I have a trick question, I am in week three and solving systems of linear equations with 2 variables using either the elimination method or the substitution method and then graphing. My problem on the test is missing the second variable, how do I create new equations containing them so that I can solve and graph? My equations are: 4x = 8 and 5y = 15
Stacey Warren - Expert brainly.com
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it's too easy. That's why you are having trouble. x =2 and y = 3
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yes, waaaaaaaaaaaaaaaaaaaaaaaaaay to easy
that is the intersection of the two lines. One is vertical and one is horizontal.
you don't need to eliminate anything. you can reduce the left-hand side of both equations through division to solve for the variables.
4x = 8
x = 2
5y = 15
y = 3
(2,3) is the cross
that is too easy, I have to have other points on the graph
There is only one point of intersection.
Yes, I know, but in our text, it states that : Because we are considering linear equations in two variables, the equation x = 3 is equivalent to x + 0(y) = 3. Now we can see that any value of y can be used, but the x value must always be 3. Therefore, some of the solutions are (3, 0), (3, 1), (3, 2), (3, − 1), and (3, −2).
just plot both lines then;
This is what I need to set up somehow
Nothing is that simple in week 3 of college algebra
you have two lines. 4x = 8 and 5y = 15
equivalently, 4x + 0y = 8 and 0x + 5y = 15
so for 4x = 8 you'd have the points ..., (2,-1), (2,0), (2,1), ... that is every point on the line x = 2
same for 5y = 15: ...,(-1,3), (0,3), (1,3), ... every poitn on the line y = 3
plot them and find the intersection i.e. (2,3). college algebra is a joke of a class so don't think this is too easy.
OMG, it took me 12 hours to do 8 problems, this is only my first question, I'm dying here