Please help with the slope problem...I don't need answers, I want to know how to solve it step by step. thank you...see the problem in attached file.
do you know the slope-intercept form of the equation of a line? \(\Large y=mx+b?\)
no that's the point I want to know
so \(\Large y=mx+b\) is the slope-intercept form of the equation of a line, where \(m\) is the slope and \(b\) is the y-intercept in your case, you are given the x-int (-5,0) and y-int (0,-2) you already have the y-int, so just plug in the y-int to the equation you will have \(\Large y=mx-2\) so we just need the slope \(m\) to find the slope we use the formula: rise/run or simply \(\Large m= \frac{y_2-y_1}{x_2-x_1}\) where \((x_1,y_1)\) and \((x_2,y_2)\) are the give points substitute the points in the slope formula and solve for the slope \(\Large m= \frac{-2-0}{0-(-5)}\) you'll get??
after you get the slope, put it in the equation
I'll get \[m=-\frac{ 2 }{ 5 } or m=\frac{ 2 }{ 5 }\] with negative or positive? Or am I getting it wrong?
with the negative because 5 will be positive :) so now put the value of m in the equation
\[y=-\frac{ 2 }{ 5} x - 2\] am I understanding correct?
yes, so far you're good ;p
thank you so much!! and do I follow the same steps on this one?
like a bit opposite
for this one, i think you need to simplify this whole equation until it will be in the form y=mx+b try it, i'll check it if we got the same thing :))
okei :D
done?
almost um I got this but not sure if right \[y= -\frac{ 5 }{ 2 }\] so far but i don't know what to do in parenthesis
@Data_LG2 so did I do right? I think \[y=-\frac{ 5 }{ 2 } x-3\]
sorry late respond, i just ate my dinner that looks right (^_^)
thank you!! (^_^)
no problem :) did you understand it now?
Yes I do, that's what i like it best, people give answers but why do we need answers if we're not going to know how to solve it, once we know how to solve it'll be easier to know the answers in future (^_^)
good to hear that ^.^ everyone should be like you ;p if you need anything just ask , see ya 'round!
haha thank you! and will do! (^_^) see ya!
Join our real-time social learning platform and learn together with your friends!