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.