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Ierofanga [76]
3 years ago
11

Find the smallest positive integer solution to the following system of congruences: x = 0 (mod 5) x = 8 (mod 11) The solution is

Mathematics
1 answer:
igomit [66]3 years ago
8 0

Answer:

The smallest positive integer solution to the given system of congruences is 30.

Step-by-step explanation:

The given system of congruences is

x=0(mod5)

x=8(mod11)

where, m and n are positive integers.

It means, if the number divided by 5, then remainder is 0 and if the same number is divided by 11, then the remainder is 8. It can be defined as

x=5m

x=11n+8

5m\cong 11n+8

Now, we can say that m>n because m and n are positive integers.

For n=1,

5m=11(1)+8=19

5m=19

19 is not divisible by 5 so m is not an integer for n=1.

For n=2,

5m=11(2)+8

5m=30

m=6

The value of m is 6 and the value of n is 2. So the smallest positive integer solution to the given system of congruences is

x=5(6)=30

Therefore the smallest positive integer solution to the given system of congruences is 30.

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2х+5y-3z-=14   <u>let's call this </u><u>equation 1.</u>

x-2y=-12   <u>let's call this </u><u>equation 2.</u>

-x+Зу-2z =13  <u>let's call this </u><u>equation 3.</u>

<u>USING </u><u>EQUATION 1</u><u> AND </u><u>EQUATION 3</u><u>.</u>

2х+5y-3z-=14   (EQUATION 1)

-x+Зу-2z =13  (EQUATION 3)

<u>LET'S GET RID OF </u><u>z</u><u>, BY MAKING THE COEFFICIENT OF </u><u>z</u><u> IN THE TWO EQUATION THE SAME. WE WILL MULTIPLY </u><u>EQUATION 1</u><u> BY </u><u>2</u><u> AND </u><u>EQUATION 3</u><u> BY </u><u>3</u><u>.</u>

2х+5y-3z-=14   (EQUATION 1) * 2

-x+Зу-2z =13  (EQUATION 3) * 3

4х+10y-6z-=28   <u>let's call this </u><u>equation 4.</u>

-3x+9у-6z =39  <u>let's call this </u><u>equation 5.</u>

<u>TO GET RID OF </u><u>z</u><u>, WE WILL HAVE TO SUBTRACT </u><u>EQUATION 5</u><u> FROM </u><u>EQUATION 4</u><u> PLEASE TAKE NOTE OF THE SIGNS (-) (+).</u>

4х+10y-6z-=28   (EQUATION 4)

- (-3x+9у-6z =39) (EQUATION 5)

(4x - 3x) + ((+10y) - (+9y)) + ((-6z) - (-6z)) = (28 - 39)

x + y + 0 = -11

x + y = -11  <u>let's call this </u><u>equation 6.</u>

<u>USING </u><u>EQUATION 2</u><u> AND </u><u>EQUATION 6</u><u>, LET'S FIND </u><u>x</u><u> AND </u><u>y</u><u>.</u>

x-2y=-12  (EQUATION 2)

x + y = -11  (EQUATION 6)

<u>x</u><u> has the same coefficient in both equations, which is </u><u>1</u><u>. Let's get rid of </u><u>x</u><u> so we can find the value of </u><u>y</u><u>.</u>

<u>We will subtract </u><u>equation 6</u><u> from </u><u>equation 2</u><u>. Take note of the signs.</u>

x-2y=-12  (EQUATION 2)

- (x + y = -11)  (EQUATION 6)

(x - x) + ((-2y) - (+y)) = ((-12) - (-11))

0 + -3y = -1

-3y = -1

<u>Divide both sides by </u><u>-3</u><u>.</u>

-3y/-3 = -1/-3

y = 1/3

<u>LET'S SUBSTITUTE THE VALUE OF </u><u>y</u><u> INTO</u><u> EQUATION 2</u><u> TO GET THE VALUE OF </u><u>x</u><u>.</u>

x-2y=-12  (EQUATION 2)

x-2(1/3)=-12  

x -2/3 = -12

<u>Add </u><u>2/3</u><u> to both sides.</u>

x -2/3 + (2/3) = -12 + (2/3)

x + 0 = -34/3

x = -34/3

<u>LET'S SUBSTITUTE THE VALUES OF </u><u>x</u><u> and </u><u>y</u><u> INTO</u><u> EQUATION 1</u><u> TO GET THE VALUE OF </u><u>z</u><u>.</u>

2х+5y-3z-=14   (EQUATION 1)

2(-34/3)+5(1/3)-3z-=14  

-68/3 + 5/3 - 3z = 14

-21 - 3z = 14

<u>Add </u><u>21</u><u> to both sides.</u>

-21 + (21) - 3z = 14 + (21)

0 - 3z = 35

-3z = 35

<u>Divide both sides by </u><u>-3</u><u>.</u>

-3z/-3 = 35/-3

z = -35/3 <em>or</em>  

Therefore, x = -34/3, y = 1/3, and z = -35/3

Please thank, rate 5 stars, and give brainliest. Thank you.

6 0
3 years ago
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