straight line graphs 5 marker

[ashen snow]

need a bit of help

part b

l1: y=3/5x + 6 (given in the question)
l2: y=-5/3 x + 40/3

equated them:
3/5x + 6 = -3/5x + 40/3

got x to be 55/9

so we know point C is (55/9, y)

not sure if I did the algebra correctly to get the 55/9, pls correct me if its wrong

so I gotta sub the 55/9 into any one of the equations since they both pass point C

i'll go for y=3/5x + 6

y= 3/5(55/9) + 6
= 29/3

so ig point C = (59/9, 29/3)

then I gotta use 1/2(b x h) to get ABC

oh ye I forgot to mention:
l1 cuts the x axis when y=0

to get the x intercept:
0= 3/5(x) + 6
-6 = 2/5x therefore x = -15
point A = (-15, 0)

can some1 pls tell me if I've messed up anywhere? cuz i think I may have

<@&791435371564892232>

[sinful ferry]

when u use calc and equate the l₁ and l₂ u gave

it gives 55/17

[ashen snow]

ohh

is that where I went wrong?

oh shi I didnt flip

for my coefficients of x I put 3/5 AND -3/5

shouldve been -5/3 right

[sinful ferry]

ooh

[ashen snow]

yeah?

[sinful ferry]

yes that must've been it

[ashen snow]

other than that am I good to go?

[sinful ferry]

lemme read the rest

[ashen snow]

kk ty

yeah yk that (55,17), I can use that in any one of the 2 equations to get my y value right

[sinful ferry]

ur method for finding A is right but u made a small mistake somewhere w algebra, but when u redo it you'll get it

[sinful ferry]

yes

[ashen snow]

0 = 3/5x + 6

[sinful ferry]

yes

[ashen snow]

yea so I got A

[sinful ferry]

u got A as (-15,0) last time which wasn't right

[ashen snow]

oh

I gotta do it again?

lemme try

[sinful ferry]

yes

[ashen snow]

lmao I took it as 2/5

l1

ye A = -10

(-10, 0)

[sinful ferry]

yes

[ashen snow]

cool ty, I can do the rest

thank u

[sinful ferry]

okok

nppp