Solve the equations algebraivally, test the solution for validity, and graph the system to check your number of solutions.
x^2-2y^2=2
xy=2
so y = 2/x
x^2 - 2 (2/x)^2 = 2
x^2 - 8 /x^2 = 2
x^4 - 2 x^2 = 8
let z = x^2
z^2 - 2 z - 8 = 0
(z -4)(z+2)= 0
z = +4 or z = -2
so
x^2 = +4 or x^2 = -2
x = +2 then y = 1
x= -2 then y = -1
x = 2 i then y = 2/2i = -i
x = -2i then y = +i
put these all back in to check
What, no ideas of your own at all? This isn't even calculus; just algebra.
add them up to get
x^2+xy-2y^2 = 4
(x+2y)(x-y) = 4
just by inspection, (2,1) is a solution, and since both graphs are symmetric about (0,0), (-2,-1) is also a solution.
Algebraically, I guess you could just substitute y=2/x and get
x^2 - 2(4/x^2) = 2
x^4 - 8 = 2x^2
x^4-2x^2+1 = 9
(x^2-1)^2 = 9
x^2 = 1±3
x = ±2, ±√2 i
To solve the equations algebraically, we'll use the substitution method.
Step 1: Solve the second equation for x.
xy = 2
x = 2/y
Step 2: Substitute the value of x into the first equation.
(2/y)^2 - 2y^2 = 2
4/y^2 - 2y^2 = 2
Step 3: Multiply both sides of the equation by y^2 to eliminate the denominator.
4 - 2y^4 = 2y^2
Step 4: Rearrange the equation.
2y^4 + 2y^2 - 4 = 0
Step 5: Factor the equation if possible.
We can factor out a 2 from the equation:
2(y^4 + y^2 - 2) = 0
Now we have a quadratic expression in terms of y^2:
(y^2 + 2)(y^2 - 1) = 0
Step 6: Set each factor equal to zero and solve for y.
y^2 + 2 = 0
y^2 = -2
There are no real solutions for this equation.
y^2 - 1 = 0
y^2 = 1
y = ±1
Step 7: Substitute the values of y back into the original equation to find x.
When y = 1:
x = 2/1
x = 2
When y = -1:
x = 2/-1
x = -2
The solutions to the system of equations are (x, y) = (2, 1) and (x, y) = (-2, -1).
To test the solutions for validity, substitute these values back into the original equations:
For (x, y) = (2, 1):
2^2 - 2(1^2) = 2
4 - 2 = 2
2 = 2 (valid)
2 * 1 = 2 (valid)
For (x, y) = (-2, -1):
(-2)^2 - 2(-1)^2 = 2
4 - 2 = 2
2 = 2 (valid)
-2 * -1 = 2 (valid)
Both solutions are valid.
To graph the system, plot the two points (2, 1) and (-2, -1) on a coordinate system. The points should lie on the intersection of the two curves.
To solve the equations algebraically, we will use the method of substitution.
1. Start with the second equation: xy = 2.
Solve for x by dividing both sides of the equation by y: x = 2/y.
2. Substitute this value of x into the first equation:
(2/y)^2 - 2y^2 = 2.
Simplify the equation: 4/y^2 - 2y^2 = 2.
3. Multiply both sides of the equation by y^2 to eliminate the denominator:
4 - 2y^4 = 2y^2.
4. Rearrange the equation to have zero on one side:
2y^4 + 2y^2 - 4 = 0.
To solve this equation, we can use factoring or the quadratic formula:
Using factoring:
2(y^4 + y^2 - 2) = 0.
2(y^2 + 2)(y^2 - 1) = 0.
Applying zero product property:
y^2 + 2 = 0 or y^2 - 1 = 0.
y^2 = -2 or y^2 = 1.
For y^2 = -2, there are no real solutions since the square of any real number cannot be negative.
For y^2 = 1, we have two possible solutions: y = 1 and y = -1.
5. Substitute the values of y back into the equation x = 2/y to find the corresponding x-values:
For y = 1, x = 2/1 = 2.
For y = -1, x = 2/(-1) = -2.
So, the solutions to the system of equations are: (x, y) = (2, 1) and (x, y) = (-2, -1).
Next, let's test the solution for validity by substituting the values of x and y back into the original equations:
For the first equation: x^2 - 2y^2 = 2.
Substituting (2, 1):
(2)^2 - 2(1)^2 = 4 - 2 = 2 (valid).
Substituting (-2, -1):
(-2)^2 - 2(-1)^2 = 4 - 2 = 2 (valid).
The solutions are valid for the first equation.
For the second equation: xy = 2.
Substituting (2, 1):
(2)(1) = 2 (valid).
Substituting (-2, -1):
(-2)(-1) = 2 (valid).
The solutions are valid for the second equation as well.
Finally, let's graph the system:
The graph of the first equation, x^2 - 2y^2 = 2, represents a hyperbola.
The graph of the second equation, xy = 2, represents a rectangular hyperbola.
To graph the system, plot the points (2, 1) and (-2, -1), which are the solutions we found earlier. Connect the points on each graph. You will find the two curves intersect at these points.
The graph will show that the system has two solutions, corresponding to the two points of intersection.