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3 years ago

Solve the system of linear equations. separate the x- and y- values with a coma.

Mathematics

1 answer:

kobusy [5.1K]3 years ago

4 0

Solve the following system:
{7 x - 3 y = 37 | (equation 1)
{14 x + 11 y = 23 | (equation 2)
Swap equation 1 with equation 2:
{14 x + 11 y = 23 | (equation 1)
{7 x - 3 y = 37 | (equation 2)
Subtract 1/2 × (equation 1) from equation 2:
{14 x + 11 y = 23 | (equation 1)
{0 x - (17 y)/2 = 51/2 | (equation 2)

Multiply equation 2 by 2/17:
{14 x + 11 y = 23 | (equation 1)
{0 x - y = 3 | (equation 2)

Multiply equation 2 by -1:
{14 x + 11 y = 23 | (equation 1)
{0 x+y = -3 | (equation 2)
Subtract 11 × (equation 2) from equation 1:
{14 x+0 y = 56 | (equation 1)
{0 x+y = -3 | (equation 2)

Divide equation 1 by 14:
{x+0 y = 4 | (equation 1)
{0 x+y = -3 | (equation 2)
Collect results:
Answer: {x = 4
{y = -3

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Explanation:
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Answer:

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Step-by-step explanation:

Angle "a" and 30° are vertical angles, hence equal.

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3 years ago

Find the value of the discriminant for <img src="https://tex.z-dn.net/?f=7x%5E%7B2%7D%20%2B5x%2B1%3D0" id="TexFormula1" title="7

kondor19780726 [428]

Answer:

No real roots

Step-by-step explanation:

Given

7x² + 5x + 1 = 0 ← in standard form

with a = 7, b = 5, c = 1

To determine the nature of the roots use the discriminant

Δ = b² - 4ac

• If b² - 4ac > 0 then roots are real and distinct

• If b² - 4ac = 0 then roots are real and equal

• If b² - 4ac < 0 then the roots are not real

Here

b² - 4ac = 5² - (4 × 7 × 1) = 25 - 28 = - 3

Thus the 2 roots are not real

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Step-by-step explanation:

If you need any explanation, we can communicate normally

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