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. . .

3x + y -1. The two congruent angles of this isosceles triangle are: (3x +2)degrees and (5y - 3)degrees. What is the value of x and y?

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The two equal sides of an isosceles triangle are 2x+3y-5 and 3x+y-1.

So 2x+3y -5 = 3x+y -1.Or

Subtract 2x+3y - 1 from both sides:

x-2y = - 4........(1).

The given equal angles are 3x+2 and 5y -3. Or

3x+2 = 5y-3.

3x-5y = -3-2 = -5.

3x-5y = -5.....(2)

From (1) x = 2y -4. Put x = 2y-4 in (2):

3(2y-4) - 5y = = -5

6y -12 -5y = -5

6y-y = 12 -5 = 7

y = 7.

Put y= 10 in (1) : x - 2*14 = -4

x = 14-4 = 10.

Therefore x= 10 and y = 7.

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**

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