# How to solve 5(3-x)+1=)3(x4)?

Welcome to my article 5(3-x)+1=)3(x4) 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 5(3-x)+1=)3(x4) read and understand it carefully till the end.

Let us know how to solve this question 5(3-x)+1=)3(x4).

First write the question on the page of the notebook

## 5(3-x)+1=)3(x4)

Let us first solve the question.

\displaystyle 5\text{x3-5x}x+1=3\text{x}x+4\text{x}3

\displaystyle 15-5x+1=3x+12

put the variable amount together and the constant amount together

\displaystyle 15+1-12=3x+5x

\displaystyle 16-12=8x

\displaystyle 8x=16-12

\displaystyle 8x=4

\displaystyle x=\frac{4}{8}

The numerator and denominator of 4 \displaystyle x=\frac{1}{2}

So the absolute solution of this article 5(3-x)+1=)3(x4) is \displaystyle x=\frac{1}{2} .

## See other similar examples.

Example no 1.

### 8(2-x)-6=6(x+1)+2+9x

Let’s know the solution of this article.——

8(2-x)-6=6(x+1)+2+9x

The first one is on Costak.

\displaystyle 8\text{x}2-8\text{x}x-6=6x\text{x+6x1+2+9x}

\displaystyle 16-8x-6=6x+6+2+9x

Putting the variable amount together and the constant amount together.

\displaystyle 16-6-6-2=6x+9x+8x

\displaystyle 16-14=23x

\displaystyle 2=23x

\displaystyle \frac{2}{{23}}=x

or \displaystyle x=\frac{2}{{23}}

### Example no 2.

7(6x)+2(5+x)=(8x+1)+2(2(8=x)

Let’s know the solution of this article.——

\displaystyle 7(6+x)+2(5=x)=(8x=1)+2(8+x)

In the given article, let us first solve the cost.

\displaystyle 7\text{x}6+7\text{x}x+2\text{x}5+2\text{x}x=8x+1+2\text{x8+2x}x

\displaystyle 42+7x+10+2x+=8x=1=16+2x

After this, by keeping the variable amount together and the constant amount together, we solve it by doing addition and subtraction.

\displaystyle 47+10-1-16=-7x-2x+8x+2x

Now let’s solve the + sign together and the – sign together.

57-17=-9x=10x

Since – and – are joined together.