Question

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

Answer 1

2y=-5+8
y=-5/2+8/2
y=-5/2+4
y-intercept = 4

Answer 2

ax+by=c

c/a = xint

c/b = yint

Answer 3

to find the x=intercept get the x alone

Answer 4

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!

Answer 5

thnks for the help it is all clear now but im still nervous bout my final exam tomorw before our next unit

Answer 6

:) 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

Answer 7

Nice (: I'll keep that trick in mind.

Answer 8

Good luck, tay1497! Just keep practicing problems and you'll do great.

Answer 9

i hope so xishem

Answer 10

now how do i find all those for equations like x=1??

Answer 11

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.

Answer 12

now how do i write that in y=mx+b form?

Answer 13

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?

Answer 14

Oh, wait are you asking specifically about the second equation? Let me do that one...

Answer 15

and first lol i thnk im an air head

Answer 16

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.

Answer 17

k thnks for the help gotta c if i can do this now lol

Answer 18

hey xishem can you help me with my hw? idk how to reply to you in private or if i can lol

Answer 19

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.

Answer 20

thnks