solve the system by addition method
7x^2+y^2=49
7x^2-y^2=49
14 x^2 = 2*49
x^2 = 7
x = +/- 7
using x = +7
7*49 + y^2 = 49
y^2 = -6 *49 no real solution
using x = -7
same answer, no real solution
from Damon's
x^2 = 7
next line:
x = ± √7
subbing that into the first
49 + y^2 = 49
y = 0
you have an ellipse intersecting a hyperbola.
They meet at (√7,0) and (-√7,0)
will the answer look like
{(0,7),(0,-7)} or {(0,ratical 7),(0,-ratical 7)}
no, only the two points
(sqrt 7,0) and (-sqrt 7,0) work in both equations.
your solution of {(0,7),(0,-7)} works only in the first equation
{(0,ratical 7),(0,-ratical 7)} doesn't work for either one of them.
the hyperbola has the x-axis as its axis of symmetry, opens up 'sideways' and never crosses the y-axis.
BTW, does √7 show up as
the square root sign of 7 on your computer ?
To solve the given system of equations by the addition method, we'll add the equations together to eliminate one of the variables. Let's go step by step:
Step 1: Write down the equations:
7x^2 + y^2 = 49 ---- (Equation 1)
7x^2 - y^2 = 49 ---- (Equation 2)
Step 2: To eliminate one variable, we'll add the two equations together:
(7x^2 + y^2) + (7x^2 - y^2) = 49 + 49
Simplifying the left side of the equation, the y^2 terms cancel out:
14x^2 = 98
Step 3: Divide both sides of the equation by 14 to solve for x^2:
14x^2 / 14 = 98 / 14
x^2 = 7
Step 4: Take the square root of both sides of the equation to solve for x:
√(x^2) = √7
x = ±√7
So the solutions for x are x = √7 and x = -√7.
Step 5: Plug the value of x into one of the original equations to solve for y.
Let's use Equation 1:
7(√7)^2 + y^2 = 49
49 + y^2 = 49
y^2 = 0
Step 6: Take the square root of both sides of the equation to solve for y:
√(y^2) = √0
y = 0
So the solution for y is y = 0.
Therefore, the solution to the given system of equations is (x, y) = (±√7, 0).