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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
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.02y=5+8 y=5/2+8/2 y=5/2+4 yintercept = 4

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1ax+by=c c/a = xint c/b = yint

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0to find the x=intercept get the x alone

Xishem
 4 years ago
Best ResponseYou've already chosen the best response.0Well, the xintercept is where the line intercepts the xaxis, and therefore when y = 0. Alternatively, the yintercept is where the line intercepts the yaxis, and therefore when x = 0. So, by plugging in 0 into the equation for either x or y, we can find the yintercept and xintercept respectively... For the first equation...\[5x+2y=8 \rightarrow 5(0)+2y=8 \rightarrow y=4\]Thus, the yintercept is (0, 4). Now for the xintercept...\[5x+2(0)=8 \rightarrow x=\frac{8}{5}\]Thus, the xintercept is (8/5, 0). Now try this with the other equation. @amistre64: Neat trick!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thnks for the help it is all clear now but im still nervous bout my final exam tomorw before our next unit

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1:) 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
 4 years ago
Best ResponseYou've already chosen the best response.0Nice (: I'll keep that trick in mind.

Xishem
 4 years ago
Best ResponseYou've already chosen the best response.0Good luck, tay1497! Just keep practicing problems and you'll do great.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now how do i find all those for equations like x=1??

Xishem
 4 years ago
Best ResponseYou've already chosen the best response.0Well, 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 yintercept is (1, 0). Now, for the xintercept we want to plug in 0 for x...\[x=1 \rightarrow 0\neq1\]Since that statement is never true, there is no xintercept.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now how do i write that in y=mx+b form?

Xishem
 4 years ago
Best ResponseYou've already chosen the best response.0Well, the 'b' in the standard slopeintercept equation stands for the yintercept. Since you know at least one point on the line (either of the intercepts) and a value for the yintercept, 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 slopeintercept form, substituting only the values for m and b to get...\[y=4x+4\]Does that make sense?

Xishem
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, wait are you asking specifically about the second equation? Let me do that one...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and first lol i thnk im an air head

Xishem
 4 years ago
Best ResponseYou've already chosen the best response.0For vertical lines (which is what x=1 is), the slopeintercept form is...\[x=b\]Where b is the xintercept. Therefore, "x=1" is already in slopeintercept form.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0k thnks for the help gotta c if i can do this now lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hey xishem can you help me with my hw? idk how to reply to you in private or if i can lol

Xishem
 4 years ago
Best ResponseYou've already chosen the best response.0Good 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.
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