Below a three systems of equations,
Were asked to determine the type of solution.
The types of solutions are given below:
a. One unique solution.
B. No solution
C. Infinitely many solutions.
Remember a solution has any ordered pair
X [indistict] Y that would satisfy both equations.
We will take a look at these systems graphically,
as well as algebraically.
Let's begin by graphing the first system
on the coordinate plane.
I've already set this up to save some time,
here as a graph of negative X minus two Y equals one.
And here's the graph of the second equation,
three X plus six Y equals negative three.
And notice how the two lines are on top of each other
or the two lines coincide,
which means the two lines have an infinite number
of points in common, and therefore
the system has an infinite number of solutions.
So the first system would be,
C. Infinitely many solutions.
Let's also take a look at this algebraically.
Let's use the method of elimination or addition
and try to solve the system.
When using this method,
we want the X terms or Y terms to be opposites,
so that when we add the equations together,
the X terms and Y terms simplify out.
So looking at the X terms, notice how,
if this first X term in the first equation
was negative three X,
when we added it to the second equation,
the X term would simplify out.
So let's multiply both sides of the first equation by three,
leave the second equation the same,
and then add the two equations together.
So the first equation after
multiplying both sides by three would be
negative three X minus six Y equals three.
The second equation remains the same.
Now we add the equations together,
but notice how when we do this,
both the X terms and Y terms are opposites,
and therefore when adding the left sides,
the sum would be zero,
on the right the sum is also a zero,
Leaving us with zero equals zero.
So whenever solving algebraically
and both variables simplify out
and we're left with a true equation like this,
zero equals zero is always true.
This indicates we have infinite solutions.
And now graphed the second system.
Again, I've already set this up.
The first question is still
negative X minus two Y equals one.
And the second equation is now
three X minus six y equals nine.
Notice the two lines intersect in one point,
which indicates we have one unique solution.
The solution is the point of intersection,
which we can see here is the ordered pair,
One comma negative one.
So because the system has one unique solution,
the answer is A.
But again,
let's go ahead and solve this algebraically as well,
using the same method of elimination or addition.
So again, let's multiply the first equation by three,
and the second equation we'll leave the same.
This will the X terms opposites.
So multiplying the first equation by three.
Again, we have negative three X minus six Y equals three.
The second equation remains the same.
Adding the equations together.
The X terms do simplify out because they are opposites.
Adding the Y terms,
we get negative 12 Y equals
on the right we have positive 12,
to solve for Y we divide both sides by negative 12.
Simplifying, we have Y equals negative one.
to find the corresponding X value.
We have to sub negative one for Y into one of the equations.
Let's sub negative one for Y in the first equation,
which gives us
negative X minus two times negative one equals one.
Simplifying, We have negative X plus two equals one,
subtracting two on both sides,
gives us negative X equals negative one,
dividing both sides by negative one,
we have X equals positive one.
So again,
we already know we have one unique solution,
where X is one and Y is negative one.
And now let's graph the third system.
First equation is still negative X minus two Y equals one.
The second equation is now three X plus six y equals zero.
And notice how we're,
now we have two parallel lines,
and since parallel lines never intersect,
they don't have any points in common,
which indicates the system has no solution.
So at the last system is B.
no solution.
And again, let's try to solve this algebraically.
So again,
we'll multiply the first equation by three
and keep the second equation the same.
Similar to the first system we adding the X and Y terms.
The sum is zero,
but this time on the right we have
three plus zero, which is three.
So again, both variables simplified out,
but now we're left with a false equation.
Since zero doesn't equal three.
This indicates the system has no solution.
I hope you found this helpful.