What is the intersection point of the equations : $4x+2y+1=0$ and $12x+4y+8=0$?

Question

What is the specific $x$ value and $y$ value for this set of equation? Since these lines are parallel I am not able to find the $x$ and $y$ values. But does that imply that this equations are solution less? When I try method of elimination it results into $3-8=-5=0$, but how is that possible?

Answer 1

Writing your equations in the form $$4x+2y+1=0$$ $$3x+y+2=0$$ Multiplying the second equation by $-2$ and adding to the first we get $$-2x-3=0$$ Can you proceed?

Answer 2

$4x=-1-2y$ so using elimination: $3(-1-2y)+4y+8=-3-6y+4y+8=-2y+5=0$. Find $y$.

Answer 3

Since $0\times x$ where $x$ is any number is still zero, you can do the following:

$$4x+2y+1=3\times(4x+2y+1)=3\times0=12x+6y+3$$

You now have two equations that are equal to $0$ which both also have a $12x$ term in them. Try to proceed from here.

Answer 4

Multiply the first equation on both sides by $-3$ then add it to the second equation. The result is $-2y = -5$, which gives $y =\frac{5}{2}$. Then plug this number into either of the two original equations and solve for $x$.

Answer 5

scale first by 3, cancelling the x's on subtraction. leaves you with $2y-5=0$ or $y=2.5$ .Input that into the second, and you get $12x+18=0$ or $x=-1.5$