dave13 wrote:

Bunuel wrote:

SOLUTION

Which of the following lines in the xy-plane does not contain any point with integers as both coordinates?

(A) y = x

(B) y = x + 1/2

(C) y = x + 5

(D) y = x*1/2

(E) y = x/2 + 5

The answer is y=x+1/2, since for any integer value of x, y becomes integer and a half: .

Answer: B.

Bunuel, bro

how are you

would be great if you could visualize solution

have a great weekend

dave13 , I am no Bunuel.

Maybe I can help anyway.

The equations are in the form that you note above, y = mx + b

If there is no +b, it just means that b = 0

(we could still write option A, e.g., as y = (1)x + 0, in which m= 1 and b= 0)

If the equation of a line does not have a (+ b), then it passes through the origin

because both the x- and y- intercepts are 0.

We could manipulate that equation: set x = 0, then y=0, find the two intercepts: are the coordinates integers?

It's faster just to plug values in for x and find y. Bunuel noticed that in B,

every time we plug an integer in for x, we cannot get an integer for y.

That fact plays out below.

• Approach #1: Find exceptions to NEVER to eliminate answersWe essentially have a

never question:

which of these lines represented by these equations

will never give (integer, integer) for any point (x, y) on the line ?

With "never," if we can find one instance in which (x,y) is (integer, integer)

from the equation in an answer option, then we can eliminate that answer.

To get rid of an answer, "input" an integer for x into the equation.

Pick strategically for x because we are also trying to make y an integer.

A) y = xLet x = 0. Then y = 0

Let x = 1. Then y = 1.

We now have two points

(0,0) and

(1,1) in which x and y are both integers. Eliminate A.

B) y = (x + \(\frac{1}{2}\)) = (x + 0.5)Let x = 0. Then y = 0.5

Let x = 1. Then y = 1.5

Let x = 2. Then y = 2.5

Let x = -1. Then y = (-1 + \(\frac{1}{2}\)) = - \(\frac{1}{2}\) = -0.5

When I plug in an integer for x, I cannot seem to get an integer for y.

I have (0,

0.5), (1,

1.5), (2,

2.5) and (-1,

-0.5). Hold this answer.

C) y = x + 5 Let x = 0. Then y = 5

Let x = 1. Then y = 6

We have two points on that line:

(0, 5) and

(1, 6).Both (x, y) coordinates are integers. We need them not to be integers. Eliminate C.

D) y = x * \(\frac{1}{2}\) => pick values for x that are integers divisible by 2

(I have to use an integer for x, and I am trying to make y an integer, too, so that I can eliminate this answer).

Let x = 0. Then y = 0

Let x = 2. Then y = (2 * \(\frac{1}{2}\)) = 1

Let x = 4. The y = (4 * \(\frac{1}{2}\)) = 2

We have three points on the line:

(0,0) and

(2,1) and

(4,2)All of those (x,y) coordinates are integers. Eliminate D.

(E) y = \(\frac{x}{2}\) + 5. (Integer + integer) = integer,

so to get the first value, x, an integer, I need to make x divisible by 2.

Let x = 0. Then y = (\(\frac{x}{2}\) + 5) = (0 + 5) = 5

Let x = 2. Then y = (\(\frac{2}{2}\) + 5) = (1 + 5) = 6

Let x = 4. Then y = (\(\frac{4}{2}\) + 5) = (2 + 5) = 7

Points on this line:

(0,5) and

(2, 6) and

(4, 7). All integers. Eliminate E.

The answer must be B.

I couldn't think of any integral (integer) value of x

to which I could add \(\frac{1}{2}\) and get an integer for y.

It makes sense; if x must be an integer, adding \(\frac{1}{2}\) to an integer

is never going to get me another integer

for the value of y. Instead I will get 1.5 or 2.5 or 3.5 and so on.

• Approach #2: T-charts (and graphing)You wanted a visual.

Okay.

Same as above, only I drew little T-charts.

I pick an integer for x or a value that will give me an integer for x,

I run it through the equation, and I get a y-value.

LHS of T-chart: choose and input

integers for the variable x, then "run" the number through the equation

RHS of T-chart: the output is the value of y

In the diagram below, the graphs of the lines give a visual representation

of the equation of the line. Make a T-chart. Plot two or three coordinate pairs. Connect the coordinates.

(We could, though, look only at the T-charts. We can see in those charts

whether x and y are both integers.)

Answer B yields y-values

that are not integers.If x is an integer, y cannot be.

The points on the line indicated by the equation in B

will never both be integers.

Hope that helps.

Attachments

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