Please help me... I need help on simultaneous equations...

the problem is...
y=2x-1
x^2+y^2=29

Please help me I need to Understand this concept!!!!
Please and thank you!

y=2x-1 ....eq 1
x^2+y^2=29.....eq 2
substitute for y from eq1 in eq2
x^2 + (2x - 1)^2 = 29
x^2 + 4x^2 - 4x +1 =29
5x^2 - 4x -28 = 0
(5x - 14)(x + 2) = 0
x = 14/5
or x = -2.
Substitute for x in eq 1
x = 14/5 gives y = 28/5 - 1 = 23/5
x = -2 gives y = -4 -1 = -5

The first equation tells us that y = 2x−1

So in second equation, replace y with 2x−1, then solve for x:

x² + y² = 29
x² + (2x−1)² = 29
x² + 4x² − 4x + 1 = 29
5x² − 4x − 28 = 0
5x² + 10x − 14x − 28 = 0
5x (x + 2) − 14 (x + 2) = 0
(x + 2) (5x − 14) = 0

x = −2, x = 14/5

Now we just find values of y for each value of x using original equation:
y = 2x − 1

x = −2 --------> y = 2(−2) − 1 = −5
x = 14/5 -----> y = 2(14/5) − 1 = 28/5 − 5/5 = 23/5

Solutions are: (−2, −5) and (14/5, 23/5)

Mαthmφm

substitute y in second equation from first equation

x^2+(2x-1)^2=29 expand to
x^2+4x^2-4x+1=29 or
5x^2-4x-28=0 solve for x by quadratic formula
x=[4+-sqr(16+560)]/10
x=-2 or 2.8 then
y=-5 or 4.6