Question

IF IN THE graph of two - variable linear eq. l and m are intersect each other then...❓

answer with full explanation fast....

Answer 1

A graph of the two lines y = 3x and x + 2y = 4 shows us that the lines intersect, meaning that there is a single (x, y) value that satisfies both equations. ... To check this finding, we can compare the slopes of the equations. To make that easy, we'll rearrange the equations into slope-intercept form, or y = mx + b.

If both variables are eliminated and you are left with a true statement, this indicates that there are an infinite number of ordered pairs that satisfy both of the equations. In fact, the equations are the same line. Solve for x and y.

I hope it's helpful for u.....¯\_(ツ)_/¯

Answer 2

Answer:

Explanation:

Okay, let’s look at it this way: when does a line pass through the origin? The line represents possible values of x and y that satisfy the equation.

So, when a line passes through the origin, it passes through the coordinates (0,0). x = 0, y = 0. So, let’s model this with the equation ax + by + c = 0. Sub in x = 0 and y = 0 to the equation and we find that 0 + 0 + c = 0. Clearly, c = 0.

So, with this simple explanation, I hope you understand when a line does pass through the origin.

Now, let’s look at when a line doesn’t pass through the origin. This is when c is not equal to 0. Hence, when x = 0, y cannot equal 0; c + by = 0, and we know that c is not 0. If y is 0, then we get c = 0… where c is not 0. Ehh. Thus, you can see that a line does not pass through the origin when c is not equal to 0 by ehat is hope is a simple explanation. You don’t need to know how to prove it, I presume, but that’s not too hard either.

Oops, I realised that I just assumed you were talking about linear graphs. For quadratic graphs, the reasoning is similar. For graphs of the form y = ax^2, the minimum/maximum point of the graph will be the origin. For graphs of the form y = ax^2 + bx, it will pass through the origin but the line of symmetry will be different. For graphs of the form y = ax^2 + bx + c (you know, where c is not zero) , the graph will not pass through the origin because the maximum/ minimum point is actually raised or lowered by c units.

please read it carefully

hope it helps

:)