Question

I don't know what this is talking about

Find the vaule of a b and c that will garuntee that the system has exactly one solution

x+2y = 3
ax+by = c

Answer 1

howd you do that?

Answer 2

first I'm going to put both in y=mx+b form

\[y=\frac{-x}{2}+\frac{3}{2}\]

\[y=\frac{-ax}{b}+\frac{c}{b}\]

Parallel lines would give us no solution. Parallel lines are lines with same slope and different y-intercepts.

Same lines would give us infinitely many lines. These are lines with same slope and same y-intercept.

Any other types would give us one intersection.

So this is what we want:
\[\frac{-1}{2} \neq \frac{-a}{b} \]

Answer 3

For example if a =5 and b=2 and c= whatever like 1
We have what we want above
\[\frac{-1}{2} \neq \frac{-5}{2}\]
so we would have the following two lines
\[y=\frac{-x}{2}+\frac{3}{2}\]
\[y=\frac{-5x}{2}+\frac{1}{2}\]

To show you that this does have one intersection I will solve the system
\[\frac{-x}{2}+\frac{3}{2}=\frac{-5x}{2}+\frac{1}{2}\]
first I multiply both sides by 2
\[-x+3=-5x+1\]
\[5x-x=1-3 =>4x=-2 => x=\frac{-2}{4}=\frac{-1}{2}=> y=\frac{\frac{-1}{2}}{2}+\frac{3}{2}=\frac{-1}{4}+\frac{6}{4}=\frac{5}{4}\]
This is just one possible set of values that a,b,c can be

Answer 4

wow ok thanks

Answer 5

you just don't want -1/2 to equal -a/b

Answer 6

so choosing a=2 and b=4 would be wrong
do you see why?

Answer 7

yes

Answer 8

because 2/4 reduces to 1/2

Answer 9

they are the same fraction