MAT 1302 Unit VII-Solving Systems of Equations Part II Q4-5

5 days ago

English subtitle

Solving Systems of Linear Equations, Part
Two.
Solve the system by the method of your choice.
Here we have two equations, two linear equations,
and we can either solve this by substitution
or the addition method.
I'm going to solve this using the addition
method.
First, I want to rewrite the top equation
so that all the letters are on the left side
and then the constant terms are on the right
side.
So, that just means that I'm going to subtract
the y from both sides.
So, I get 4x minus y equals a negative seven.
Now both equations are written in the correct
form, so I can go ahead and use my addition
method.
Now I want to multiply an equation by a value
so that when I add the equations together,
the x's go away or the y's go away.
If I multiply the second equation by a negative
one, when I add the x's together, they will
be eliminated.
So, I'm going to multiply the second equation
by a negative one and this gives me a negative
4x plus y equals a negative two and remember
I multiplied every single term by a negative
one.
The top equation stays the same and now we
can add the equations together.
When we do that, we get 4x minus 4x, which
is 0, but then I also have minus y plus y,
which is also 0.
So, both of my variables disappeared and I'm
left with 0 equals negative 7 plus negative
2.
This is a negative nine.
Now, zero equals negative nine is not a true
statement.
So, since zero equals negative nine is not
true, the solution is no solution and we can
say that the solution set is
the empty set.
Solve the system by the method of your choice.
Here we have two systems of linear equations
and I'm going to solve this system using the
substitution method.
That's because the top equation is already
solved for x, so I can just go ahead and plug
in my expression for x into the second equation.
So, I'll take the second equation and I'm
going to substitute a for y plus 1 for x and
now I'm going to solve this equation.
The first step is to distribute the 3 to both
terms within parentheses.
We have 12y plus 3 and then minus 12y equals
3.
I'll combine like terms on the left side.
So, I have a positive 12y and a negative 12y.
These cancel because 12y minus 12y is 0.
So, I'm just left with a 3 on the left side
and a 3 on the right side.
Now, 3 equals 3 is a true statement.
That means that my solution is infinitely
many solutions.
That's the answer to my problem and if we
were to graph this system of equations, we
would see that these two equations are actually
the same line, so that's why there are infinitely
many solutions.