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Solve this linear quadratic system

y = 3x + 4
y + x² = 0

1 Answer

5 votes

Answer:

x = -1, y = 1

Explanation:

Solve the following system:

{y = 3 x + 4 | (equation 1)

y = x + 2 | (equation 2)

Express the system in standard form:

{-(3 x) + y = 4 | (equation 1)

-x + y = 2 | (equation 2)

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

{-(3 x) + y = 4 | (equation 1)

0 x + (2 y)/3 = 2/3 | (equation 2)

Multiply equation 2 by 3/2:

{-(3 x) + y = 4 | (equation 1)

0 x + y = 1 | (equation 2)

Subtract equation 2 from equation 1:

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

0 x + y = 1 | (equation 2)

Divide equation 1 by -3:

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

0 x + y = 1 | (equation 2)

Collect results:

Answer: {x = -1, y = 1

User Max Banaev
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