How to solve 1/(x+1)+2/(x+2)=4/(x+4)?

Welcome to my article 1/(x+1)+2/(x+2)=4/(x+4). This question is taken from the simplification lesson.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
For complete information on how to solve this question 1/(x+1)+2/(x+2)=4/(x+4), read and understand it carefully till the end.

Let us know how to solve this question 1/(x+1)+2/(x+2)=4/(x+4).

First write the question on the page of the notebook

1/(x+1)+2/(x+2)=4/(x+4)

solve it by writing it above

\displaystyle \frac{1}{{(x+1)}}+\frac{2}{{(x+2)}}=\frac{4}{{(x+4)}}

By writing this way, it is easy to do addition, subtraction, multiplication and division.

After this, let’s see the mind of the people of this article.
If the denominator has the same value, then its numerators are added together.
Here the value of the Haro is sky, so the least common factor of these Haro is solved by finding out.

Finding the least common common denominator of the denominator (x + 1 ),(x + 2 ) gives (x ^ 2+3x +2 ).

Now the term \displaystyle \frac{1}{{(x+1)}}[/latex ] divides the denominator (x^ 2 +3 x +2 ) by dividing( x+2 ) to get Let’s multiply ( x+2 ) the numerator over ( x+1 ) by 1.</p> <p></p> <p>Similarly, dividing the term \displaystyle \frac{2}{{(x+2)}} in (x ^ 2 +3x+2) gives (x+1) Multiply (x+1) over the denominator of (x+2) in the numerator 2 .
In this way we have to do –

\displaystyle \frac{{1\text{x}(x+2)+2(x+1)}}{{(x+1)(x+2)}}=\frac{4}{{x+4}}

\displaystyle \frac{{x+2+2x+2}}{{2{{x}^{2}}+x+2}}=\frac{4}{{x+4}}

Now we add the same variable quantity or constant quantity together.

\displaystyle \frac{{3x+4}}{{{{x}^{2}}+3x+2}}=\frac{4}{{x+4}}

Remaining = is placed between the two terms, to solve it, we do slant multiplication.
On multiplying, the solution will be as follows –

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\displaystyle (3x+4)(x+4)=({{x}^{2}}+3x+2)\text{x}4

\displaystyle 3x(x+4)+4(x+4)=4({{x}^{2}})+4(3x)+4(2)

\displaystyle 3{{x}^{2}}+12x+4x+16=4{{x}^{2}}+12x+8

Add or subtract quadratic with quadratic.

\displaystyle 3{{x}^{2}}+16x+16=4{{x}^{2}}+12x+8

Now we will transpose the terms –

\displaystyle 3{{x}^{2}}-4{{x}^{2}}+16x-12x=16-8=0

\displaystyle -{{x}^{2}}+4x+8=0

Can be deleted from both the sides by common, after doing this we will get this –

\displaystyle {{x}^{2}}-4x-8=0

\displaystyle {{x}^{2}}-8x+4x-8=0

\displaystyle x(x-8)+4(x-2)=0

\displaystyle x(x-8)+4(x-2)=0

Then find the value of x.
If , x -8 = 0
So x = 8 Answer like this,
x = 2 answer ,x = -4 answer
So the answer to this question will be 8, 2, – 4

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To see more similar examples read this question –

8/(x -2 )+6 /(x -3)=2/3 .

Welcome to my question 8/(x -2 )+6 /(x -3)=2/3 .
This article is placed in a quadratic equation

Let us know the solution of this question-

8/(x -2 )+6 /(x -3)=2/3

First of all we have to write the article on the page of the copy,In this way, the method of solving the question will be –

\displaystyle \frac{8}{{(x-2)}}+\frac{6}{{(x-3)}}=\frac{2}{3}

Let’s solve like this above question.

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\displaystyle \frac{{8(x-3)+6(x-2)}}{{(x-2)(x-3)}}=\frac{2}{3}[/ltex]</p> <p></p> <p> \displaystyle \frac{{8x-24+6x-12}}{{{{x}^{2}}-3x-2x+6}}=\frac{2}{3}

\displaystyle \frac{{14x-36}}{{{{x}^{2}}-5x+6}}=\frac{2}{3}

Remaining two sides have a = between the two sides, so solve it by multiplying slantly -

On slant multiplication, the solution will be as follows -

\displaystyle 3(14x-36)=2({{x}^{2}}-5x+6)

\displaystyle 42x-108=2{{x}^{2}}-5x+6

\displaystyle 42x-2{{x}^{2}}+5x-108-6=0

\displaystyle 37x-2{{x}^{2}}-114=0

\displaystyle 2{{x}^{2}}-37x+114=0

\displaystyle 2{{x}^{2}}-18x-19x+114=0

\displaystyle 2x(x-9)-19(x-6)=0

\displaystyle (x-9)(x-6)(x-19)=0

After this find the value of x

If ,x -9=0
so x = 9

Similarly,
x = 9, 6, 19/2 Answer

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