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anonymous

  • 4 years ago

i need to find the y and x intercept of linear equatioons like 5x+2y=8 and x=1, im cluless

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  1. anonymous
    • 4 years ago
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    2y=-5+8 y=-5/2+8/2 y=-5/2+4 y-intercept = 4

  2. amistre64
    • 4 years ago
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    ax+by=c c/a = xint c/b = yint

  3. anonymous
    • 4 years ago
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    to find the x=intercept get the x alone

  4. Xishem
    • 4 years ago
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    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!

  5. anonymous
    • 4 years ago
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    thnks for the help it is all clear now but im still nervous bout my final exam tomorw before our next unit

  6. amistre64
    • 4 years ago
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    :) 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

  7. Xishem
    • 4 years ago
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    Nice (: I'll keep that trick in mind.

  8. Xishem
    • 4 years ago
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    Good luck, tay1497! Just keep practicing problems and you'll do great.

  9. anonymous
    • 4 years ago
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    i hope so xishem

  10. anonymous
    • 4 years ago
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    now how do i find all those for equations like x=1??

  11. Xishem
    • 4 years ago
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    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.

  12. anonymous
    • 4 years ago
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    now how do i write that in y=mx+b form?

  13. Xishem
    • 4 years ago
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    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?

  14. Xishem
    • 4 years ago
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    Oh, wait are you asking specifically about the second equation? Let me do that one...

  15. anonymous
    • 4 years ago
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    and first lol i thnk im an air head

  16. Xishem
    • 4 years ago
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    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.

  17. anonymous
    • 4 years ago
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    k thnks for the help gotta c if i can do this now lol

  18. anonymous
    • 4 years ago
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    hey xishem can you help me with my hw? idk how to reply to you in private or if i can lol

  19. Xishem
    • 4 years ago
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    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.

  20. anonymous
    • 4 years ago
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    thnks

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