Graph the linear equation. find three points that solve the equation, then plot on the graph. -5x -3y = -7

-5x - 3y = -7
-3y = 5x - 7
y = -5/3x + 7/3so there is a slope of -5/3and a y int of 7/3 (or 2 1/3)

the x int can be found by subbing in 0 for y and solving for x
0 = -5/3x + 7/3
5/3x = 7/3
x = 7/3 * 3/5
x = 7/5 (or 1 2/5)

so lets put this in standard form and find some points on ur line that are not fractions.
y = -5/3x + 7/3
5/3x + y = 7/3...multiply everything by 3
5x + 3y = 7this is standard form

5x + 3y = 7subbing in 2 for x
5(2) + 3y = 7
10 + 3y = 7
3y = 7 - 10
3y = -3
y = -3/3
y = -1so one solution is (2,-1)

5x + 3y = 7subbing in -1 for x
5(-1) + 3y = 7
-5 + 3y = 7
3y = 7 + 5
3y = 12
y = 12/3
y = 4so another point on ur line is (-1,4)

5x + 3y = 7subbing in -4 for x
5(-4) + 3y = 7
-20 + 3y = 7
3y = 7 + 20
3y = 27
y = 27/3
y = 9 and another point on ur line is (-4,9)

another one is (3,-3)

thats gonna be all the even numbers that u will get on ur graph..
(2,-1), (-1,4), (-4,9) , (3,-3)

so to break it down, ur y int (where the line crosses the y axis) is at (0,7/3) or (0,2 1/3)ur x int (where ur line crosses the x axis) is at (7/5,0) or
(1 2/5,0)and then there are the points on ur line (2,-1),(-1,4),(-4,9), and (3,-3)

Top right since it passes vertical line test