x^2 + 9y^2 = 37
x - 2y = -3
x = 2y-3, so we have
(2y-3)^2 + 9y^2 = 37
4y^2 - 12y + 9 + 9y^2 = 37
13y^2 - 12y - 28 = 0
(y-2)(13y+14) = 0
so, solutions are (1,2) and (-67/3,-14/13)
for 2nd equation, we get x = 2y-3
thus plugging into first equation gets
(2y-3)^2+9y^2 = 37.
4y^2-12y+9+9y^2 = 13y^2-12y+9=37 = 13y^2-12y-28 = 0.
Solving with the quadratic formula gives us: 2,-14/13
I assume you are solving.
from the 2nd: x = 2y-3
sub into the 1st
(2y-3)^2 + y^2 - 37 = 0
4y^2 - 12y + 9 + 9y^2 - 37 = 0
13y^2 - 12y - 28 = 0
(y-2)(13y + 14) = 0
y = 2 or y = -14/13
if y = 2, then x = 2-3 = -1
if y = -14/13, x = -28/13 - 3 = -67/13
points of intersection:
(-1,2) and (-67/13 , -14/13)
arrgghhh!
for first point:
y = 2, then x = 4-3 = 1 , just like Steve had above
point is (1,2)
so, is this a typo fest, or what?
Good job, Knights.
Hope you can dig out the correct solutions, Princess!
To solve this system of equations, you can use the method of substitution or elimination. Here, I will show you how to solve it using the method of substitution.
1. Start with the second equation: x - 2y = -3.
Solve this equation for x in terms of y:
x = 2y - 3.
2. Substitute the expression for x from step 1 into the first equation:
(2y - 3)^2 + 9y^2 = 37.
3. Expand the squared term on the left side:
4y^2 - 12y + 9 + 9y^2 = 37.
4. Combine like terms:
13y^2 - 12y + 9 = 37.
5. Move all terms to one side to get a quadratic equation:
13y^2 - 12y + 9 - 37 = 0.
6. Simplify the equation:
13y^2 - 12y - 28 = 0.
7. Solve the quadratic equation. You can either factor it or use the quadratic formula.
Factoring:
Since the equation doesn't factor easily, we will use the quadratic formula.
The quadratic formula is given by:
y = (-b ± sqrt(b^2 - 4ac))/(2a).
For our quadratic equation, a = 13, b = -12, and c = -28.
Substituting the values into the quadratic formula, we have:
y = (-(-12) ± sqrt((-12)^2 - 4 * 13 * -28))/(2 * 13)
= (12 ± sqrt(144 + 1456))/(26)
= (12 ± sqrt(1600))/(26)
= (12 ± 40)/(26)
So, we have two possible solutions:
y1 = (12 + 40)/26 = 52/26 = 2
y2 = (12 - 40)/26 = -28/26 = -1.0769 (approx.)
8. Substitute the values of y back into the second equation to find the corresponding x-values:
For y1 = 2:
x - 2(2) = -3
x - 4 = -3
x = -3 + 4
x = 1.
For y2 = -1.0769 (approx.):
x - 2(-1.0769) = -3
x + 2.1538 = -3
x = -3 - 2.1538
x = -5.1538 (approx.).
9. The solutions to the system of equations are:
(x1, y1) = (1, 2)
(x2, y2) = (-5.1538, -1.0769) (approx.)