Solve using Elimination or Substitution:

4x^2+9y^2=36
x+3y=3

Ideal for substitution

from the 2nd : x = 3 - 3y

sub into 1st

4(3-3y)^2 + 9y^2 = 36
36 - 72y + 36y^2 + 9y^2 = 36
45y^2 - 72y = 0
5y^2 - 8y = 0
y(5y-8) = 0
y = 0 or y = 8/5

if y = 0 , x = 3

if y = 8/5 , then x = 3-24/5 = -9/5

intersection points (3,0) and (-9/5, 8/5)

Thanks for the help.

To solve the given system of equations using the method of elimination or substitution, we can start by rearranging the second equation to solve for x in terms of y.

Equation 2: x + 3y = 3

Rearranging Equation 2:

x = 3 - 3y

Now, we can substitute this expression for x in the first equation:

Equation 1: 4x^2 + 9y^2 = 36

Substituting x = 3 - 3y into Equation 1:

4(3 - 3y)^2 + 9y^2 = 36

Expanding the squared term:

4(9 - 18y + 9y^2) + 9y^2 = 36

Distributing:

36 - 72y + 36y^2 + 9y^2 = 36

Combining like terms:

36 + 45y^2 - 72y = 36

Rearranging to put it in standard quadratic form:

45y^2 - 72y = 0

Now, we can factor out the common factor of y:

y(45y - 72) = 0

Setting each factor equal to zero:

y = 0 or 45y - 72 = 0

Solving the second equation:

45y - 72 = 0
45y = 72
y = 72/45
y = 8/5

So, we have found two potential solutions for y: y = 0 and y = 8/5.

Substituting these values back into Equation 2 to find the corresponding values for x:

For y = 0:
x + 3(0) = 3
x = 3

For y = 8/5:
x + 3(8/5) = 3
x + 24/5 = 3
x = 3 - 24/5
x = 15/5 - 24/5
x = -9/5

Thus, the system of equations has two solutions: (x, y) = (3, 0) and (x, y) = (-9/5, 8/5).