ 4x + 3y - 12 = 0

Answer :-

$\sf 3x + 4y = 1 \: \: \: \: \: \: \: \: -i$

$\sf 4x + 3y - 12 = 0\: \: \: \: \: \: \: \: -ii$

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Multiplying the equation i by 4 :-

➩ $\sf 4(3x + 4y = 1)$

➩ $\sf 12x + 16y = 4\: \: \: \: \: \: \: \: -iii$

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Multiplying the equation ii by 3 :-

➩ $\sf 3(4x + 3y - 12) = 0$

➩ $\sf 12x + 9y - 36 = 0\: \: \: \: \: \: \: \: -iv$

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Subtracting equation iv from equation iii :-

➩ $\sf 12x + 16y - 12x - 9y + 36 = 4$

➩ $\sf -7y + 36 = 4$

➩ $\sf 7y = 32$

➩ $\boxed{\sf y = \dfrac{32}{7}}$

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Substituting the value of y in equation i :-

➩ $\sf 3x + 4y = 1$

➩ $\sf 3x + 4 \big( \dfrac{32}{7}\big) = 1$

➩ $\sf 21x + 128 = 7 \times 1$

➩ $\sf 21x = 7 - 128$

➩ $\sf 21x = -121$

➩ $\boxed{\sf x = -\dfrac{121}{21}}$