WHY can we use the substitution method to solve linear equations?

you want to find the solution that is common to both linear equations....
ex:  y = 2x  and x - 3y = 15
you want to know what ordered pair (x,y) makes both equations true at the same time.
If you graphed these two equations, you would see that this is the point where they intersect.

so using substitution....
the second equation becomes x - 3(2x) = 15
x - 6x = 15...... -5x = 15 ..... x = -3

now if you replace x with -3 in either equation...
y = 2(-3) = -6   and -3 - 3y = 15 ...
-3y = 18......y = -6
so the ordered pair (-3 , -6) is the solution for both linear equations.........

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I take it you mean simultaneous equations.

The reason is that each equation represents a relationship between two unknowns

eg

x + y = 7
this has an infinite set of solutions (1,6) (7,-1) etc

but at the same time x and y have to satisfy another relationship

eg

x-y=1

so the solution satisfies both relationships

ie x+y=7 and x-y=1 at the same time

from the first equation x=7-y

but we know that x-y=1 so:

to satisfy the second equation (7-y) - y = 1  (our substitution method)

hence our result (4,3)

hope thats clear

Assume you have the following equations
2x + y = 4 and
x + 2y = 5

Since you have two unknowns in a equation, you can not solve readily the unknowns.  But if you solve for the value of one in terms of the other and substitute  in the other equation, the result is an equation in one unknown, which is readily solvable.

If we use equation 2 and solve for x;
x = 5 - 2y
Substitute to Equation 2
2(5 - 2y) + y = 4  ------------- you have just y in this equation
10 - 4y + y = 4
-3y = 4 - 10 = -6
y = 2 and
x = 5 - 2y = 5 - 2(2) = 1

An equation by its very nature tells you about something that is equal.  Now, if you have something on the left equal to something on the right of the equals sign you can do various operations without losing the truth of the equation.

e.g.  suppose you have x + y = 3
You can add or subtract something (a constant or a variable) to both sides to arrive at another true statement,
e.g.  subtract 3 gives: x + y - 3 = 3 - 3, so x + y - 3 = 0
You can also multiply or divide both sides by the same amount
e.g. multiply by 2 gives: 2(x + y) = 2 x 3, so 2x + 2y = 6

Then with any of the equations that you can write in this manner you can substitute the expression on the left for the expression on the right into another (simultaneous) equation, because you know they are equal.

e.g. suppose you have two equations (that are true at the same time, or simultaneously, i.e. with the same values of x & y):
x + y = 3  and  x + 2y = 5
from the first you can say that x + y - y = 3 - y, so x = 3 - y
Knowing this to be true you can substitute into the second:
x + 2y = 5, so (3 - y) + 2y = 5, so 3 + y = 5, so y = 5 - 3,
so y = 2

You now know the value of one unknown and can substitute for this into either equation, e.g. the first:
x + y = 3, so x + 2 = 3, so x + 2 - 2 = 3 - 2, so x = 1

i.e. x = 1 and y = 2


Check back through the examples with these values so that you're happy that all the operations (provided that the same operation is applied to BOTH sides of an equation) leave you with a true statement (i.e. another equation).
.

Suppose we have two linear equations of
y = ax + b and y = cx + d.

A linear equation can be seen as a straight line in Cartesian coordinates. And we all know, that if there are two lines, there are only three possibility of its manner: intercept each other in precisely one point (x,y), have no point of interception, and intercept in infinite points (the two lines are affix).

The first condition met when a >< c, the second one occur when a = c and b >< d, and the third condition satisfied of course when a = c and b = d.

Since there are no pairs of (x,y) satisfy the second condition and there are infinite pairs of (x,y) that satisfy for the third conditions, I assume the question is when the first condition occur.

The answer for the first condition is one (and ONLY one) pair of (x,y) which satisfy both linear equations.

To find a pair of (x,y) which satisfy both equation first we have to find pairs of (x,y) which satisfy the first linear equation. Suppose the pairs that satisfy the first equation are (x, ax + b) for every x. Then we have to find the pair that satisfy the second linear equation from the pairs of (x, ax + b).
To satisfy the second equation, we must have ax + b = cx + d (because only pairs from the first result is taken, where y = ax + b)
Thus we have x = (d ? b)/(a ? c) as the answer.

That?s why we called it ?substitution? method. Because we substitute the y in the second linear equation with ?specific? y which satisfy the first linear equation.

Sorry if my answer is not satisfying. I'm not good at explaining.