x^2-2xy=3...1
3x^2-y^2=26...2
From #1, y = (x^2-3)/(2x)
so, in #2,
3x^2 - ((x^2-3)/(2x))^2 = 26
Or, assuming x≠0,
(11x^4-98x^2-9)/(4x^2) = 0
(x-3)(x+3)(11x^2+1) = 0
x = -3 or 3 or complex values.
hdyhgfjj
To solve the system of equations 1 and 2, we can use the method of substitution or elimination.
Let's solve it using the method of substitution:
From equation 1, we can rearrange it to find x:
x^2 - 2xy = 3
x^2 = 2xy + 3
x = √(2xy + 3)
Now substitute this value of x into equation 2:
3x^2 - y^2 = 26
3(√(2xy + 3))^2 - y^2 = 26
3(2xy + 3) - y^2 = 26
6xy + 9 - y^2 = 26
6xy - y^2 = 17 ----> Equation 3
Now we have two equations:
Equation 1: x = √(2xy + 3)
Equation 3: 6xy - y^2 = 17
We can solve equation 3 for y in terms of x:
y^2 - 6xy = 17
y^2 = 17 + 6xy
y = √(17 + 6xy)
Now we substitute this value of y into equation 1:
x = √(2x(√(17 + 6xy))^2 + 3)
Simplifying this equation will give us the solution for x.
To solve the system of equations 1 and 2, we'll use the method of substitution:
Step 1: Solve equation 1 for x.
Starting with equation 1: x^2 - 2xy = 3
Rearrange the equation to solve for x:
x^2 = 2xy + 3
Take the square root of both sides:
x = √(2xy + 3)
Step 2: Substitute the expression for x in equation 2.
Now, substitute the value of x into equation 2:
3(√(2xy + 3))^2 - y^2 = 26
Simplify and expand:
3(2xy + 3) - y^2 = 26
6xy + 9 - y^2 = 26
6xy - y^2 = 17
Step 3: Solve the resulting equation for y.
Now, we have a single equation with one variable (y). Rearrange the equation:
y^2 - 6xy + 17 = 0
Step 4: Solve the quadratic equation for y.
At this point, you can either try factoring the quadratic equation or use the quadratic formula to find the value(s) of y.
Using the quadratic formula, where a = 1, b = -6x, and c = 17, we have:
y = (-b ± √(b^2 - 4ac)) / 2a
Solve for y:
y = (-(-6x) ± √((-6x)^2 - 4(1)(17))) / (2(1))
y = (6x ± √(36x^2 - 68)) / 2
y = 3x ± √(9x^2 - 17)
Step 5: Substitute the values of y back into equation 1 to solve for x.
Now, substitute the two values of y back into equation 1 (x^2 - 2xy = 3) and solve for x for both cases:
Case 1: y = 3x + √(9x^2 - 17)
x^2 - 2x(3x + √(9x^2 - 17)) = 3
Simplify and expand:
x^2 - 6x^2 - 2x√(9x^2 - 17) = 3
-5x^2 - 2x√(9x^2 - 17) = 3
Rearrange the equation:
2x√(9x^2 - 17) = -5x^2 - 3
Square both sides of the equation:
4x^2(9x^2 - 17) = 25x^4 + 30x^2 + 9
Expand and simplify:
36x^4 - 68x^2 = 25x^4 + 30x^2 + 9
Subtract 25x^4 and 30x^2 from both sides:
11x^4 - 98x^2 - 9 = 0
Case 2: y = 3x - √(9x^2 - 17)
x^2 - 2x(3x - √(9x^2 - 17)) = 3
Simplify and expand:
x^2 - 6x^2 + 2x√(9x^2 - 17) = 3
-5x^2 + 2x√(9x^2 - 17) = 3
Rearrange the equation:
2x√(9x^2 - 17) = 5x^2 + 3
Square both sides of the equation:
4x^2(9x^2 - 17) = 25x^4 + 30x^2 + 9
Expand and simplify:
36x^4 - 68x^2 = 25x^4 + 30x^2 + 9
Subtract 25x^4 and 30x^2 from both sides:
11x^4 - 98x^2 - 9 = 0
At this point, you can solve the resulting quartic equations by factoring, using numerical methods, or by other techniques to find the values of x.