anonymous
  • anonymous
i need to find the y and x intercept of linear equatioons like 5x+2y=8 and x=1, im cluless
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
2y=-5+8 y=-5/2+8/2 y=-5/2+4 y-intercept = 4
amistre64
  • amistre64
ax+by=c c/a = xint c/b = yint
anonymous
  • anonymous
to find the x=intercept get the x alone

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Xishem
  • Xishem
Well, the x-intercept is where the line intercepts the x-axis, and therefore when y = 0. Alternatively, the y-intercept is where the line intercepts the y-axis, and therefore when x = 0. So, by plugging in 0 into the equation for either x or y, we can find the y-intercept and x-intercept respectively... For the first equation...\[5x+2y=8 \rightarrow 5(0)+2y=8 \rightarrow y=4\]Thus, the y-intercept is (0, 4). Now for the x-intercept...\[5x+2(0)=8 \rightarrow x=\frac{8}{5}\]Thus, the x-intercept is (8/5, 0). Now try this with the other equation. @amistre64: Neat trick!
anonymous
  • anonymous
thnks for the help it is all clear now but im still nervous bout my final exam tomorw before our next unit
amistre64
  • amistre64
:) the proof is: \[ax+by=c\] \[\frac{ax}{c}+\frac{by}{c}=\frac{c}{c}\] \[\frac{x}{c/a}+\frac{y}{c/b}=1\] when x=0, y = c/b when y = 0, x = c/a
Xishem
  • Xishem
Nice (: I'll keep that trick in mind.
Xishem
  • Xishem
Good luck, tay1497! Just keep practicing problems and you'll do great.
anonymous
  • anonymous
i hope so xishem
anonymous
  • anonymous
now how do i find all those for equations like x=1??
Xishem
  • Xishem
Well, if you want to use amistre64's trick, you can put it into standard form and work from there...\[1x+0y=1\]If you want to use my method, you'll do the same thing...\[y_{intercept} \rightarrow 1x+0(0)=1 \rightarrow x=1\] Thus the y-intercept is (1, 0). Now, for the x-intercept we want to plug in 0 for x...\[x=1 \rightarrow 0\neq1\]Since that statement is never true, there is no x-intercept.
anonymous
  • anonymous
now how do i write that in y=mx+b form?
Xishem
  • Xishem
Well, the 'b' in the standard slope-intercept equation stands for the y-intercept. Since you know at least one point on the line (either of the intercepts) and a value for the y-intercept, you can substitute in those values and solve for m. For the first equation, the process would be something like this... \[y=mx+b\]\[4=m(0)+b \rightarrow b=4\]Now we would write the equation in slope-intercept form, substituting only the values for m and b to get...\[y=4x+4\]Does that make sense?
Xishem
  • Xishem
Oh, wait are you asking specifically about the second equation? Let me do that one...
anonymous
  • anonymous
and first lol i thnk im an air head
Xishem
  • Xishem
For vertical lines (which is what x=1 is), the slope-intercept form is...\[x=b\]Where b is the x-intercept. Therefore, "x=1" is already in slope-intercept form.
anonymous
  • anonymous
k thnks for the help gotta c if i can do this now lol
anonymous
  • anonymous
hey xishem can you help me with my hw? idk how to reply to you in private or if i can lol
Xishem
  • Xishem
Good luck, tay1497 (: If you need any more help, don't hesitate to ask. @Inhale: Just post your questions as normal and I'll try to help.
anonymous
  • anonymous
thnks

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