Question

If x + y = 10 and x - y = 4, find the value of
${x}^{2} - y {}^{2}$
​

Answer 1

Solution

Given:-

=> x + y = 10 (i)

=> x - y = 4 (ii)

Now (i) + (ii) the equation

=> x + y + x - y = 10 + 4

=> 2x = 14

=> x = 14/2

=> x = 7

Now Put the Value on (ii)Eq

=> x - y = 4

=> 7 - y = 4

=> -y = 4-7

=> y = 3

So we get x = 7 and y = 3

So we have to find x² + y²

=> x² + y²

=> 7² + 3²

=> 49 + 9

=> 58

Answer

=> 58

Answer 2

We know,

$a^2 - b^2 = (a + b)(a - b)$

∴ $x^2 - y^2 = (x + y)(x - y)$

⇒ $x^2 - y^2 = (10)(4)$

⇒ $x^2 - y^2 = 40$.

So, if x + y = 10 and x - y = 4, then the value of (x² - y²) will be equal to 40.

ALITER:-

∵ $x + y = 10$ __(1)

and $x - y = 4$ __(2)

By subtracting Equation (2) from. (1),

i.e. $(x + y) - (x - y) = 10 - 4$

⇒ $2y = 6$

⇒ $y = 3$

Now, putting the value of y in any equation, say in (1),

$x + 3 = 10$

⇒ $x = 10 - 3 = 7$

Easily we can get, $x^2 - y^2 = (7)^2 - (3)^2 = 49 - 9 = 40$

More identities:-

• $(a \pm b)^2 = a^2 \pm 2ab + b^2$
• $(a \pm b)^3 = a^3 \pm b^3 \pm 3ab(a \pm b)$
• $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ac)$
• $(x + a)(x + b) = x^2 + (a + b)x + ab$
• $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac)$
• $x^2 + y^2 = (x + y)^2 - 2ab$
• $a^3 \pm b^3 = (a \pm b)(a^2 \pm ab + b^2)$
• $4ab = (a + b)^2 - (a - b)^2$
• $2(a^2 + b^2) = (a + b)^2 + (a - b)^2$.

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