Solve the system algebraically
2x^2-5y^2=1
x^2+5y^2=68
Please show work!!!
Intersection of a hyperbola with an ellipse, cool
2 x^2 - 5 y^2 = 1
2 x^2 + 10y^2 = 136
--------------------subtract
-15 y^2 = -135
y^2 = 9 so y = +/-3
x^2 = 68-45 = 23 so x = +/-sqrt(23)
(sqrt 23 , 3)
(sqrt 23 , -3)
(-sqrt 23 , 3 )
(-sqrt 23 , -3)
How about just adding them as they sit
3x^2 = 69
x^2 = 23
x = ±√23
plug x^2 = 23 back into the 2nd
23 + 5y^2 = 68
5y^2 = 45
y^2 = 9
y = ±3
so 4 solutions:
(√23,3) , (√23, -3) , (-√23 , 3) , and (-√23 , -3)
To solve the system algebraically, we can use the method of substitution.
Step 1: Solve one equation for one variable in terms of the other variable. Let's solve the second equation for x^2:
x^2 = 68 - 5y^2
Step 2: Substitute the value of x^2 from the second equation into the first equation:
2(68 - 5y^2) - 5y^2 = 1
Step 3: Simplify and solve for y:
136 - 10y^2 - 5y^2 = 1
-15y^2 = -135
y^2 = 9
y = ±3
Step 4: Substitute the values of y back into the equation obtained in step 1 to find the corresponding values of x:
For y = 3:
x^2 = 68 - 5(3)^2
x^2 = 68 - 45
x^2 = 23
x = ±√23
For y = -3:
x^2 = 68 - 5(-3)^2
x^2 = 68 - 45
x^2 = 23
x = ±√23
Therefore, the solutions to the system algebraically are:
(x, y) = (√23, 3), (-√23, 3), (√23, -3), (-√23, -3)
To solve the given system of equations algebraically, we can use the method of substitution. Here's how you can do it step-by-step:
1. Start with the system of equations:
2x^2 - 5y^2 = 1 ...(Equation 1)
x^2 + 5y^2 = 68 ...(Equation 2)
2. Solve one equation for one variable. Let's solve Equation 2 for x^2:
x^2 = 68 - 5y^2
3. Substitute the expression you obtained for x^2 in Equation 1:
2(68 - 5y^2) - 5y^2 = 1
4. Simplify the equation:
136 - 10y^2 - 5y^2 = 1
136 - 15y^2 = 1
5. Rearrange the equation to isolate the variable:
-15y^2 = 1 - 136
-15y^2 = -135
6. Divide both sides of the equation by -15 to solve for y^2:
y^2 = (-135) / (-15)
y^2 = 9
7. Take the square root of both sides to solve for y:
y = ± √(9)
y = ± 3
8. Substitute the values of y back into Equation 2 to solve for x^2:
For y = 3:
x^2 + 5(3)^2 = 68
x^2 + 45 = 68
x^2 = 23
For y = -3:
x^2 + 5(-3)^2 = 68
x^2 + 45 = 68
x^2 = 23
9. Take the square root of both sides to solve for x:
x = ± √(23)
10. Therefore, the solutions to the system of equations are:
(x, y) = (± √(23), 3) and (± √(23), -3)
Note: Make sure to check the answers by substituting them into the original equations to ensure they satisfy both equations.