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Darina [25.2K]
3 years ago
11

Select all the systems of equations that have exactly one solution.

Mathematics
1 answer:
ahrayia [7]3 years ago
8 0

Answer:

b) y = 3x + 1

y = x+1

Step-by-step explanation:

Select all the systems of equations that have exactly one solution.

a)

y = 3x + 1

y = 3x + 7

y - 3x = 1.... Equation 1

y - 3x = 7...... Equation 2

We solve using Elimination

We Subtract Equation 2 from 1

(y - y ) - (3x - 3x) = 7 - 1

0 = 6

b) y = 3x + 1...... Equation 1

y = x +1...... Equation 2

x = y - 1

We substitute y - 1 for x in Equation 1

y= 3(y - 1) + 1

y = 3y - 3 + 1

y - 3y = -3 + 1

-2y = -2

y = -2/-2

y = 1

Solving for x

x = y - 1

x = 1 - 1

x = 0

x = 0, y = 1,

The Equations in b) have only one solution

c) x+y=10

x = 10 - y

2x+2y=20

We substitute

2(10 - y) + 2y = 20

20 - 2y + 2y= 20

-2y + 2y = 20 - 20

0 = 0

It has no solution

d) x+y=10.... Equation 1

x+y=12..... Equation 2

We solve using Elimination

We Subtract Equation 2 from 1

x - x + y - y = 12 - 10

0 = 2

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There are 6^3 = 216 possible outcomes of rolling these 3 dice. Let's count the number of possible rolls that meet the criteria b < y < r, manually.
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 r = 3, y = 2, b = 1 is the only possibility for r=3. So n = 1
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2 years ago
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The correct option is:

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