What are the steps to solving simultaneous equations using substitution?

[blackfriars]

hi could anyone help with this one as i do not know the steps to take . as far as i know you can only use substitution
4y^2-3x^2=1
x-2y=1

 

[skeeter]

4y^2-3x^2=1
x-2y=1

line intersecting a hyperbola ...

$x = 2y+1 \implies 4y^2 - 3(2y+1)^2 = 1$

can you solve the resulting quadratic in $y$?

 

[blackfriars]

hi skeeter, thanks for the reply i have used desmos and geo gebra and the intersection points ere (-1,-1)and (0,-0.5)
now i was able to get the (-1,-1) but for the quadratic i was getting (-1/2,-3/4)
this is what i did

-8y^2-12y-4=0
----------------------
-4

2y^2+3y+1=0

using the formula a=2,b=3,c=-4
the intersections i should have got were (-1,-1) and (0,-0.5)
hope you can help to show where i went wrong thanks

 

[MarkFL]

Let's start where skeeter left off:

\(\displaystyle 4y^2-3(2y+1)^2=1\)

Expand squared binomial:

\(\displaystyle 4y^2-3\left(4y^2+4y+1\right)=1\)

Distribute the -3:

\(\displaystyle 4y^2-12y^2-12y-3=1\)

Collect like terms:

\(\displaystyle -8y^2-12y-4=0\)

Divide through by -4:

\(\displaystyle 2y^2+3y+1=0\)

This is what you have, but when you went to apply the quadratic formula, you used $c=-4$ instead of $c=1$.

Factor:

\(\displaystyle (2y+1)(y+1)=0\)

Hence:

\(\displaystyle y\in\left\{-1,-\frac{1}{2}\right\}\)

Now with $x=2y+1$, this gives us the solutions:

\(\displaystyle (x,y)=(-1,-1),\,\left(0,-\frac{1}{2}\right)\)