We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

Suppose you graphed every single point of the form $(2t + 3, 3-3t)$ . For example, when $t=2$ , we have $2t + 3 = 7$ and $3-3t = -3$, so $(7,-3)$ is on the graph. Explain why the graph is a line, and find an equation whose graph is this line.

In order for this graph to be a line, we need to verify two things: first, that all the points on the graph are on the proposed line and second, that all points on the proposed line are on the graph. Be sure to deal with both.

Thanks! HALP:)

Guest Aug 22, 2018

edited by
Guest
Aug 22, 2018

#1**+2 **

Suppose you graphed every single point of the form $(2t + 3, 3-3t)$ . For example, when $t=2$ , we have $2t + 3 = 7$ and $3-3t = -3$, so $(7,-3)$ is on the graph. Explain why the graph is a line, and find an equation whose graph is this line.

we have a graph defined by the locus of all points (2t+3, 3-3t)

so

x= 2t +3 (1a) and y=3-3t (2a)

I am going to sove these simultaneously in order to eliminate t.

3x=6t+9 (1b) and 2y=6-6t (2b)

add equations 2a and 2b

**3x+2y=15**

This is the equation of a line.

All points on this line will satisfy the given condtion and no point that is not on this line will satisfy it.

Melody Aug 25, 2018