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3 years ago

5x - y = -7 4x + 2y = – 14

Mathematics

2 answers:

Kobotan [32]3 years ago

8 0

Answer:

x = -2, y = -3

Step-by-step explanation:

5x - y = -7

4x + 2y = – 14

Multiply the first equation by 2

2(5x - y) = 2*-7

10x -2y = -14

Add this to the second equation to eliminate y

10x -2y = -14

4x + 2y = – 14

---------------------------

14x = -28

Divide by 14

14x/14 = -28/14

x = -2

Now find y

4x+2y = -14

4*-2 +2y = -14

-8+2y = -14

Add 8 to each side

2y = -6

Divide by 2

2y/2 = -6/2

y = -3

shtirl [24]3 years ago

4 0

Answer:

$\boxed{\sf \ \ x=-2, \ y=-3 \ \ }$

Step-by-step explanation:

Hello,

I assume that you want to solve this system of two equations

(1) 5x - y = -7

(2) 4x + 2y = -14

We will multiply (1) by 2 and add to (2) so that we can eliminate the terms in y

2*(1)+(2) gives

10x - 2y + 4x + 2y = -7*2 -14 = -14 - 14 = -28

<=>

14x = - 28 we can divide by 14 both parts

x = -28/14 = -2

and then we replace x in (1)

5*(-2)-y=-7

-10-y=-7 add 7

-10-y+7=0

-3-y=0 add y

-3 = y

which is equivalent to y = -3

do not hesitate if you have any question

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Answer:

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General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Factoring

Calculus

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Product Rule: $\frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Chain Rule: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Quotient Rule: $\frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

-y - 2x³ = y²

Rate of change of tangent line at point (-1, -2)

Step 2: Differentiate Pt. 1

Find 1st Derivative

Implicit Differentiation [Basic Power Rule]: $-y'-6x^2=2yy'$
[Algebra] Isolate y' terms: $-6x^2=2yy'+y'$
[Algebra] Factor y': $-6x^2=y'(2y+1)$
[Algebra] Isolate y': $\frac{-6x^2}{(2y+1)}=y'$
[Algebra] Rewrite: $y' = \frac{-6x^2}{(2y+1)}$

Step 3: Differentiate Pt. 2

Find 2nd Derivative

Differentiate [Quotient Rule/Basic Power Rule]: $y'' = \frac{-12x(2y+1)+6x^2(2y')}{(2y+1)^2}$
[Derivative] Simplify: $y'' = \frac{-24xy-12x+12x^2y'}{(2y+1)^2}$
[Derivative] Back-Substitute y': $y'' = \frac{-24xy-12x+12x^2(\frac{-6x^2}{2y+1} )}{(2y+1)^2}$
[Derivative] Simplify: $y'' = \frac{-24xy-12x-\frac{72x^4}{2y+1} }{(2y+1)^2}$

Step 4: Find Slope at Given Point

[Algebra] Substitute in x and y: $y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(-1)^4}{2(-2)+1} }{(2(-2)+1)^2}$
[Pre-Algebra] Exponents: $y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(1)}{2(-2)+1} }{(2(-2)+1)^2}$
[Pre-Algebra] Multiply: $y''(-1,-2) = \frac{-48+12-\frac{72}{-4+1} }{(-4+1)^2}$
[Pre-Algebra] Add: $y''(-1,-2) = \frac{-36-\frac{72}{-3} }{(-3)^2}$
[Pre-Algebra] Exponents: $y''(-1,-2) = \frac{-36-\frac{72}{-3} }{9}$
[Pre-Algebra] Divide: $y''(-1,-2) = \frac{-36+24 }{9}$
[Pre-Algebra] Add: $y''(-1,-2) = \frac{-12}{9}$
[Pre-Algebra] Simplify: $y''(-1,-2) = \frac{-4}{3}$