finding points?

Question

how do i find the points

2x-y=7
3x^2+y^2=21

Answer 1

2x-y=7
y=2x-7
3x^2+(2x-7)^2=21
3x^2+4x^2-28x+49-21=0
7x^2-28x+28=0
dividing by 7
x^2-4x+4=0
(x-2)^2=0
x=2
when x=2 substituting in y=2x-7
y=4-7=-3
since there is only one point of intersection the line is tangential tothe curve at (2,-3)

Answer 2

Finding the points .....you mean to solve for x and y ....OK.

Given equations are 2x - y = 7 and 3x"2 + y"2 = 21.
From the first equation, y = 2x - 7
Substitute this in the 2nd equation i.e. replace y with 2x - 7
3x"2 + (2x - 7)"2 = 21
3x"2 + 4x"2 - 28x +49 = 21
7x"2 - 28x + 28 =0
x"2 - 4x +4 = 0
(x - 2)"2 = 0
x - 2 = 0
x = 2, y = 2x - 7 = 4 - 7 = -3.

Answer 3

A geometric interpretation of the above question goes as follows:

Where does the line 2x-y = 7 and the ellipse 3x^2 + y^2 = 21 intersect? Your answer is either going to be 0, 1, or 2 points, 0 being they don't intersect at all, 1 being they intersect at one point (like at the edge of the ellipse), and 2 being they intersect at two points. They can't intersect at more points given the nature of a line itself.

All we have to do is modify the first equation in terms of x or y. We'll use y, and then we get:

y = 2x - 7

We substitute this new value of y into the second equation.
3x^2 + (2x - 7)^2 = 21
3x^2 + (4x^2 - 28x + 49) = 21
7x^2 - 28x + 28 = 0
x^2 - 4x + 4 = 0
(x-2)^2 = 0
x-2 = 0
x = 2

Now, we plug this value into the first equation:
2(2) - y = 7
y = -3

Therefore, the point they cross at is (2,-3), and they only cross at one point.

Answer 4

from 1st eq: y = 2x-7
substitute in eq2:

3x^2 + (2x-7)^2 =21
simplify:

3x^2 + 4x^2 +49 -28x-21=0
7x^2-28x +28 =0
solve the quadratic eq for roots of x.
then substitute each in y =2x-7 to get y.