A question?

Question

Here is a Simultaneous Equation, I made up...

2x + y = 1
4x^2 +y^2 = 5
I solved it!... But can you?
Show your working....
10 points for correct answer...

This is NOT my homework..

Answer 1

First, we express the first equation in terms of x.

y = 1 - 2x

Then we replace y in the second equation with this.

4x^2 + (1 - 2x)^2 = 5
4x^2 + (1 - 4x + 4x^2) = 5
4x^2 + 1 - 4x + 4x^2 = 5
8x^2 - 4x - 4 = 0
2x^2 - x - 1 = 0
(2x + 1)(x - 1) = 0

x = -1/2, 1

Now, in order to solve for y, we just individually plug these into the first equation.

For x = -1/2:

2(-1/2) + y = 1, therefore y = 0

For x = 1:

2(1) + y = 1, therefore y = -1

So our solutions are:
x = -1/2, y = 0
x = 1, y = -1

Answer 2

2x + y = 1
4x² +y² = 5

=>
y = 1 - 2x
4x² +y² = 5

=>
4x² +(1 - 2x)² = 5
4x² + 4x² - 4x + 1 = 5
8x² - 4x - 4 = 0
2x² - x - 1 = 0

x = [1±(1²+4*2*1)^¹/2] / (2*2)
x = [1±9^¹/2] / 4

x' = -1/2 ... or ... x'' = 1

₢

Answer 3

y= 1-2x
4x^2 +(1-2x)^2 = 5
8x^2 - 4x - 4 = 0
quadratic formula = [-(-1) +/- √(1-4(2)(-1))]/2(2)
x= -1/2 or 1

Using the values of x for y
y= 1-2x
y= 0 or -1/2

Answer 4

2x+y=1
y=1-2x

Substitute into the second equation:
4x^2 + (1-2x)^2 = 5
4x^2 + 1- 4x + 4x^2 = 5

rearrange the equation:
8x^2- 4X - 4 = 0

Divide by 4:
2x^2 - x - 1 = 0

Solving this equation:
(2x+1)(x-1) = 0

Then:
2x+1 = 0 ----------------------- x= -1/2
y=1-2x = 2

OR
x-1 = 0 -------------------------x=1
y=1-2x = -1

Answer 5

I failed math three times in high school. This is all random nonsense to me. So, to answer your question, no, I cannot solve it.