What is the graph of the equation? -5x+y=-3 I don't know what to do, help?
You can give any arbitrary number to "x" and solve for "y", for at least two points on the graph, if you join the points you'll obtain the graph of that line.
I'm sorry could you explain just a bit more? if thats possible
I can use -5 for x right? But how would i go about solving y.. Thank you btw I'm sorry im so confused haha
Say for instance, We consider x=-5, then: \[-5(-5)+y=-3\] Now, we just operate: \[25+y=-3\] \[y=-28\] So, therefore, the point "w", as we can call it, belongs in the line whose equation is given, if you represent this point on the cartesian plane, and repeat the process for another value of "x", then join the points, you'll obtain the graph of the line. \[w(-5,-28)\]
-5x + y = -3
another more direct approach let y be the subject
Have you learned the form y = mx + b
Yes.
y = 5x -3
when x = 0 y = -3
I added 5x to both sides :-)
Would it be positive or negative?
Does that make sense?
Yes it does, I have 2 answer choices that could work i just need to figure out if its negative or positive
5x + (-5x) = 0
ok i'll tell a simpler way
please
What do you want to know is positive or negative?
The line on the graph
The slope is positive because the 5 in 5x is positive.
if u have something like this- ax+by=c then just divide all over by c \[\frac{ ax }{ c }+\frac{ by }{ c }=\frac{ c }{ c }\]\[\frac{ x }{ \frac{ c}{ a }}+ \frac{ y }{ \frac{ c }{ b } }=1\] calculate c/a and c/b and (c/a, 0)=point where the graph cuts x axis (0,c/b)=point where graph cuts y axis so basically just identify a,b,c and then find the points then join them and thats ur graph
Do you understand what slope is?
@mickey1513 did u get my method?
So, its going to be positive
yay it was right. thank you i get it now all of you deserve medals
well thr best way was like what owl said u jst take some random value of x and get y and plot that point then u take another random value of x and put it in equation nd get y and pot this point and then join these 2 points
Thanks for asking :-)
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