Does anyone understand Substitution in Algebra?

Okay. Here goes.

First, take the fact that y is 4x, and "Substitute" it into the other equation as y.

So  x + 4x = 5.

Combine the like terms of x + 4x.

So  5x = 5.

Then, divide each side by 5.

So x = 1!

Then, go back around and put the x=1 into one of the equations to find y.

So y = 4(1).

Then simplify to y = 4.

So your answer is (1, 4)!!

y = 4x
x + y = 5

first to solve for x you substitute 4x from the first equation in for y in the second equation
x + y = 5
x + (4x) = 5
5x = 5
x = 1

then to solve for y you can use either equation and substitute the value you got when you solved for x and substitute it in for x
x + y = 5
(1) + y = 5
y = 4

y = 4x
y = 4(1)
y = 4

x = 1 and y = 4 and there is only one solution because when you solved for both x and y there was only one answer and not multiple

Substitution is the easier of the two methods for solving simultaneous equations, as the name suggest it requires the replacing of an unknown in terms of the other.

In this example you already have y in terms of x, that is equation(1) y= 4x

sub this into equation(2) or more simply put replace y with 4x in equation(2)

Now we have equation(3) x+4x = 5

solve to get x

5x =5
x=1

now sub this value in to equation(1)
that is y= 4(1)
            = 4

now state the answer x=1 and y=4

Follow this pattern when solving all simultaneous equations and you should not have any problems.

Regarding no Solutions and infinite Solutions you may run into this as you do more advance problems where the powers of one of the unknowns is greater than 1. when solving these you would eventually form a quadratic equation if it cannot be solved then there is no Solution to your problem.
Concerning infinite solutions when ever you have to divide by 0 you will end up with infinite Solutions.
However it is hardly likely that you would run into this.