2x + y = 1, x^2 + y^ 2 = 1

Question

simultaneous problem non - linear equation plssss tell me d solution...

Answer 1

2x + y = 1

x^2 + y^2 = 1 (This is the Unit Circle for Trig.)

x^2 + (1 - 2x)^2 = 1

x^2 + 1 - 4x + 4x^2 = 1

5x^2 - 4x = 0

x(5x - 4) = 0

x = 0 or x = 4/5

y = 1 or y = 1 - 2(4/5) = -3/5

Solutions are (0,1) and (0.8,-0.6).

check:

2x + y = 1      √              √

x^2 + y^2 = 1  √       0.64+0.36 √

 

See GeoGebra sketch here:

http://www.wyzant.com/resources/files/260655/unit_circle_line

Answer 2

Hi again Rodela;

2x + y = 1, x^2 + y^ 2 = 1

Let's take the first equation and isolate y...

y=1-2x

Let's take the second equation and replace y with (1-2x)...

x^2 + y^ 2 = 1

x²+(1-2x)²=1

Let's FOIL the (1-2x)(1-2x)...

FIRST...(1)(1)=1

OUTER...(1)(-2x)=-2x

INNER...(-2x)(1)=-2x

LAST...(-2x)(-2x)=4x²

1-2x-2x+4x²

1-4x+4x²

x²+1-4x+4x²=1

5x²-4x+1=1

The 1 on both sides cancels...

5x²-4x=0

x(5x-4)=0

Either or both...the x=0 and/or the (5x-4)=0

5x-4=0

5x=4

x=4/5

 

Let's establish the value of y using both original equations...

x=0

2x + y = 1........x^2 + y^ 2 = 1

[(2)(0)]+y=1....(0)²+y²=1

y=1.................y²=1

.......................√y²=√1

.......................y=1, y=-1

We are going to disregard y=-1 because it does not correspond to the other equation.

(0,1)

 

x=4/5

2x + y = 1............x^2 + y^ 2 = 1

[(2)(4/5)]+y=1....(4/5)²+y²=1

(8/5)+y=1..........(16/25)+y²=1

(8/5)+y=5/5.......(16/25)+y²=25/25

y=-3/5.............y²=9/25

........................√y²=√(9/25)

.........................y=3/25, y=-3/5

We are going to disregard y=+3/5 because it does not correspond to the other equation.

(4/5,-3/5)