Answer :
you have to put the equation in slope-intercept form, which is y=mx+b, so you have to subract 10x from both sides to get 6y=-10x+100, then divide all the terms by 6 to get y=-5/3x+16 2/3. Then all you do is graph 16 2/3 on the y axis and from that point do the slope, which is -5/3, so you have to go down on the y axis 5, and over to the right 3, and put a point. Then you just repeat this to get a few more points to be able to draw a line. (I hope this helps! sorry if i confused you even more, I'm not the best at explaining things, but hopefully I cleared things up a bit for you)
[tex]standard\ equation\ linear :\\y=ax+b\\\\10x + 6y =100\ \ |\ subtract\ 10x\ to\ both\ sides \\\\6y=-10x+100\ \| \ divide \ each \ term \ by \ 2\\\\3y=-5x+50| \ divide \ each \ term \ by \ 3 \\\\y=-\frac{5}{3}x+\frac{50}{3}[/tex]
[tex]x=1 \ \ \to\ y=-\frac{5}{3}*1+\frac{50}{3} =-\frac{5}{3} +\frac{50}{3}= \frac{45}{3}=15\\\\ x=10 \ \ \to\ y=-\frac{5}{3}*10+\frac{50}{3}=-\frac{50}{3} +\frac{50}{3}=0[/tex]
[tex]x=1 \ \ \to\ y=-\frac{5}{3}*1+\frac{50}{3} =-\frac{5}{3} +\frac{50}{3}= \frac{45}{3}=15\\\\ x=10 \ \ \to\ y=-\frac{5}{3}*10+\frac{50}{3}=-\frac{50}{3} +\frac{50}{3}=0[/tex]
