Can you explain the Substitution method in algebra to me?

Substitution just means you replace a variable with some other expression or value in order to solve an equation.  

If you are referring to using the substitution to solve a system of two linear equations, you write one of the equations so that it is solved for one of the variables and then substitute that expression into the other equation in the system.  This will give you one equation with one variable.

Solve the new equation.  You will get a numeric value for the variable.  Now substitute that number for the variable into either equation, and solve for the other variable.  You than have the solution for the system.

You are exchanging just y (for ex.) for what it equals in the second equation. So if y= x +3 and  y= x+5, the problem would be x+3=x+5. Combine and solve from there, but take the answer and multiply (with the real problems) the answer you receive with one of the equations and get x or y or whatever your variable is.

I hope this makes sense!!! Good Luck!!

The best way is to learn by example.

say our system is:
x+y=4
x-y=2

first solve one equation in terms of one variable
y=4-x
Then substitute this y-value in the next equation:
x-(4-x)=2

solve for x
2x=6
x=3

Then substitute this x-value in one of the equations to find y.
y=4-(3)=1

The solution is (3,1)

lets say  x = 2 + n

and we have an equation:  4x + 6x = 5

by 'substituting' x into the equation we just replace where ever there is an x with 2 + n
so  4x + 6x = 5  becomes  4(2+n) + 6(2+n) = 5

you would then open out the brackets and solve n.

Do you mean substitution for simultaneous equations?

Let's say you have these two equations:
3x+y=19
2x+4y=12

You need to solve for both x and y, but first, choose whichever variable you want to solve for first.
For this example, I'd solve for x first because there's a really easy isolation of y in the first equation.

So, go ahead and isolate whichever variable you're not solving for in one of the equations.
y=19-3x
Now, in the second equation, you can substitute this for y (or whatever variable) so that you only have one variable to solve for in your other equation.
2x+4(19-3x)=12
Solve for the variable.
2x+76-12x=12
-10x+76=12
-10x=-64
x=64/10=32/5

Once you have the value of one equation, use what you substitued for the other to solve for the other.
So, I'm gonna use y=19-3x.
y=19-3(32/5)
y=19-96/5
y=-1/5

So, my answer is (32/5, -1/5), but that's the process.

If for an example you were given the problem x^6 + x^3 + 12 = 0 you would solve this by substituting y in place x^3 so:

y^2 + y + 12 = 0

y^2 + 4y + 3y + 12 = 0

y (y + 4) + 3 (y + 4) = 0

(y + 3) (y + 4) = 0

1) y = -4

2) y = -3

But since you have substituted y as x^3 therefore you'll repalce them again to find the value of x so:

1) x^3 = -4

x = (-4)^1/3

x = -4/3

2) x^3 = -3

x = (-3)^1/3

x = -1


I hope this helps/

I think you are asking about the substitution method for solving linear equations

if we have two equations
a x + b y = c : eqn 1
d x + e y = f : eqn 2

where a, b, c, d, e, and f are constants

The substitution method solved one of the equations for either x or y for example

eqn 1 can be solved for x
x = (c - by)/a

we can then eliminate the x variable in eqn 2 by SUBSTITUTING the value of x

d x + e y = f
d( (c - by)/a ) + ey = f
dc - bd y + ae y = af
(ae - bd) y = af - dc
y = ( af - dc)/ (ae -bd)