How do you graph #y=2/3x+5/7#?

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Answer 1 · 2017-10-05T12:29:07

See a solution process below:

First, solve for two points which solve the equation and plot these points:

First Point: For #x = 0#

#y = 0 + 5/7#

#y = 5/7# or #(0, 5/7)#

Second Point: For #x = 4#

#y = (2/3 * 4) + 5/7#

#y = 8/3 + 5/7#

#y = (7/7 xx 8/3) + (3/3 xx 5/7)#

#y = 56/21 + 15/21#

#y = 71/21 or #(4, 71/21)#

We can next graph the two points on the coordinate plane:

graph{(x^2+(y-(5/7))^2-0.025)((x-4)^2+(y-(71/21))^2-0.025)=0 [-10, 10, -5, 5]}

Now, we can draw a straight line through the two points to graph the line:

graph{(y - (2/3)x - (5/7))(x^2+(y-(5/7))^2-0.025)((x-4)^2+(y-(71/21))^2-0.025)=0 [-10, 10, -5, 5]}