Question

please provide this question solution​

Answer 1

Answer:

which question is this.........

Answer 2

Answer:

Given : x+y=13x+y=13 and xy=\frac{25}{4}xy=

4

25

To find : The value of x-y ?

Solution :

Let x+y=13x+y=13 ......(1)

xy=\frac{25}{4}xy=

4

25

......(2)

Substitute the value of x from (1) and put in (2),

(13-y)y=\frac{25}{4}(13−y)y=

4

25

13y-y^2=\frac{25}{4}13y−y

2

=

4

25

y^2-13y+\frac{25}{4}=0y

2

−13y+

4

25

=0

4y^2-52y+25=04y

2

−52y+25=0

(y-0.5)(y-12.5)=0(y−0.5)(y−12.5)=0

y=0.5,12.5y=0.5,12.5

Substitute in (1),

When y=0.5, x+0.5=13x+0.5=13

x=13-0.5=12.5x=13−0.5=12.5

When y=12.5, x+12.5=13x+12.5=13

x=13-12.5=0.5x=13−12.5=0.5

Therefore, The values are (0.5,12.5) and (12.5,0.5).

Now, Substitute in x-yx−y

When, x=0.5 and y=12.5

x-y=0.5-12.5=-12x−y=0.5−12.5=−12

When, x=12.5 and y=0.5

x-y=12.5-0.5=-12x−y=12.5−0.5=−12