maths a/s simultaneous equation?

x - 2y = 1 (solve by using substitution)
x^2 + y^2 = 29

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

x^2 + y^2 = 29
(2y + 1)^2 + y^2 = 29
(2y + 1)(2y + 1) + y^2 = 29
4y^2 + 2y + 2y + 1 + y^2 - 29 = 0
5y^2 + 4y + 1 - 29 = 0
5y^2 + 4y - 28 = 0
5y^2 + 14y - 10y - 28 = 0
(5y^2 + 14y) - (10y + 28) = 0
y(5y + 14) - 2(5y + 14) = 0
(5y + 14)(y - 2) = 0

5y + 14 = 0
5y = -14
y = -14/5 (-2.8)

y - 2 = 0
y = 2

x - 2y = 1
x - 2(-14/5) = 1
x + 28/5 = 1
x = 1 - 28/5
x = 5/5 - 28/5
x = -23/5 (-4.6)

x - 2y = 1
x - 2(2) = 1
x - 4 = 1
x = 1 + 4
x = 5

∴ [x = -23/5 (-4.6) , y = -28/5 (-14/5)] , [x = 5 , y = 2]

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

x^2 + y^2 = 29
(1 + 2y)^2 + y^2 = 29
1 + 4y^2 + 4y + y^2 = 29
5y^2 + 4y - 28 =0
5y^2 -10y + 14y - 28 = 0
5y(y - 2) + 14(y-2) = 0
(5y+14)(y - 2) = 0
=> 5y + 14 = 0
y = -14/5

OR

y-2 = 0
y = 2

Use the quadratic formula to make it simpler

yes 5y^2 +4y -28 is correct.
then you solve it as a quadratic equation.
write it as 5y^2 + 14y -10y -28 = 0.
=> y(5y+14) -2(5y + 14) = 0.
=> (y-2)(5y + 14) = 0.
=> y=2 or y=-14/5 are the 2 solutions.
Hope you get it.

x = 2y + 1
(2y + 1)² + y² = 29
4y² + 4y + 1 + y² = 29
5y² + 4y - 28 = 0
(5y + 14)(y - 2) = 0
y = - 14/5 , y = 2

x + 28/5 = 1
x = - 23/5

x - 4 = 1
x = 5

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

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

then sub in the 2 values of y into the first equation to get the x values... you should have two sets of answers in the end

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

replace x=1+2y in the equation x^2+y^2=29
we get
(1+2y)^2+y^2=29
1+4y+4y^2=29
4y^2+4y-28=0

it looks about right. try using the quadratic equation to solve for y.