Solve simultaneously the equations: x-y=1 x2+2y2=1

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To solve for x in the equation 9^((x−3)) × 81^((1−x)) = (1)/(27^x), we need to express all terms with the same base, which is 3. Step 1: Andika namba zote kama 'powers' za 3. Tunajua kuwa 9 = 3², 81 = 3⁴, na 27 = 3³. Hivyo, equation inakuwa: (3²)^((x−3)) × (3⁴)^((1−x)) = (1)/((3³)^x); Step 2: Tumia kanuni ya 'power of a power', (a^m)ⁿ = a^(mn). 3^(2(x−3)) × 3^(4(1−x)) = (1)/(3^(3x)); 3^(2x−6) × 3^(4−4x) = (1)/(3^(3x)); Step 3: Tumia kanuni ya kuzidisha 'powers' zenye 'base' sawa, a^m × aⁿ = a^(m+n), kwa upande wa kushoto. 3^((2x−6) + (4−4x)) = (1)/(3^(3x)); 3^(−2x−2) = (1)/(3^(3x)); Step 4: Andika upande wa kulia kwa kutumia 'negative exponent', (1)/(aⁿ) = a⁻ⁿ. 3^(−2x−2) = 3^(−3x) Step 5: Kwa kuwa 'bases' ni sawa, tunalinganisha 'exponents'. −2x−2 = −3x Step 6: Tatua kwa x. −2x + 3x = 2; x = 2 Jibu la mwisho ni: x=2

Solve simultaneously the equations: x−y=1 x2+2y2=1

Answer

Step-by-step solution

Andika namba zote kama 'powers' za 3.

Tumia kanuni ya 'power of a power', $(a^m)^n = a^{mn}$ .

Tumia kanuni ya kuzidisha 'powers' zenye 'base' sawa, $a^m \times a^n = a^{m+n}$ , kwa upande wa kushoto.

Andika upande wa kulia kwa kutumia 'negative exponent', $\frac{1}{a^n} = a^{-n}$ .

Kwa kuwa 'bases' ni sawa, tunalinganisha 'exponents'.

Tatua kwa $x$ .

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