# Algebra Solutions Recommended By British Professors

It has become a norm of the class in every country that students fear math in general. And the reason may lie in the way they are taught in classes. It is customary for students of all backgrounds from every school, college, or university to study math as part of their curriculum. Topics like Pre-algebra and Algebra are something that everyone should know as it is required in a lot of places in life.

One of the biggest pros of studying math, particularly algebra, is that it will never go to waste. Therefore, if you are a student who is scared of algebra, this article will teach you basic algebra solutions that can come in handy to you on any day.

There are many British professors that recommend various solutions to algebraic problems. Learning these can help someone not just in his academic coursework but also in various other sectors of life. Let’s explore a few of those solution tricks and hacks.

## Solutions to Linear Equations with Integral Coefficients

Let us consider the case of x+2=8. As we can clearly see, it is a linear equation with coefficients that are integral.

First, we will need to decide on which side the x will be – either on the left of the equal sign or to the right. Then, putting other constants on the opposite side of the equal sign, we can add or subtract to find the solution of x. In this particular case, we can write: x+(2-2) = (8-2), which gives us the solution that is x=6. That way, simply by manipulating addition or subtraction operations and even multiplication and division, we can find solutions to linear equations.

It is very common for first-timers to seek help with algebra and the usual algebra problems. However, if it is pre algebra help that one might require to find the algebra solution, there are several online resources that can be helpful. Besides that, those methods described below are the ideal ones for him.

## Solutions to Linear Equations with Fractional Coefficients

First, we will need to put all fractional terms containing x on one side of the equal sign, and the rest are put on the other side. After some additions or subtractions of the fractional terms, we will end up with single terms on each side. Then lastly, the algebraic solution will be given by multiplying the reciprocal of that term to each side to expose the real value of x.

Let us consider the equation: x/3+x/4 = 5/6. Now, after a brief addition, we stumble upon the x term, which now lies with: 7x/12=5/6. Now, a simple multiplication on each side by the reciprocal of 7/12 yields: x=(5/6)*(12/7) and we get the solution: x=10/7.

Quadratic equations have the form: ax2+bx+c=0, where a, b, and c are constants of the single variable x. There is a very easy and simple way to find the solution to this kind of second-degree equation, recommended by British education as part of their math course.

The best practice is to employ a method called ‘middle term break’, where you break the equation as a factor or multiplied terms to find quick solutions.

Let us consider the case of: 6×2 + 19x + 10 = 0. Here, the middle term can be broken to write: 6×2 + 4x + 15x + 10 = 0. Or, with further decrease in terms, we can write: 2x(3x + 2) + 5(3x+2) = 0 or, (3x + 2)(2x+5) = 0. Now from such factored terms, we get two solutions by comparing it against zero on the right. x = (-2/3) and (-5/2). This is a common way to find the solutions to quadratic equations.

Another great and fancy way to find a solution to these kinds of equations is to use a direct formula when the variable is squared, i. e., the independent variable is powered to 2. For that, we use the famous quadratic formulas: x = (-b2 + sqrt (b2 – 4ac))/2a and x = (-b2 – sqrt (b2 – 4ac))/2a. We can a paired value always because of x being squared or having power two. If the (b2 – 4ac) term produces a negative value, we get an imaginary value. Hence, we can obtain a complex conjugate of an imaginary number from that. That will constitute our solution to the second-degree quadratic equation.

Let’s consider the case: 3×2 – 5x – 4 = 0. Now, a = 3, b = -5 and c = -4. Therefore, by putting these in two quadratic formulas, we get the solution: x = (-(-5)2 + sqrt ((-5)2 – 4(3)(-4)))/2(3) = (5 + sqrt (73))/6 and similarly, from the other one with minus, we get: x = (-(-5)2 – sqrt ((-5)2 – 4(3)(-4)))/2(3) = (5 – sqrt (73))/6. Therefore, if you are reading a math book in a library or somehow involve yourself in research work, these algebra solutions will always come to your aid.

## Solutions to Problems by Cross Multiplying

There is another way one can solve algebraic problems when they are in fractions. Let’s consider the case: (2x/5) = (3/4). By employing the method of cross multiplying, we can also find the value of x. This way, we can cross product from each side to the other side of the equal sign and get the solution for x, which will be: x = (5/2)(3/4) = 15/8. Like others, it is also a recommended way to solve fraction linear equations.

## Conclusion

The key lesson that we can take away from this is that the British professors, over time, developed many techniques and formulas to find algebraic solutions. The real hero of the story is the tricks and hacks that can help you tackle algebra in a much more definitive way. Gone are the days when math is feared. You can now figure out a way to learn solutions to linear equations with a single unknown, quadratic equations by middle-term break method or by quadratic formula, and cross-multiplying method. Therefore, you should definitely give these algebra solutions a read that many professors recommend to use.