Solve the system x^2+y^2=2 2x-y=1

Expert Answers
hala718 eNotes educator| Certified Educator

x^2 +y^2 =2.....(1)

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

We will use the substitution method:

2x-y=1 ==> y=2x-1

Now substitute in (1)

x^2 + (2x-1)^2 =2

x^2+ 4x^2 -4x +1 =2

5x^2 -4x -1 =0

Factorize:

(5x+1)(x-1)=0

x1= -1/5  ==> y1= 2x1-1 = 2(-1/5)-1= -7/5

x2= 1   ==> y2= 2y2-1= 2(1)-1= 1

The solution are the points: (-1/5,-7/5) and (1,1)

giorgiana1976 | Student

We'll solve the second equation for y and, after that, we'll substitute the result in y, into the first equation.

So, we'll solve the system using the technique of substitution.

To solve the second equation for y, we'll subtract first 2x from both sides:

-y = -2x+1

Now, we'll multiply both sides by -1:

y = 2x-1

Wr'll substitute y by 2x-1 in the first equation of the system:

x^2 + (2x-1)^2 = 2

We'll expand the square:

x^2 + 4x^2 - 4x + 1 - 2 = 0

5x^2 - 4x - 1 = 0

We'll apply the quadratic formula:

x1 = [4+sqrt(16+20)]/10

x1 = (4+6)/10

x1 = 1

x2 = (4-6)/10

x2 = -2/10

x2 = -1/5

Now, we'll substitute 1 for x in y=2x-1

y = 2-1

y = 1

We'll substitute x = -1/5 for x in y=2x-1

y = -2/5 - 1

y = -7/5

The solutions of the system are:

{(1,1);(-1/5,-7/5)}

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