Consider the equation 7x + 3y = 42.

Question

Part 1: On your own paper, graph this equation using the slope-intercept method. In the space provided, explain, in words, each step of the procedure you used. Make sure to use complete sentences and correct grammar. (3 points)
Part 2: On your own paper, graph this equation using the intercepts method. In the space provided, explain, in words, each step of the procedure you used. Make sure to use complete sentences and correct grammar. (3 points)

Answer 1

I am not certain of your question about this. I will be happy to critique your thinking.

Answer 2

I just need to show the work from getting the equation 7x + 3y = 42 into slope- intercept form. Then I need the step by step work for graphing it in intercepts form.

Answer 3

Your equation is perfect for finding intercepts.

let x=0 , (put your little finger over the 7x term), then y = 14
let y = 0, ( put your little finger over the 3y term), then x = 6
so you have 2 nice points, (0,14) and (6,0)

to change to slope-yintercept form
3y = -7x + 42
y = (-7/3)x + 14

Answer 4

Thank you!

Answer 5

1. 7X + 3Y = 42.
Convert the std form Eq to the slope-interceptform:
7X + 3Y = 42,
3Y = -7X + 42,
Y = -7/3X + 14, (Y = mx + b).
Our 1st point will be at the Y-int
where X is always 0.
P1(0 , 14).

Since our slope is neg. DECREASE
b by 7:
y = 14 - 7 = 7.
X = 3 = The denominator of the slope.
P2(3 , 7).

GRAPH: P1(0 , 14) , P2(3 , 7)

2. 7X + 3Y = 42
At the y-int., X = 0.
Calculate Y:
7(0) + 3Y = 42
3Y = 42
Y = 14.
P1(0 , 14).

At the X-INT, Y = 0.
Calculate X:
7X + 3(0) = 42,
7X = 42,
X = 6.
P2(6 , 0).

GRAPH: P1(0 , 14) , P2(6 , 0).