#### Solve simultaneous linear equations:

(i) x(x + z) – y (y + z)

(ii) a^{12}x^{4} – a^{4}x^{12}

**Solution:**

(i) x(x + z) – y (y + z)

x^{2} + xz – y^{2} – yz

Rearrange the above terms we get,

x^{2} – y^{2} + xz – yz

We know that, (a^{2} – b^{2}) = (a + b) (a – b)

So, (x + y) (x – y) + z(x – y)

(x – y) (x + y + z)

(ii) a^{12}x^{4} – a^{4}x^{12}

a^{12}x^{4} – a^{4}x^{12}

Take out common in both terms,

a^{4}x^{4} (a^{8} – x^{8})

a^{4}x^{4}((a^{4})^{2} – (x^{4})^{2})

We know that, (a^{2} – b^{2}) = (a + b) (a – b)

a^{4}x^{4} (a^{4} + x^{4}) (a^{4} – x^{4})

a^{4}x^{4} (a^{4}_{ }+ x^{4}) ((a^{2})^{2} – (x^{2})^{2})

a^{4}x^{4}(a^{4} + x^{4}) (a^{2} + x^{2}) (a^{2} – x^{2})

a^{4}x^{4} (a^{4} + x^{4}) (a^{2} + x^{2}) (a + x) (a – x)

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