3y+4x=3 x+3y=-6
I sort of already solved it. I used standard form. (The one where you have a table and do x intercept and y intercept. For y int, x=0 and for x int, y=0) Ax=By=C
I got x=3/4 and y=1; for the second one, I got x=-6 and y=2. What did I do wrong?

step-by-step explanation:

through just a quick inspection, one can see that the graph of this cubing function starts in quadrant iii and ends in quadrant i.   when x = 0, y = 3, so the y-intercept is (0, 3).

if graphs are part of the problem statement, share them.   .

the area of a triangle is determined by the formula: base x height * 1/2

you'll need to assign a base to the triangle. based on the vertices given, the simplest base would be the line segment from (-2,1) to (2,1). the distance between any two numbers is found by just subtracting them. here, you should subtract the x values to find the distance of this side.

2 - (-2) = 4.

so the distance (aka the length of the base) is 4. the height of the triangle is the distance between the base and the highest point on the triangle which is (3,4). we should subtract the y values this time to find the height/distance.

4 - 1 = 3

so the height is 3.

base = 4

height = 3

hope this if needed hmu