for 10 points answer this by solving this?

Question

x^2+y^2=2
x+y=1
solve this by using substitution
because i am stuck with this

Answer 1

x² + y² = 2
x + y = 1

Solve for y:
y = -x + 1

Substitute into first equation:
x² + (-x + 1)² = 2
x² + x² - 2x + 1 = 2
2x² - 2x + 1 - 2 = 0
2x² - 2x - 1 = 0

a = 2, b = -2, c = -1

Quadratic formula:
...... 2 ± sqrt( (-2)² - 4*(2)*(-1) )
x = --------------------------------------
........... ........... ...4

...... 2 ± sqrt( 4 + 8 )
x = -----------------------
...... ........... 4

...... 2 ± sqrt( 12 )
x = --------------------
.... ........... 4

...... 2 ± 2√3
x = -----------
........... 4

...... 1 ± √3
x = -----------
........... 2

x = (1 + √3) / 2
y = (1 - √3) / 2

(or vice versa).

Double-checking:
x = (1 + √3) / 2
y = (1 - √3) / 2

x + y = (1 + √3 + 1 - √3) / 2
x + y = 2 / 2
x + y = 1

x = (1 + √3) / 2
x² = (1 + 2√3 + 3) / 4
x² = (4 + 2√3) / 4
x² = 1 + √3/2

y = (1 - √3) / 2
y² = (1 - 2√3 + 3) / 4
y² = (4 - 2√3) / 4
y² = 1 - √3/2

x² + y² = 1 + √3/2 + 1 - √3/2
x² + y² = 2

So the answers are:
x = (1 + √3) / 2
y = (1 - √3) / 2

or

x = (1 - √3) / 2
y = (1 + √3) / 2

If you calculate these out, you'll get two points:
( -0.3660254, 1.3660254 )
and
( 1.3660254, -0.3660254 )

I've graphed the functions and the intersection points below:

Answer 2

First, solve the 2nd equation for either x or y. I like using x in my equations, so I will solve for y.

y = 1 - x

Now substitute that into the 1st equation.

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

Distribute (use FOIL method) to multiply the binomials together.

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

Subtract 2 from both sides

2x^2 - 2x - 1 = 0

This trinomial is not factorable, so we should use the quadratic formula.

x = [2 +/- sqrt(4 - 4*2*-1)] / 2*2
x = [2 +/- sqrt(4 + 8)] / 4
x = [2 +/- sqrt(12)] / 4
x = [2 +/- 2sqrt(3)] / 4
x = (1 +/- sqrt3)/2

Then substitute in to solve for y.

y = 1 - x
y = 1 - (1+/- sqrt3)/2
y = +/- sqrt3/2

Answer 3

X²+Y² = 2
X+Y = 1 ….Y = (1-X)
X²+(1-X)² = 2
X²+ ( 1 – 2X + X²) = 2
X² + 1 – 2X + X² -2 =0
2X² - 2X -1 = 0 use quadratic formula
X= (-1±√3)/2

Answer 4

Here is the answer:

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

(x+y)^2 = (1)^2 = 1
=> x^2 + y^2 + 2xy = 1
=>2+2xy = 1
=>2xy = 1-2 = -1
=>xy = -1/2
(x-y)^2 = x^2+y^2-2xy
=> (x-y)^2 = 2-2(-1/2) = 3

so, we have
(x+y)^2 = 1
(x-y)^2 = 3

=> x+y = 1^(1/2) = 1 --(equation 1)
=>x-y = 3^(1/2) = 1.732 (square root of 3) --(equation 2)

adding equation 1 and equation 2 gives the value of x
2x = 2.732
=> x= 1.366; y = 1-1.366 = -0.366
x=1.366 and y = -0.366

that's the final answer.

Answer 5

x^2+y^2=2
x+y=1 ------> y = 1 -x

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

2x^2 -2x -1 =0

Discriminant = D = 4 +4(1)(2) = 12

x1 = (2 +2sqrt(3))/4 = 1/2 + sqrt(3) /2
x2 = 1/2 - sqrt(3) /2

y1 = 1 -x1 = 1/2 - sqrt(3) /2
y2 = 1 -x2 = 1/2 + sqrt(3) /2