Full solution

Q. given that

xy = 5

and

x + y = 7

, find the value of

(x-y)^2

Given Equations: We are given two equations: $\newline$ $1$ . $xy = 5$ $\newline$ $2$ . $x + y = 7$ $\newline$ We need to find the value of $(x-y)^2$ .
Using Identity: To find $(x-y)^2$ , we can use the identity $(x-y)^2 = x^2 - 2xy + y^2$ .
Substitute $xy$ : We already know the value of $xy$ , which is $5$ . So we can substitute this into our identity. $\newline$ $(x-y)^2 = x^2 - 2\cdot5 + y^2$ $\newline$ $(x-y)^2 = x^2 - 10 + y^2$
Square Second Equation: We don't have the values of $x^2$ and $y^2$ directly, but we can square the second given equation $(x + y = 7)$ to find $x^2 + 2xy + y^2$ . $\newline$ $(x + y)^2 = 7^2$ $\newline$ $x^2 + 2xy + y^2 = 49$
Substitute $2xy$ : We can substitute the value of $2xy$ from step $3$ into this new equation. $\newline$ $x^2 + 2\cdot5 + y^2 = 49$ $\newline$ $x^2 + 10 + y^2 = 49$
Solve for $x^2 + y^2$ : Now we can solve for $x^2 + y^2$ by subtracting $10$ from both sides of the equation. $\newline$ $x^2 + y^2 = 49 - 10$ $\newline$ $x^2 + y^2 = 39$
Substitute back into $(x-y)^2$ : We can now substitute the value of $x^2 + y^2$ back into the expression for $(x-y)^2$ . $(x-y)^2 = x^2 - 10 + y^2$ $(x-y)^2 = 39 - 10$ $(x-y)^2 = 29$