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

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singup | Student, Grade 11 | (Level 2) eNoter

Posted August 26, 2012 at 10:32 AM via web

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Solve for x y equations

x+y^1/2=2

2x+(y+3)^1/2=4

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Top Answer

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted August 26, 2012 at 10:57 AM (Answer #1)

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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).`

Top Answer

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mjripalda | High School Teacher | (Level 1) Senior Educator

Posted August 26, 2012 at 12:52 PM (Answer #2)

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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.                                

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