Mathematics, 03.11.2020 14:00 caudhdi11721

3.a. Solve the following equations.

Answers: 2

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Answer from: rsnyder

1. Solving $\frac{1}{3}(2x-1)-\frac{1}{4}(2x+1)=\frac{1}{12}(2-x)$ we get x=3

2. Solving $\frac{x-2}{4}+\frac{2x-1}{4}=x-\frac{1}{2}$ we get x=-1

Step-by-step explanation:

We need to solve the equations:

1. $\frac{1}{3}(2x-1)-\frac{1}{4}(2x+1)=\frac{1}{12}(2-x)$

Expanding the brackets

$\frac{2x}{3}-\frac{1}{3} -\frac{2x}{4}-\frac{1}{4} =\frac{2}{12}-\frac{x}{12} \\\frac{2x}{3}-\frac{1}{3} -\frac{x}{2}-\frac{1}{4} =\frac{1}{6}-\frac{x}{12}$

Taking LCM on left side we get 12

$\frac{2x*4-1*4-x*6-1*3}{12}= \frac{1}{6}-\frac{x}{12}\\\frac{8x-4-6x-3}{12}= \frac{1}{6}-\frac{x}{12}\\\frac{8x-6x-3-4}{12}= \frac{1}{6}-\frac{x}{12}\\\frac{2x-7}{12}= \frac{1}{6}-\frac{x}{12}\\\frac{2x}{12}-\frac{7}{12}= \frac{1}{6}-\frac{x}{12}$

Add 7/12 on both sides

$\frac{2x}{12}-\frac{7}{12}+\frac{7}{12} = \frac{1}{6}-\frac{x}{12}+\frac{7}{12}\\\frac{x}{6}=\frac{1*2-x+7}{12}\\ \frac{x}{6}=\frac{-x+9}{12} \\\frac{x}{6}=-\frac{x}{12} +\frac{9}{12}\\ \frac{x}{6}=-\frac{x}{12} +\frac{3}{4}$

Adding x/12 on both sides and simplifying

$\frac{x}{6}+\frac{x}{12} =-\frac{x}{12} +\frac{3}{4}+\frac{x}{12}\\\frac{x*2+x}{12}=\frac{3}{4}\\\frac{3x}{12}= \frac{3}{4}\\$

Multiply both sides by 12/3

$\frac{3x}{12}*\frac{12}{3}=\frac{3}{4}*\frac{12}{3} \\Simplifying:\\x=3$

So, solving $\frac{1}{3}(2x-1)-\frac{1}{4}(2x+1)=\frac{1}{12}(2-x)$ we get x=3

2. $\frac{x-2}{4}+\frac{2x-1}{4}=x-\frac{1}{2}$

Taking LCM and solving:

$\frac{x-2+2x-1}{4}=\frac{2x-1}{2}\\\frac{x+2x-1-2}{4}=\frac{2x-1}{2}\\\frac{3x-3}{4}=\frac{2x-1}{2}\\$

Cross multiplying

$2(3x-3)=4(2x-1)\\6x-6=8x-4\\Combining\ like\ terms \ :\\6x-8x=-4+6\\-2x=2\\x=-1$

So, solving $\frac{x-2}{4}+\frac{2x-1}{4}=x-\frac{1}{2}$ we get x=-1

Answer from: Quest

The correct answer would be the function g(x) increased at a faster rate

Answer from: Quest

(2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

step-by-step explanation:

simplify the following:

x/x^2 + 3 x + 2 - (x (x + 1))/(x + 2)

combine powers. x/x^2 = x^(1 - 2):

x^(1 - 2) + 3 x + 2 - (x (x + 1))/(x + 2)

1 - 2 = -1:

x^(-1) + 3 x + 2 - (x (x + 1))/(x + 2)

put each term in 1/x + 3 x + 2 - (x (x + 1))/(x + 2) over the common denominator x (x + 2): 1/x + 3 x + 2 - (x (x + 1))/(x + 2) = (x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)):

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2))

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)) = ((x + 2) + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2)):

(2 + x + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

3 x^2 (x + 2) = 3 x^3 + 6 x^2:

(2 + x + 3 x^3 + 6 x^2 + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

2 x (x + 2) = 2 x^2 + 4 x:

(2 + x + 6 x^2 + 3 x^3 + 2 x^2 + 4 x - x^2 (x + 1))/(x (x + 2))

-x^2 (x + 1) = -x^2 - x^3:

(-x^2 - x^3 + 3 x^3 + 6 x^2 + 2 x^2 + 4 x + x + 2)/(x (x + 2))

grouping like terms, 3 x^3 - x^3 + 6 x^2 + 2 x^2 - x^2 + 4 x + x + 2 = (3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2:

((3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

3 x^3 - x^3 = 2 x^3:

(2 x^3 + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

6 x^2 + 2 x^2 - x^2 = 7 x^2:

(2 x^3 + 7 x^2 + (x + 4 x) + 2)/(x (x + 2))

x + 4 x = 5 x:

answer: (2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

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