Question

Solving simultaneous linear equations. Why is simply equating the y values sufficient to give us the point of intersection??

Answer 1

Because it is 2d chart, we have equations of straight lines, so it can intersect only on one location

Answer 2

Equating the y's means finding the x for which each line has the same y value. This is the x-coordinate of the point of intersection.

Answer 3

Line equations represent a relationship between the two variables; any point of the line has the property that if you substitute x and y into the equation then the equation holds true, and if you do the same for any point not on the line then the equation is false.

When two lines intersect this gives you a point geometrically, but for the equations it means that both must be true at the same time. If the two lines are y = f(x) and y = g(x), then these are simultaneously true, so f(x) = g(x). Any solution of x here will correspond to a point of intersection. Since we're working with straight lines here, f(x) and g(x) are linear functions and so this equation of x is linear, meaning there can only be 0, 1, or infinitely many solutions.

Answer 4

Let's take your example.  Maybe it would help to give the two lines different labels for their y-values, so

y1 = x + 1

y2 = 2x - 1

x is indeed a variable, and as we consider the values x can take, we can calculate the corresponding y on each line, eg if I pick x = 1:

y1 = 1 + 1 =2    -> so this line goes through (1,2)

y2 = 2 * 1 - 1 = 1   -> this line goes through (1,1)

If I pick x = 3:

y1 = 4

y2 = 5

I notice that between these two points, y1 has gone from being higher than y2 to being lower than y2.  So naturally, we wonder if there is a point where the two lines meet, ie a particular value of x with a corresponding y1 and a corresponding y2 with the property that y1 and y2 are the same.  

So, we are asking, is there a value of x such that x+1 has the same value as 2x-1.  This will mean y1 = y2, and we could call that common value y.

Clearly, we can find the answer to our question by solving x + 1 = 2x-1.

Yes, y = x+1 expresses a relationship between two variables, and y = 2x-1 expresses another relationship between two variables.  But x and y can't satisfy both those relationships at the same time, except for a particular pair of values of (x,y).  When we asked to solve these equations simulatneously, we are being asked to find that pair which satisfies both relationships at the same time.

Answer 5

A function is a collection of ordered pairs (x,y) where for any x in the domain, there is only y paired with it. A graph is a convenient way to visualize the (x,y) pairs as coordinates in 2 dimensional plane.

If the graphs of two functions intersect, then they must share some pair (x,y) in common. In you example, equation 1 is the set of points (x, x+1) and equation 2 is the set of points (x,2x-1). If those two graphs intersect, then there must be some point (x,y) shared by both. Let’s assume that happens when x=a. Then the point is represented as (a,a+1) on the first graph and (a,2a-1) on the second graph. But if they are the same point, then the x and y coordinates must be equal. Setting the x values equal gives a = a, which isn’t very interesting. But setting the y values equal gives a+1 = 2a-1, from which get a=2. Plugging that into either equation gives a y value of 3. So, the graphs intersect at (2,3) because the point (2,3) is on both graphs. Typically, we just skip the part where we say that x=a and just work directly with the variables c and y.

Answer 6

Two points are equal if and only if their x coordinates are equal and their y coordinates are equal. So you're not _just_ setting the y's equal to each other; you're also implicitly setting the x's equal to each other too. So at the end, you only have one x value.

I think a big part of the confusion is that we're using the same letters for both lines, which is ambiguous. If I instead write the first line as y1 = a * x1 + b, and the second line as y2 = c * x2 + d, you are absolutely right that setting the y's equal is not enough, because we get y1 = y2 ==> a * x1 + b = c * x2 + d, which could be true for any y if we put in different values for x1 and x2. But if we say that x1 must equal x2, then we've locked ourselves into one variable.