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2 years ago

Hi how do I solve this simultaneous equation

Mathematics

1 answer:

WITCHER [35]2 years ago

8 0

Answer:

M (-3, -5/2)

N (3, -1)

Step-by-step explanation:

Solve the first equation for x.

4y = x − 7

x = 4y + 7

Substitute into the second equation.

x² + xy = 4 + 2y²

(4y + 7)² + (4y + 7)y = 4 + 2y²

Simplify.

16y² + 56y + 49 + 4y² + 7y = 4 + 2y²

18y² + 63y + 45 = 0

2y² + 7y + 5 = 0

Factor.

(y + 1) (2y + 5) = 0

y = -1 or -5/2

Plug back into the first equation to find x.

x = 4(-1) + 7 = 3

x = 4(-5/2) + 7 = -3

M (-3, -5/2)

N (3, -1)

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Please help! acellus

ch4aika [34]

Answer:

The number that belongs in the green box is equal to 909.

General Formulas and Concepts:
Algebra I

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Trigonometry

[Right Triangles Only] Pythagorean Theorem:
$\displaystyle a^2 + b^2 = c^2$

a is a leg
b is another leg
c is the hypotenuse

Step-by-step explanation:

Step 1: Define

Identify given variables.

a = 30

b = 3

c = x

Step 2: Find x

Let's solve for the general equation that allows us to find the hypotenuse:

[Pythagorean Theorem] Square root both sides [Equality Property]:
$\displaystyle \begin{aligned}a^2 + b^2 = c^2 \rightarrow c = \sqrt{a^2 + b^2}\end{aligned}$

Now that we have the formula to solve for the hypotenuse, let's figure out what x is equal to:

[Equation] Substitute in variables:
$\displaystyle \begin{aligned}c & = \sqrt{a^2 + b^2} \\x & = \sqrt{30^2 + 3^2}\end{aligned}$
Evaluate:
$\displaystyle \begin{aligned}c & = \sqrt{a^2 + b^2} \\x & = \sqrt{30^2 + 3^2} \\& = \boxed{ \sqrt{909} } \\\end{aligned}$

∴ the hypotenuse length x is equal to √909 and the number under the square root, our answer, is equal to 909.

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Learn more about Trigonometry: brainly.com/question/27707750

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Topic: Trigonometry

3 0

2 years ago