How many solutions does the system below have?

1 answer:

4 0

Answer: 1 solution

An augmented matrix for the system is
[2,6,12;1,1,2] where we will call the first two columns A and the third b. A is linearly independent since it is row equivalent to the identity matrix, which tells us that for any b there will be a unique solution (x,y).

You might be interested in

Write the equation of the line that passes through (−3,1) and (2,−1) in slope-intercept form

Alex787 [66]

Answer:

$y=-\frac{2}{5}x-\frac{1}{5}$

Step-by-step explanation:

The equation of a line is y = mx + b

Where:

m is the slope

b is the y-intercept

First, let's find what m is, the slope of the line.

Let's call the first point you gave, (-3,1), point #1, so the x and y numbers given will be called x1 and y1.

Also, let's call the second point you gave, (2,-1), point #2, so the x and y numbers here will be called x2 and y2.

Now, just plug the numbers into the formula for m above, like this:

$m = -\frac{2}{5}$

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

$y=-\frac{2}{5}x + b$

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

(-3,1). When x of the line is -3, y of the line must be 1.
(2,-1). When x of the line is 2, y of the line must be -1.

Now, look at our line's equation so far: $y=-\frac{2}{5}x + b$ . $b$ is what we want, the - $-\frac{2}{5}$ is already set and x and y are just two 'free variables' sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (-3,1) and (2,-1).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!

You can use either (x,y) point you want. The answer will be the same:

(-3,1). y = mx + b or $1=-\frac{2}{5} * -3 + b$ , or solving for b: $b = 1-(-\frac{2}{5})(-3)$ . $b = -\frac{1}{5}$ .
(2,-1). y = mx + b or $-1=-\frac{2}{5} * 2 + b$ , or solving for b: $b = 1-(-\frac{2}{5})(2)$ . $b = -\frac{1}{5}$ .