simultaneous equation with x squared?

1-2x=x^2-3x-5

x^2-x-6=0

(x-3)(x+2)=0

x=3 or x=-2

apply this to the 1st equation

when x=3 y=1-2(3)=-5

when x=-2 y=1-2(-2)=5

answer x=3,y=-5 or x=-2,y=5

You have to substitute the linear equation into the quadratic.

Since y is common to both equations, you can write this:

1-2x=x^2-3x-5

and from this:

x^2-x-6=0

Then solve this as you would any other quadratic.

(x-3)(x+2)=0

Then:

x-3=0

x=3

and x+2=0

x=-2

Substitute these values into the linear equation:

y=1-2x

y=1-2(3)

y=-5

and y=1-2(-2)

y=5

so you get the two solutions, which should be written as pairs:

x=3 and x=-2

y=-5 y=5

from the given 2 equations we have

x^2 -3x-5=1 - 2x

=>x^2-x-6=0

=>x^2-3x+2x-6=0

=>(x-3)(x+2)=0

therefore either x=2 or x=3

when x= - 2 then from first equation we have y=1 - 2*(-2)= 5

and when x=3 then from first equation we have y=1 - 2*3= - 5

y = 1 - 2x --------1)

y = x^2 - 3x - 5 ----------2)

Substitute eqn 1) in eqn 2)

1 - 2x = x^2 - 3x - 5

x^2 - 3x + 2x - 5 - 1 = 0

x^2 - x - 6 = 0

x^2 - 3x + 2x - 6 = 0

x(x - 3)+2(x - 6) = 0

(x - 3) (x + 2) = 0

x = 3 or x = -2

Substitute these two values in eqn 1)

When x = 3

y = 1 - 2x = 1 - 2(3) = 1 - 6 = -5

When x = -2

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

(x,y) = (3,-5) or (x,y) = (-2,5) -----------Answer

as requested, first step only !

y = y which is -

x^2 - 3x - 5 = 1 - 2x

then combine like terms and solve or factor the quadratic