If x + y = 6 and x - y = 4, find

(i) x² + y²

(ii) xy.

(i) We know that,

(x + y)² = x² + y² + 2xy …..(i)

(x - y)² = x² + y² - 2xy ….(ii)

Adding eqn. (i) and (ii) we get,

(x + y)² + (x - y)² = x² + x² + y² + y² + 2xy - 2xy = 2x² + 2y².

⇒ 2x² + 2y² = (x + y)² + (x - y)².

∴ x² + y² = $\dfrac{(x + y)^2 + (x - y)^2}{2}$

Substituting values we get,

$\Rightarrow x^2 + y^2 = \dfrac{(6)^2 + (4)^2}{2} \\[1em] = \dfrac{36 + 16}{2} \\[1em] = \dfrac{52}{2} \\[1em] = 26.$

Hence, x² + y² = 26.

(ii) We know that,

(x + y)² = x² + y² + 2xy …..(i)

(x - y)² = x² + y² - 2xy ….(ii)

Subtracting eqn. (ii) from (i) we get,

(x + y)² - (x - y)² = x² - x² + y² - y² + 2xy - (-2xy) = 4xy.

⇒ (x + y)² - (x - y)² = 4xy.

∴ xy = $\dfrac{(x + y)^2 - (x - y)^2}{4}$

Substituting values we get,

$xy = \dfrac{6^2 - 4^2}{4} \\[1em] = \dfrac{36 - 16}{4} \\[1em] = \dfrac{20}{4} \\[1em] = 5.$

Hence, xy = 5.