# How to solve 3 1/4 = 1/2 + x with answer?

Welcome to my article 3 1/4 = 1/2+x. 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 3 1/4 = 1/2+x, read and understand it carefully till the end.

Let us know how to solve this question 3 1/4 = 1/2+x.

First write the question on the page of the notebook

## 3 1/4 = 1/2 + x

We see that mixed fractions are available in the given question. First of all convert this mixed fraction into a simple fraction.

\displaystyle 3\frac{1}{4}=\frac{1}{2}+x

\displaystyle \frac{{4\text{x}3+1}}{4}=\frac{1}{2}+x

\displaystyle \frac{{12+1}}{4}=\frac{1}{2}+x

\displaystyle \frac{{13}}{4}=\frac{1}{2}+x

reverse the numbers

\displaystyle \frac{{13}}{4}-\frac{1}{2}=x

We are seeing that each of these posts are different.
So we solve by finding the least common factor of their denominator.

\displaystyle \frac{{13\text{x}1-1\text{x}2}}{4}=x

\displaystyle \frac{{13-2}}{4}=x

\displaystyle \frac{{11}}{4}=x

Hence the exact solution of this question is 11/4.

example number 1

## 2x^-11x+15=0

\displaystyle 2{{x}^{2}}-11x+15=0

To solve such a problem, the formulas of the quadratic equation \displaystyle a{{x}^{2}}+bx+c=0 compare.
Comparing a=2 , b=-11, c=15
since we know that
D=(b ^2 -4ac )
Where D = Distributed
Now substituting the values of terms in D=(b ^2 -4ac )
[(-11)^2 -4x2x15 ]
121-120
1
Then ,

\displaystyle \sqrt{D}=\sqrt{1}

\displaystyle \sqrt{D}=1

Let the original be \displaystyle \alpha and\beta then \displaystyle \alpha =\frac{{-b=\sqrt{D}}}{{2a}} The formula for \displaystyle \alpha makes sense.

\displaystyle \alpha =\frac{{11+1}}{{2\text{x}2}}

\displaystyle \frac{{12}}{4}

\displaystyle =3

Similarly, extracting the value of \displaystyle \beta gives \displaystyle \beta =\frac{5}{2} .

So the answer to this question is 3 and 5/2

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## Example no 2

\displaystyle \frac{{{{{(3)}}^{{n+4}}}-{{{(6\text{x}3)}}^{{n+1}}}}}{{{{{(3)}}^{{n+2}}}}}

Solve this question as follows

\displaystyle \frac{{{{{(3)}}^{{n+4}}}-{{{(2\text{x}2\text{x}3)}}^{{n+1}}}}}{{{{{(3)}}^{{n+2}}}}}

\displaystyle \frac{{{{{(3)}}^{{n+4}}}-{{{(2\text{x}{{\text{3}}^{2}})}}^{{n+1}}}}}{{{{{(3)}}^{{n+2}}}}}

\displaystyle \frac{{{{{(3)}}^{{n+4}}}-{{{(2\text{x3})}}^{{n+1}}}}}{{{{{(3)}}^{{n+2}}}}}

In this \displaystyle {{{{(\text{3})}}^{{n+1}}}} if required.

\displaystyle \frac{{{{{(3)}}^{{n+4}}}-[{{3}^{2}}\text{2 }!!]!!\text{ }}}{{{{{(3)}}^{{n+2}}}}}

\displaystyle {{{{(3)}}^{{n+2}}}} Delete when down and up.