# Solve for x y equations x+y^1/2=2 2x+(y+3)^1/2=4

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### 2 Answers

Let,

EQ1: `x+sqrty=2 ` EQ2: `2x+sqrt(y+3)=4`

To solve for x and y, let's use elimination method of system of equations.

Let's try to eliminate the variable x. To do so, multiply EQ1 by 2.

`2(x+sqrty)=2*2`

`2x+2sqrty=4`

Then, subtract it from EQ2.

`2x+sqrt(y+3)=4`

`(-)` `2x+2sqrty` `= 4`

`---------------`

`0x + sqrt(y+3)-2sqrty=0`

`sqrt(y+3)=2sqrty`

To remove the square root, square both sides by 2.

`(sqrt(y+3))^2=(2sqrty)^2`

`y+3=4y`

`-3y=-3`

`y=1`

Then, substitute y=1 to either EQ1 or EQ2.

`x+sqrty=2`

`x+sqrt1=2`

`x+1=2`

`x=1`

Since the given equations have a radical expression, it is necessary to substitute the solution (1,1) to both EQ1 and EQ2 to verify

EQ1: `1+sqrt1=2 ` EQ2: `2*1+sqrt(1+3)=4`

`1+1=2` `2+2=4`

`2=2` (True) `4=4` (True)

**Since it satisfies both equations, hence (1,1) is the solution to the given system of equations. **

**Sources:**

You should consider the first equation and you may write y in terms of x such that:

`sqrt y = 2 - x => y = (2-x)^2 => y = 4 - 4x + x^2`

You should consider the second equation and you need to isolate the square root to the left side such that:

`sqrt(y+3) = (4-2x) => y + 3 = (4-2x)^2 => y = (4-2x)^2 - 3`

You need to equate the first and the second equations such that:

`4 - 4x + x^2 = (4-2x)^2 - 3 => 4 - 4x + x^2 = 16 - 16x + 4x^2 - 3`

`3x^2- 16x + 4x + 16 - 3 - 4 = 0`

`3x^2 - 12x + 9 = 0 => x^2 - 4x + 3 = 0`

Using quadratic formula yields:

`x_(1,2) = (4+-sqrt(16-12))/2 => x_(1,2) = (4+-sqrt4)/2`

`x_(1,2) = (4+-2)/2 => x_1 = 3 ; x_2 = 1`

You need to substitute 1 for x in equation `y = (4-2x)^2 - 3` such that:

`y = (4-2)^2 - 3 => y = 1`

You need to substitute 3 for x in equation `y = (4-2x)^2 - 3` such that:

`y = (4-2*3)^2 - 3 => y = 4 - 3 = ` 1

**Hence, evaluating the solutions to the system of equations yields `(1,1)` and `(3,1).` **