Find the solution set of the equation
{x^2+y^2=34
{x+y=2
just substitute y=2-x and you have
x^2 + (2-x)^2 = 34
2x^2 - 4x - 30 = 0
x^2 - 2x - 15 = 0
now it should be clear ...
y = -x +2
x^2 + (-x +2)^2 = 34
x ^2 + x^2 -2x -2x +4 = 34
2x^2 -4x + 4 = 34
2x^ -4x + 4 -34 = 34-34
2x^2 -4x - 30 = 0
2(x^2 -2x -15) =0
2(x-5)(x+3) =0
x -5 =0
x =5
x+ 3 = 0
x =-3
y = -3 ,5
(5,-3), (-3,5)
To find the solution set of the given system of equations, we can use the method of substitution or elimination.
Method 1: Substitution
Step 1: Solve one equation for one variable in terms of the other.
From the second equation, we can solve for y as y = 2 - x.
Step 2: Substitute the expression for y into the other equation.
Substituting y = 2 - x into the first equation, we get:
x^2 + (2 - x)^2 = 34
Simplifying the equation gives:
x^2 + 4 - 4x + x^2 = 34
2x^2 - 4x + 4 = 34
2x^2 - 4x - 30 = 0
Step 3: Solve the quadratic equation.
Factoring out a common factor of 2 gives:
2(x^2 - 2x - 15) = 0
(x^2 - 2x - 15) = 0
(x - 5)(x + 3) = 0
So, x - 5 = 0 or x + 3 = 0
Solving for x, we have x = 5 or x = -3.
Step 4: Substitute the values of x back into the equation y = 2 - x to find the corresponding values of y.
If x = 5, then y = 2 - 5 = -3. So one solution is (5, -3).
If x = -3, then y = 2 - (-3) = 5. So another solution is (-3, 5).
Therefore, the solution set of the given system of equations is {(5, -3), (-3, 5)}.
Method 2: Elimination
Step 1: Multiply the second equation by -1 and add it to the first equation to eliminate y.
(x^2 + y^2) + (-x - y) = 34 + (-2)
Simplifying, we get:
x^2 - x + y^2 - y = 32
Step 2: Rewrite the equation in terms of one variable.
Rearrange the equation to get:
x^2 - x + y^2 - y - 32 = 0
Step 3: Complete the square to factorize the quadratic equation.
We can complete the square for x^2 - x and y^2 - y separately.
x^2 - x can be written as (x - 0.5)^2 - 0.25
y^2 - y can be written as (y - 0.5)^2 - 0.25
So, the equation becomes:
(x - 0.5)^2 - 0.25 + (y - 0.5)^2 - 0.25 - 32 = 0
(x - 0.5)^2 + (y - 0.5)^2 - 0.5 - 32 = 0
(x - 0.5)^2 + (y - 0.5)^2 = 32.5
Step 4: Determine the solutions by solving the equation derived from elimination.
The equation obtained above is in the form (x - a)^2 + (y - b)^2 = r^2, which represents a circle centered at (a, b) with a radius of r.
Comparing it with our equation, we can conclude that the given system of equations has no real solutions. The equation represents a circle with a radius of sqrt(32.5) centered at (0.5, 0.5), and the line y = 2 - x is tangent to the circle.
Therefore, the solution set of the given system of equations is empty, i.e., there are no real solutions.