Solve the following, then check your answer by graphing.
x^2+4y^2=25
y^2-x=1
Eq1: X^2 + 4Y^2 = 25.
Eq2: Y^2 - X = 1.
Y^2 = X + 1.
In Eq1, substitute
Eq1: X^2 + 4Y^2 = 25.
Eq2: Y^2 - X = 1.
Y^2 = X + 1.
In Eq1, substitute X+1 for Y^2:
X^2 + 4(X+1) = 25,
X^2 + 4X + 4 = 25,
(X+2)^2 = 25,
Take sqrt of both sides:
X+2 = +-5,
X = X = 5-2 = 3.
X = -5-2 = -7.
Solution set: (3,+-2), (-7,+-sqrt6i).
To solve the system of equations, we'll use the substitution method.
Let's solve the second equation for y^2:
y^2 - x = 1
y^2 = x + 1
Now we'll substitute this value of y^2 in the first equation:
x^2 + 4(y^2) = 25
x^2 + 4(x + 1) = 25
x^2 + 4x + 4 = 25
x^2 + 4x - 21 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, factoring is not straightforward, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
Here, a = 1, b = 4, and c = -21. Substituting these values into the formula:
x = (-4 ± √(4^2 - 4(1)(-21))) / 2(1)
x = (-4 ± √(16 + 84)) / 2
x = (-4 ± √100) / 2
x = (-4 ± 10) / 2
Simplifying:
x1 = (-4 + 10) / 2 = 6/2 = 3
x2 = (-4 - 10) / 2 = -14/2 = -7
Now, let's find the corresponding values of y^2 using the second equation:
y^2 = x + 1
For x = 3:
y^2 = 3 + 1 = 4
y1 = ± √4 = ±2
For x = -7:
y^2 = -7 + 1 = -6
This equation has no real solutions because the square root of a negative number is not a real number.
Therefore, the solution to the system of equations is:
(x, y) = (3, 2) and (3, -2)
Now, let's graph these equations to check our solutions.