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eduard
3 years ago
13

Find ten solutions to the linear equation to the linear equation 1/2x+y=5

Mathematics
1 answer:
yuradex [85]3 years ago
7 0

Answer:

(0,5), (2,4), (4,3), (6,2), (8,1), (10,0), (12,-1), (14,-2), (16,-3), (18,-4), and (20,-5)

Step-by-step explanation:

The given linear equation is :

\frac{1}{2} x + y = 5

Or

y =  -  \frac{1}{2} x + 5

Any ordered pair that satisfies this equation is a solution.

When x=0,

y =  -  \frac{1}{2}  \times 0 + 5 = 5

(0,5) is a solution

When x=2,

y =  -  \frac{1}{2}  \times 2+ 5 = 4

(2,4) is a solution.

When x=4,

y =  -  \frac{1}{2}  \times 4+ 5 = 3

(4,3) is a solution;

When x=6:

y =  -  \frac{1}{2} \times 6+ 5 = 2

(6,2).

When x=8,

y =  -  \frac{1}{2}  \times 8 + 5 = 1

Another solution is (8,1)

When x=10,

y =  -  \frac{1}{2}  \times 10 + 5 = 0

(5,0) is a solution

When x=12,

y =  -  \frac{1}{2}  \times 12+ 5 =  - 1

(12,-1) is a solution.

When x=14,

y =  -  \frac{1}{2}  \times 14+ 5 =  - 2

When x=16, we get:

y =  -  \frac{1}{2}  \times 16 + 5 =  - 3

When x=18, we get:

y =  -  \frac{1}{2}  \times 18+ 5 =  - 4

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Xelga [282]

Answer:

y = \dfrac{1}{2}x-4

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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The equation of circle in standard form is (x - 3)^2 + (y - 7)^2 = 16

<h3><u>Solution:</u></h3>

Given that circle having center point (3,7) and the radius r = 4

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