MHB What are the steps to solving simultaneous equations using substitution?

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The discussion focuses on solving simultaneous equations using substitution, specifically the equations 4y² - 3x² = 1 and x - 2y = 1. The substitution x = 2y + 1 is applied, leading to a quadratic equation in y. A participant initially misapplies the quadratic formula, using incorrect coefficients, but later corrects this to find the solutions for y. The final solutions for the system of equations are confirmed as (-1, -1) and (0, -0.5).
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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
 
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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 $y$?
 
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
 
Let's start where skeeter left off:

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

Expand squared binomial:

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

Distribute the -3:

$$4y^2-12y^2-12y-3=1$$

Collect like terms:

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

Divide through by -4:

$$2y^2+3y+1=0$$

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

Factor:

$$(2y+1)(y+1)=0$$

Hence:

$$y\in\left\{-1,-\frac{1}{2}\right\}$$

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

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