Solving simultaneous equations help?

hey mate,

(i) x + y = 15
(ii) x^2 + y^2 = 125

from (i)
x + y = 15 --> y = 15 - x
Substitute into (i)
thus,
x^2 + y^2 = 125 --> x^2 + (15 - x)^2 = 125
--> x^2 + 225 - 30x + x^2 = 125
--> 2x^2 - 30x + 100 = 0
--> x^2 - 15x + 50 = 0
Employ Quadratic formula,
x = ( 15 +- sqrt( 225 - 4(1)(50) ) )/ 2(1)
x = (15 +- sqrt( 25) )/ 2
x = (15 +- 5 ) / 2
Thus,
x = (15 + 5) / 2 = 10 or x = (15 - 5)/2 = 5
Hence our two solutions for x are,
x = 10 , 5
Recall y = 15 - x
Hence,
when x = 10 , y = 15 - 10 = 5
when x = 5 , y = 15 - 5 = 10

Thus we have two solutions (x,y) being
(10,5) , (5, 10)

Hope this helps,

David

y=15-x
x^2+x^2-30x+225=125
2x^2-30x+100
x^2-15x+50=0
x=(15+-8(225-200)^1/2)/2
x=10 y=5
x=5 y=10

x+y=15
x^2+y^2=125

x^2+y^2=125
x^2 +2xy +y^2=125 + 2xy
(x+y)^2 = 125 + 2xy
15^2 = 125 + 2xy
225 = 125 + 2xy
100 = 2xy
50 = xy

x+y = 15
x = 15 -y

50 =xy = (15 - y)y
50 = 15y - y^2
y= 5 or y = 10

if y = 5, x = 10. if y = 10, x =5.

x² + (15 - x)² = 125
x² + 225 - 30x + x² = 125
2x² - 30x + 100 = 0
x² - 15x + 50 = 0
(x - 10)(x - 5) = 0
x = 10 , x = 5
y = 5 , y = 10
(10,5) , (5,10)

x + y = 15 (solve by using substitution)
x^2 + y^2 = 125

x + y = 15
x = 15 - y

x^2 + y^2 = 125
(15 - y)^2 + y^2 = 125
(15 - y)(15 - y) + y^2 = 125
225 - 30y + y^2 + y^2 = 125
2y^2 - 30y + 225 - 125 = 0
2y^2 - 30y + 100 = 0
y^2 - 15y + 50 = 0
y^2 - 5y - 10y + 50 = 0
(y^2 - 5y) - (10y - 50) = 0
y(y - 5) - 10(y - 5) = 0
(y - 5)(y - 10) = 0

y - 5 = 0
y = 5

y - 10 = 0
y = 10

x + y = 15
x + 5 = 15
x = 15 - 5
x = 10

x + y = 15
x + 10 = 15
x = 15 - 10
x = 5

∴ (x = 10 , y = 5) , (x = 5 , y = 10)

x+y=15 ................. (i)
x^2+y^2=125 ..............(ii)
The equation can be converted into a quadratic equation as follows
By squaring (i) on both sides , we get
x² + y²+2xy = 225 ...........(iii)
Subtracting (ii) from (iii), gives
2xy = 100 Hence x= 50/y .....(iv)
Substituting (iv) in (i), gives
50/y +y=15 Simplifying , we get
50 + y² =15y
or y² − 15y + 50 =0 or (y − 5)(y −10) = 0
Hence y = 5 or 10
From (iv), when y = 5 ; x=50/y= 50/5 = 10
when y = 10; x=50/y= 50/10 =5
Hence the solution set are (5, 10) and (10, 5)

If x+y = 15 then
(x+y)^2 = 225
x^2+y^2+2xy = 225
u have x^2+y^2 = 125
thus, 125+2xy = 225
xy = 225-125/2
xy = 50

Now, y = 15-x from the eqn x+y = 15
So xy = 50 i.e. x(15-x) = 50
u'll obtain x^2-15x+50 = 0
solving this u'll get
(x-10) = 0 & x-5 = 0
So the numbers are
x=10 y=5
gud luck

x+y=15
x^2+y^2=125
(x+y)^2 = 15^2 OR (x^2+y^2) + 2xy = 225 OR 2xy = 225 -- 125 = 100
(x--y)^2 = (x+y)^2 -- 4xy = 225 -- 200 = 25 whence x -- y = 5, --5
adding x + y = 15 and x -- y = 5, --5, 2x = 20, 10 OR x = 10, 5
then y = 15 -- x = 15 -- 10 = 5 Or y = 15 -- 5 = 10
x = 10, y = 5 OR x = 5, y = 10.