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3 years ago

How do you eliminate 6x-12y=24 along with -x-6y=4

1 answer:

5 0

First you need to multiply by a factor that will givr you the same value in one of the x or y value.
(6x-12y=24)1
+(-x-6y=4)-2
6x-12y=24
=2x+12y=-8
8x=16÷8
x=2
now that you have x, substitute x by 2 in one of the eqauations
-2-6y=4
+2. +2
=-6y=6
÷-6. ÷-6
y=-1

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Find the general solution of the following differential equation. Primes denote derivatives with respect to x.(x+2y)y'=2x-yleft

kaheart [24]

Answer:

$-[ln(x^2-yx-y^2)] = K\\$

Step-by-step explanation:

Given the differential equation $(x+2y)y'=2x-y$ , this can also be written as;

$(x+2y)\frac{dy}{dx} =2x-y$

On simplification

$(x+2y)\frac{dy}{dx} =2x-y\\\\\frac{dy}{dx} = \frac{2x-y}{x+2y} \\\\let \ y = vx\\\frac{dy}{dx} = v+x\frac{dv}{dx}$

The differential equation becomes;

$v+x\frac{dv}{dx} =\frac{ 2x-vx}{x+2vx}\\\\v+x\frac{dv}{dx} = \frac{ x(2-v)}{x(1+2v)}\\\\v+x\frac{dv}{dx} = \frac{2-v}{1+2v}\\\\x\frac{dv}{dx} = \frac{2-v}{1+2v} - v\\\\x\frac{dv}{dx} = \frac{(2-v)-v(1+2v)}{1+2v}\\\\x\frac{dv}{dx} = \frac{2-v-v-2v^2}{1+2v}\\\\x\frac{dv}{dx} = \frac{2-2v-2v^2}{1+2v}$

$\frac{dx}{x} = \frac{1+2v}{2-2v-2v^2}dv\\\\integrating\ both \ sides\\\\$

$\int\limits \frac{dx}{x} = \int\limits \frac{1+2v}{2-2v-2v^2}dv\\\\lnx = \frac{1}{2} \int\limits \frac{1+2v}{1-v-v^2}dv\\\\lnx + C = -\frac{1}{2}ln(1-v-v^2)$

$C = -\frac{1}{2}ln(1-v-v^2) - lnx \\\\ -ln(1-v-v^2) - 2lnx = 2C\\\\-[ln(1-v-v^2) + lnx^2] = 2C\\\\-[ln(1-v-v^2)x^2] = 2C\\since\ v = y/x\\\\- [ln(1-y/x-y^2/x^2)x^2] = K\\\\-[ln(x^2-yx-y^2)] = K\\$

Hence the solution to the differential equation is $-[ln(x^2-yx-y^2)] = K\\$

3 0

3 years ago

What is the answer for d

Sergeeva-Olga [200]

D) y=x/5
Const= $\frac{1}{5}$

4 0

3 years ago

Evaluate. for n = 15 and t = 7

Bingel [31]

Answer:

what is the equation?

Step-by-step explanation:

3 0

3 years ago

How do a i solve and what is the answer to 7/20 + 5/8

siniylev [52]

Answer:

39/40

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Rectangle PQRS has vertices P(1, 4), Q(6, 4), R(6, 1), and S(1, 1). Without graphing, find the new coordinates of the vertices

monitta

The vertices of the rectangle after a reflection over the x-axis and then another reflection over the y-axis are:

P" (-1 , -4) , Q" (-6 , -4) , R" (-6 , -1) , S" (-1 , -1)

Step-by-step explanation:

Let us revise some transformation

If point (x , y) reflected across the x-axis , then its image is (x , -y) , the rule of reflection is rx-axis (x , y) → (x , -y)
If point (x , y) reflected across the y-axis , then its image is (-x , y) , the rule of reflection is ry-axis (x , y) → (-x , y)

∵ PQRS is a rectangle

∵ P = (1 , 4) , Q = (6 , 4) , R = (6 , 1) , S = (1 , 1)

∵ The rectangle is reflected across x-axis

- That means change the signs of the y-coordinates of all

vertices of rectangle PQRS

∴ P' = (1 , -4) , Q' = (6 , -4) , R' = (6 , -1) , S' = (1 , -1)

∵ The rectangle P'Q'R'S' is reflected across the y-axis

- That means change the signs of the x-coordinates of all

vertices of rectangle P'Q'R'S'

∴ P" = (-1 , -4) , Q" = (-6 , -4) , R" = (-6 , -1) , S" = (-1 , -1)

The vertices of the rectangle after a reflection over the x-axis and then another reflection over the y-axis are:

P" (-1 , -4) , Q" (-6 , -4) , R" (-6 , -1) , S" (-1 , -1)

Learn more:

You can learn more about the transformation in brainly.com/question/5563823

#LearnwithBrainly

3 0

3 years ago