# x^2+y^2=34 x-2y=13so far i got x=13+2y (13-2y)^2+y^2=34 169-4y^2+y^2=34 169-5y^2=34 5y^2-169-34=0

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You need to use substitution method, hence, either substitute `13 + 2y` for `x` , or substitute `(x-13)/2` for `y` .

Since the previous answer use the substitution `13 + 2y` for `x` , I will present the working method using the substitution `(x-13)/2` for `y` such that:

`x^2 + (x - 13)^2/4 = 34`

Bringing the terms to a common denominator yields:

`4x^2 + x^2 - 26x + 169 - 136 = 0`

`5x^2 - 26x + 33 = 0`

You should use quadratic formula such that:

`x_(1,2) = (26+-sqrt(676 - 660))/10`

`x_(1,2) = (26+-sqrt16)/10 => x_(1,2) = (26+-4)/10`

`x_1 = 30/10 = 3`

`x_2 = 22/10 = 11/5`

You need to substitute 3 for x in equation `y = (x-13)/2` such that:

`y = (3-13)/2 => y = -5`

You need to substitute `11/5 = 2.2` for x in equation `y = (x-13)/2` such that:

`y = (11/5 - 13)/2 => y = -27/5 = -5.4`

**Thus, evaluating the solutions to the given system of equations yields `(2.2 ; -5.4)` and `(3 ; -5).` **

`x^2 + y^2 = 34`

`x = 13 + 2y`

To solve for x and y, let's use substitution method.

From the second equation, substitute x = 13-2y to the first equation. So,

`x^2+y^2=34`

`(13+2y)^2 + y^2 = 34`

Then, expand (13+2y)^2.

`(13+2y)(13+2y) + y^2 = 34`

`169+26y+26y + 4y^2+y^2 = 34`

`169+52y + 5y^2 = 34`

Express the equation in a form `ax^2+bx + c = 0` .

`5y^2+52y+169-34=0`

`5y^2+52y +135=0`

Factor left side.

`(5y + 27)(y+5)=0`

Then, set each factor equal to zero. And solve for y.

`5y+27=0 ` and `y +5=0`

`5y=-27` `y=-5`

`y=-27/5`

To solve for x, substitute the values of y to x = 13+2y.

`y=-27/5 ` , `x=13+2(-27/5)=13-54/5=65/5-54/5 = 11/5`

`y=-5` , `x=13+2(-5)=13-10=3`

**Hence, the solutions to the system of equations are `(3, -5)` and `(11/5, -27/5)` . **