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VARVARA [1.3K]
2 years ago
9

You multiply two integers. You increase one of them by 1 and decrease the other

Mathematics
1 answer:
snow_tiger [21]2 years ago
6 0

Integers are numbers without decimals

The integers could you have multiplied are 9 and 2

Let the numbers be x and y

So, we have:

\mathbf{x \times y}

Increase x by 1 and reduce y by 1.

So, we have

\mathbf{(x +1) \times (y -1) = 6 + xy }

Open brackets

\mathbf{xy -x + y - 1 = 6 + xy}

Subtract xy from both sides

\mathbf{ -x + y - 1 = 6}

Collect like terms

\mathbf{ -x + y = 6 + 1}

\mathbf{ -x + y = 7}

Rewrite as:

\mathbf{ y -x= 7}

The above means that, the difference between the integers is 7.

There are several integers that fit the above <em>description</em>;

One possibility is: 9 and 2

Read more about products at:

brainly.com/question/3208278

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Carter draws one side of equilateral △PQR on the coordinate plane at points P(-3,2) and Q(5,2). Which ordered pair is a possible
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Step-by-step explanation:

Hey, there!!!

Let me simply explain you about it.

We generally use the distance formula to get the points.

let the point R be (x,y)

As it an equilateral triangle it must have equal distance.

now,

let's find the distance of PQ,

we have, distance formulae is;

pq =    \sqrt{( {x2 - x1)}^{2}  + ( {y2 - y1)}^{2} }

or \:  \sqrt{( {5  + 3)}^{2} + ( {2 - 2)}^{2}  }

By simplifying it we get,

8

Now,

again finding the distance between PR,

pr = \sqrt{( {x2 - x1}^{2}  + ( {y2 - y1)}^{2} }

or,

\sqrt{( {x + 3)}^{2}  + ( {y - 2)}^{2} }

By simplifying it we get,

=  \sqrt{ {x}^{2} +  {y}^{2} + 6x  -  4y + 13  }

now, finding the distance of QR,

qr =  \sqrt{( {x - 5)}^{2} + ( {y - 2)}^{2}  }

or, by simplification we get,

\sqrt{ {x}^{2} +  {y}^{2}  - 10x - 4y + 29 }

now, equating PR and QR,

\sqrt{ {x}^{2} +  {y}^{2}  + 6x  -  4y  + 13}  =  \sqrt{ {x}^{2}  +  {y}^{2} - 10x - 4y + 29 }

we cancelled the root ,

{x}^{2} +  {y}^{2} + 6x - 4y + 13 =  {x}^{2}   +  {y}^{2}   -10x - 4y + 29

or, cancelling all like terms, we get,

6x+13= -10x+29

16x=16

x=16/16

Therefore, x= 1.

now,

equating, PR and PQ,

\sqrt{ {x}^{2} +  {y}^{2}  + 6x - 4y + 13 }  =  8}

cancel the roots,

{x}^{2} +  {y}^{2}   + 6x - 4y + 13  = 8

now,

(1)^2+ y^2+6×1-4y+13=8

or, 1+y^2+6-4y+13=8

y^2-4y+13+6+1=8

or, y(y-4)+20=8

or, y(y-4)= -12

either, or,

y= -12 y=8

Therefore, y= (8,-12)

by rounding off both values, we get,

x= 1

y=(8,-12)

So, i think it's (1,8) is your answer..

<em>Hope it helps</em><em>.</em><em>.</em><em>.</em>

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