Complete the following for each system that involves one or more quadratic equations:
Solve the system algebraically
7y^2-5x^2+20x=3
5x^2+21y^2=209
eq #1 says that 7y^2=5x^2-20x+3, so plug that into eq #2 and you have
5x^2+3(5x^2-20x+3) = 209
20x^2-60x-200 = 0
x^2-3x-10 = 0
(x-5)(x+2) = 0
And I think you can take it from there, no?
yes, thanks i can :)
To solve this system algebraically, we'll use the method of substitution. First, let's solve one equation for one variable and substitute it into the other equation:
From the first equation, we have:
7y^2 - 5x^2 + 20x = 3
To isolate y, let's move the terms involving y to one side:
7y^2 = 5x^2 - 20x + 3
Next, divide through by 7 to simplify:
y^2 = (5x^2 - 20x + 3)/7
Now we can solve for y by taking the square root of both sides:
y = ±sqrt((5x^2 - 20x + 3)/7)
Now, substitute this expression for y into the second equation:
5x^2 + 21y^2 = 209
Replacing y with ±sqrt((5x^2 - 20x + 3)/7), we get:
5x^2 + 21(±sqrt((5x^2 - 20x + 3)/7))^2 = 209
Simplifying and expanding, we have:
5x^2 + 21(5x^2 - 20x + 3)/7 = 209
Multiply through by 7 to eliminate the denominator:
35x^2 + 105x^2 - 420x + 63 = 1463
Combine like terms:
140x^2 - 420x = 1463 - 63
140x^2 - 420x = 1400
Divide through by 140 to simplify:
x^2 - 3x - 10 = 0
Now, we have a quadratic equation in terms of x. We can solve this equation using factoring, completing the square, or by using the quadratic formula.
To find the solutions for x, we'll use the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -3, and c = -10.
Plugging these values into the quadratic formula, we get:
x = (-(-3) ± sqrt((-3)^2 - 4(1)(-10))) / (2(1))
Simplifying further:
x = (3 ± sqrt(9 + 40)) / 2
x = (3 ± sqrt(49)) / 2
x = (3 ± 7) / 2
Thus, we have two possible values for x:
x = (3 + 7) / 2 = 10 / 2 = 5
x = (3 - 7) / 2 = -4 / 2 = -2
Now that we have the values of x, we can substitute them back into the expression for y:
For x = 5: y = ±sqrt((5(5)^2 - 20(5) + 3)/7) = ±sqrt(44/7) = ±2
Therefore, we have two solutions for this system when x = 5:
(5, 2) and (5, -2)
For x = -2: y = ±sqrt((5(-2)^2 - 20(-2) + 3)/7) = ±sqrt(84/7) = ±4
Therefore, we have two solutions for this system when x = -2:
(-2, 4) and (-2, -4)
So the solutions to the system of equations are:
(5, 2), (5, -2), (-2, 4), and (-2, -4).