solve by using elimtation
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zrh2sfo 1 year ago
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1. x = 5, y = 13/2

2. x = 15 , y = 24

Step-by-step explanation:

Solve the following system:

{8 y + 6 x = 82 | (equation 1)

{6 y + 9 x = 84 | (equation 2)

Swap equation 1 with equation 2:

{9 x + 6 y = 84 | (equation 1)

{6 x + 8 y = 82 | (equation 2)

Subtract 2/3 × (equation 1) from equation 2:

{9 x + 6 y = 84 | (equation 1)

{0 x + 4 y = 26 | (equation 2)

Divide equation 1 by 3:

{3 x + 2 y = 28 | (equation 1)

{0 x + 4 y = 26 | (equation 2)

Divide equation 2 by 2:

{3 x + 2 y = 28 | (equation 1)

{0 x + 2 y = 13 | (equation 2)

Divide equation 2 by 2:

{3 x + 2 y = 28 | (equation 1)

{0 x + y = 13/2 | (equation 2)

Subtract 2 × (equation 2) from equation 1:

{3 x + 0 y = 15 | (equation 1)

{0 x + y = 13/2 | (equation 2)

Divide equation 1 by 3:

{x + 0 y = 5 | (equation 1)

{0 x + y = 13/2 | (equation 2)

Collect results:

Answer: {x = 5, y = 13/2

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Solve the following system:

{8 y + 3 x = 237 | (equation 1)

{3 y + 5 x = 147 | (equation 2)

Swap equation 1 with equation 2:

{5 x + 3 y = 147 | (equation 1)

{3 x + 8 y = 237 | (equation 2)

Subtract 3/5 × (equation 1) from equation 2:

{5 x + 3 y = 147 | (equation 1)

{0 x + (31 y)/5 = 744/5 | (equation 2)

Multiply equation 2 by 5/31:

{5 x + 3 y = 147 | (equation 1)

{0 x + y = 24 | (equation 2)

Subtract 3 × (equation 2) from equation 1:

{5 x + 0 y = 75 | (equation 1)

{0 x + y = 24 | (equation 2)

Divide equation 1 by 5:

{x + 0 y = 15 | (equation 1)

{0 x + y = 24 | (equation 2)

Collect results:

Answer: {x = 15 , y = 24