How would you solve for y and x(for this triangle) using the quadratic equation: An isosceles triangle's two congruent sides are 2x + 3y -5; and. ....

According to the rule, an isosceles triangle has 2  sides
whose lengths are equal and 2 angles whose measures are also
equal.

Form enunciation, the lengths of the equal sides
are:

2x + 3y -5 (1)

3x + y -1
(2)

Since they are equal, we'll put
(1)=(2):

2x + 3y -5 = 3x + y
-1

We'll subtract (2) from
(1):

2x + 3y -5 - 3x - y +1 =
0

We'll combine like terms:

x
+ 2y - 4 = 0 (3)

We'll write the second condition of the
given isosceles triangle:

3x +2 =
5y-3

We'll subtract 5y-3 both
sides:

3x+2-5y+3 = 0

We'll
combine like terms:

3x-5y + 5 = 0
(4)

 Now, for finding x and y, we have to solve the system
formed by the equations (3) and (4), resulted from the conditions of the isosceles
triangle.

x + 2y - 4 = 0

3x-5y
+ 5 = 0

We'll solve the system using elimination method and
we'll eliminate the variable y. For this reason, we'll multiply (3) by 5 and (4) by
2:

5x + 10y - 20 + 6x - 10y  + 10 =
0

We'll combine and elimnate like
terms:

11x - 10 = 0

We'll add
10 both sides:

11x =
10

x =
10/11

We'll substitute x in
(3):

10/11 + 2y - 4 = 0

-34/11
+ 2y = 0

2y =
34/11

y =
17/11