Simultaneous equations, embarassing but i really cannot see where i've gone wrong?

Question

x-2y=1
x^2+y^2=29

without a calculator i really cannot do this totally and find x and y. I found y=-0.4 plus or minus SQRT(-544) which is impossible atleast i thought.

Any ideas, i would usually get these straight away but really cannot today. This is in an AS level C1 exam from May 2005, so no calculator and it obv can be done, using the knowledge available to people at that point in the AS level course.

Answer 1

x = 2y + 1

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

x^2 + y^2 = 29, substitute for x^2 above:

(4y^2 + 4y + 1) + y^2 = 29
5y^2 + 4y + 1 = 29
5y^2 + 4y - 28 = 0

y = (-4 +- sqrt(16-4(5)(28)) / 10
y = (-4 +- sqrt( 16 + 560)) /10
y = (-4 +- 24) / 10
y = -28/10, 20/10 = -14/5, 2

Hope this helps! :)

Answer 2

From the first equation:

x = 2y+1

Replacing x into the second one:

(2y+1)² + y² = 5 y² + 4y + 1 = 29

or

5 y² + 4y - 28 = 0

Then

y = [-4 ± √(16 + 4*5*28)] / (2*5) =
= [-4 ± 24] / 10 = [-2 ± 12] / 5

y1 = 2, y2=-14/5

x1= 5, x2 =-23/5

Answer 3

solving the first equation for x : x = 2y + 1 and substitute for x in the other equation
(2y + 1)^2 + y^2 = 29
4y^2 + 4y + 1 + y^2 = 29
5y^2 + 4y - 28 = 0
(5y + 14)(y - 2) = 0
y = -14/5 or y = 2

then x = 2y + 1 gives x = -23/5 when y = -14/5
and x = 5 when y = 2

(-23/5, -14,5) and (5,2)

No idea what AS level C1 is....so hope this helped

Answer 4

x-2y=1 ---> x = 2y +1
x^2+y^2=29

(2y +1)^2 +y^2 -29 =0

4y^2 +1 +4y +y^2 -29 =0
5y^2 +4y -28 =0
D = 16 +4(28)(5) =576 = 24^2
y1 = (-4 -24)/10 = -28/10 = -14/5
y2 = (-4 +24)/10 = 2

x = 2y +1
x1 = 2(-14/5) +1 = -28/5 +1 = -23/5
x2 = 2(2) +1 =5

Solutions are :
x = -23/5 and y = -14/5
or
x = 5 and y =2

Answer 5

I get two rational numbers! Using the quadratic formula, I got 576 under the radical. This means it factors. First, let's make sure we agree on the quadratic:

substituting 1+2y for the x:

(1+2y)^2 + y^2 = 29
1 + 4y + 4y^2 + y^2 = 29

5y^2 + 4y - 28 = 0

(5y + 14)(y - 2) = 0

So, y = 14/5, or y = 2

hope this helps!

Answer 6

(2y+1)^2 + y^2 = 29
4y^2 + 4y + 1 + y^2 = 29
5y^2 + 4y -28 = 0
(5y+14)(y-2)=0

y = 2 or -14/5
x = 2y+1 = 5 or -23/5

Answer 7

x = 1 + 2y
second equation becomes
(1 + 2y)² + y² = 29
5y² + 4y - 28 = 0
(5y + 14)(y - 2) = 0
y = -14/5 or y = 2
so then
when y = -14/5 then x = -23/5
when y = 2 then x = 5

~~

Answer 8

x=1+2y replace to other equation:
(1+2y)^2 + y^2 = 29
5y^2 + 4y - 28 = 0
y=2 or y= -14/5
x=5 x= -23/5

Answer 9

put x = 1 + 2y into equation two, we get 5y^2 + 4y - 28 = 0
then we can get ( 5y + 14)(y - 2 ) = 0
so y = - 14/5 or y = 2

Answer 10

set x=2y+1
substitute 2y+1 for x
(2y+1)(2y+1) +y^2=29
use FOIL
4y^2+4y+1+y^2=29
combine like terms, and set equal to '0'
5y^2+4y-28=0
Factor
(5y+14)(y-2)=0
y=2,y=-14/5

I figure you can take it from here