Check my work. Okay so I have a system of linear equations {y - 2 = (1/3)x and 3y = - 2x - 3.
I need to get them into slope intercept form so I can graph them. Here is my answer: on the upper one we can add 2 to both sides. y - 2 + 2 = 1/3x + 2 = y = 1/3x + 2. Now we need to get the second equation into slope intercept form. So we divide both sides by three. 3y/3 = - 2x - 3/3 = y = - 2x - 1

Question

Check my work. Okay so I have a system of linear equations {y - 2 = (1/3)x and 3y = - 2x - 3.
I need to get them into slope intercept form so I can graph them. Here is my answer: on the upper one we can add 2 to both sides. y - 2 + 2 = 1/3x + 2 = y = 1/3x + 2. Now we need to get the second equation into slope intercept form. So we divide both sides by three. 3y/3 = - 2x - 3/3 = y = - 2x - 1

jim_thompson5910 · Answer

the first one is correct

the second is not

jim_thompson5910 · Answer

you need to divide every part by 3 so the second equation should be

y = -2/3x - 1