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Q. 129

Expert-verified
Found in: Page 44

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# Which of the following equations might have the graph shown? (More than one answer is possible.)$\left(a\right)2x+3y=6\phantom{\rule{0ex}{0ex}}\left(b\right)-2x+3y=6\phantom{\rule{0ex}{0ex}}\left(c\right)3x-4y=-12\phantom{\rule{0ex}{0ex}}\left(d\right)x-y=1\phantom{\rule{0ex}{0ex}}\left(e\right)x-y=-1\phantom{\rule{0ex}{0ex}}\left(f\right)y=3x-5\phantom{\rule{0ex}{0ex}}\left(g\right)y=2x+3\phantom{\rule{0ex}{0ex}}\left(h\right)y=-3x+3$

The equation which describes the given line is

$\left(b\right)-2x+3y=6\phantom{\rule{0ex}{0ex}}\left(c\right)3x-4y=-12\phantom{\rule{0ex}{0ex}}\left(e\right)x-y=-1\phantom{\rule{0ex}{0ex}}\left(g\right)y=2x+3$

See the step by step solution

## Step 1: Given information

We are given graph and several equations

## Step 2: Explanation for correct options

Consider the option

$b\right)-2x+3y=6\phantom{\rule{0ex}{0ex}}y=\frac{2}{3}x+2$

and

$c\right)3x-4y=-12\phantom{\rule{0ex}{0ex}}4y=3x+12\phantom{\rule{0ex}{0ex}}y=\frac{3}{4}x+3$

also

$e\right)x-y=-1\phantom{\rule{0ex}{0ex}}y=x+1$

Similarly

$g\right)y=2x+3$

Each and every equation has positive y-intercept similarly from the graph we can say that the line has positive y-intercept

## Step 3: Consider the equations

$\left(a\right)2x+3y=6\phantom{\rule{0ex}{0ex}}y=-\frac{2}{3}+2\phantom{\rule{0ex}{0ex}}\left(d\right)x-y=1\phantom{\rule{0ex}{0ex}}y=x-1\phantom{\rule{0ex}{0ex}}\left(f\right)y=3x-5\phantom{\rule{0ex}{0ex}}\left(h\right)y=-3x+3$

In the equation a) and h) the slopes are negative where as in option d) and f) the y-intercept is negative hence this does not match the given graph of lines

## Step 4: Conclusion

The equations that can have the given graph as

$\left(b\right)-2x+3y=6\phantom{\rule{0ex}{0ex}}\left(c\right)3x-4y=-12\phantom{\rule{0ex}{0ex}}\left(e\right)x-y=-1\phantom{\rule{0ex}{0ex}}\left(g\right)y=2x+3$