How to solve Algebra problems

Algebra is a mathematical branch that deals with numbers and symbols along with the techniques to solve them. These symbols or letters are nothing but the missing numbers, technically called ‘variables’, in the mathematics terminology.

These are called variables as their value is not fixed and changes with the change in the nature of operation or expression. Therefore, the algebraic expression can also be said to define the relationship between variables, just as sentences describe the relationship between the words.

Calculating the basic maths expression with basic addition and subtraction expression is easy. However, solving these mixed equations is not easy. Doing Algebra problems is a difficult task for the students, as they get puzzled by the added letters between the numbers in the equation and struggle through the problems for hours. To make it easy for them to do algebra, here are few tips on “How to do Algebra”. These will help students arrive at the solution in the quickest manner.

How to do Algebra

Begin with defining the Operation order –

To solve the algebra sum, firstly it is essential to know the order of operation. The order stands as - Parentheses, Exponents, Multiplication and Division, Addition and Subtraction. To remember this order, it is good to remember the Acronym “ PEMDAS.” So, when trying to solve algebra expression, always follow this order, i.e., begin with the Parentheses, if there is one in the question.

Then, look for exponents and solve them. Similarly, after solving Parentheses and Exponents, perform the multiplication, division, addition, and subtraction, in the mentioned order.

While solving the Parentheses, students shall follow the same order of operation as PEMDAS. Also, students can perform ‘division’ and ‘multiplication’ simultaneously, as, ‘Addition’ and ‘Subtraction.’ For instance, (3+10)*5 - 8/6

This equation will be solved as

Step 1 – solve parentheses

(3+10) = 13

Step 2 – solve multiplication

13*5=65

Step 3 – solve for subtraction

65 – 8/6

65-1.33 = 63.67

Deal with the negative numbers in the equation –

A negative number is a number represented on the opposite side of the positive number on the number line. There are specific rules regarding negative numbers such as,

The two negative numbers add up to become a lower number.

Two negative signs in the equation cancel outs to become positive, that is, multiplying or dividing two negatives gives positive.

However, multiplying and dividing one negative and one positive gives negative only.

Organizing the equation –

Algebra equations are generally associated with long operations. The equation can become tough and confusing if, too many operations are performed in the one-step, such as,

in the equation “ 9/3 - 5 + 3 × 4”,

do not solve 9/3 and 3*4 in the same step.

Rather, either multiply in the first step and divide in the second or, divide in the first step and multiply in the second.

This way, the equation remains organized and does not lose its operations.

Variables as “unknown valuables” –

In algebra questions, there is a mix of numbers, signs, and alphabets. Each of them has a meaning in the equation. The symbol such as, ‘π’ termed as ‘pie,’ has a value of ‘3.14159’. Similarly, the alphabets such as, ‘x,’ ‘y,’ ‘z’ representing the variables are basically numbers without any assigned values. The equations with such variables, have to be solved for arriving at their value. For Instance, it is given that, solve – 2x+5=10

Here, x is the variable and equation has to be solved for it, as below

2x+5=10

Step 1 – organize the equation by combining the constant terms on one side, such as 2x=10-

Step 2 – perform the Subtraction operation, such as 2x=5

Step 3 – solve for the value of X, such as X=5/2=2.5

Also, there can be a recurring variable, that is, the variable can appear more than once, in the equation. For example, 6x+10x-2x = 55

In such cases, the trick to solve is to simplify the same variable simultaneously. Therefore, the above example can be solved as

6x+10x-2x = 55

Step 1 – combine the variables on one side and perform the operation of addition, such as 16x-2x=55

Step 2 – Now, perform the operation of subtraction as, given by, the rule of PEDMAS, such as 14x=55

Step 3 - solve for X, X=55/14 = 3.92

However, in such cases of recurring variables- the variable with the same exponents can, only, be combined together. For instance,4x2+2x+3x=5. Here, out of the three variable terms, two have the same exponents. Hence, only these two variable terms can be combined as, 4x2+5x=5

Therefore, the students shall be conscious while simplifying the variables in the equation. The same variable with the same exponential value can be solved together. Apart from the linear equation, algebra can also be performed on the quadratic equation with the same set of rules.

Solving for operations on both the sides of the equation –

If the equation has the operation on both sides, the solving strategy will be, taking the variables on one side, while keeping the known values on another side. Always remember when a value or variable is moved from one side to another its sign gets change.

For example, there is an equation 2x+3=4x+5, then, begin with combining variable terms and numerical terms at opposite sides, such that the equation will look like 2x-4x=5-3. Observe here, previously the equation had positive ‘4x’ and ‘3’ but as the side gets reversed, the signs changes to negative.

Now, to get the answer, solve it by the rule of PEDMAS and basic maths. Therefore, step two will be operating‘ addition’ on both sides. This will give, -2x=2. This shall now be solved for x, that will be, x= -2/2=1. Here, the value of x achieved is the negative one.

Solve through ‘canceling’ –

The other approach to solve the algebra equation is to solve through canceling. An example can best explain this approach. Let’s say; there is the equation: x+5=10*4

Then this can be solved as,

Step 1 – To eliminate the numerical value, subtract that value from both the side, such as x+5-5=10*4-5

Step 2 - The subtraction will lead to elimination of value from one side, leaving only with the variable on one side, while the rest of the equation with the numerical on the other hand, such as x = 40-5

Step 3 - Solving gives, x = 35

Similarly, multiplication and division can be canceled, from the equation to arrive at the solution. For doing so, the student has to develop the equal and opposite relationship, as the rules say for multiplication and division. Such as, for solving through canceling for the equation, 6x = 14+2, the steps will be

Step 1 – Eliminating the numerical ‘6’ that is in multiplication with the variable. For this divide both the sides with ‘6’ ,such that, X = (14 + 2)/6

Step 2 – performing the operation to arrive the value of X, therefore, X= 16/6= 2.66

In a similar way, if the equation has exponents, it can also be canceled. As known, by the rules of pre-algebra and basic mathematics, the opposite of exponent is root. Therefore, for canceling the exponent, the root has to be taken. For instance, x2 = 81, x = √81 = 9

Solve though factoring –

Once, basic algebra has been learned, the students can move towards the advanced strategy that utilizes factoring techniques. It becomes very easy to solve the equation by factoring as it decomposes the complexity of the equation into simplicity.

For instance, if the equation is in the form of

“ ax + ba” - factor it to “a(x + b)”, example : 4x + 4 = 4(x + 1)

“ax2 + bx”- factor it to “cx((a/c)x + (b/c))”, example: 12y2 + 24y = 12y(y + 2)

“x2 + bx + c” - factor it to “(x + y)(x + z)”, example: x2 + 4x + 3 = (x + 3)(x + 1).

Solve through the graph –

Graphs are another tool to arrive at the solution for algebraic expression. The equation with two variables can be easily solved through a pictorial description. The graph is said to have two-axis- x-axis and y-axis. The values associated with each variable in the equation is being plotted on these axes to arrive at the solution.

Simply, plug-in, the value for either ‘x’ or ‘y’ and solve for the other. Then, plot these values on the graph on the gives points. For example, there is an expression 2x=y, than, putting x =1 will give y=2. Plot (1,2) coordinates on the respective axis to arrive at the solution.

Solve word problem with logic –

The strategy to solve the algebra-based word problems is to convert them into the equation form and plug in the basic value and check for their validity. For instance, after forming the equation, check whether the equation holds true for different values of x.

Do not strive to get an integer as an answer for algebra, always –

After solving the algebra question; it is not always necessary that you get an integer. The answer can be in the form of a decimal, fraction, or with the exponent. For example, if for the algebraic expression, the solution has been arrived as, x=73. By using the calculator, we can find the value of 73, like 1.91. However, it is advisable to write 73 as 73 only, rather than calculating it and writing it in decimal.

Equations with Inequality –

It is not necessary that every equation will have an equality sign. There can be such equations as well, where there is the sign of inequalities, such as ‘>’ ( that is, greater than) or ‘<’ ( that is, less than). This kind of equation is solved with the same set of rules. Follow the PEDMAS, organize the equation by keeping the variables on one side and constant on other, solve each operation one-by-one, don’t hassle between the two operations, and arrive at the solution. The only difference here will be instead of the sign of “=” (equality), there will sign of ‘>’ ( representing, greater than) or ‘<’ ( representing, less than).

For instance, solve 4>2x+5

Step 1 – organize the equation by bringing constant terms on one side. The equation will become – 4-5>2x

Step 2 – solve for operation “subtraction” and get, -1>2x

Step 3 – Solve for X, to get, -1/2>x or -0.5>x

Here, the solution will be the variable ‘x’ is less than -0.5. That is, X, can be -1,-2, and so on.

Solving two equations simultaneously –

Generally, the two equations are solved through graphs, by plotting their coordinates, the solution for the equation is being arrived at. However, if the equations are with basic operations, it is advisable to solve them simultaneously by performing the given operations. Such as,

Suppose it is given that,

X= 3y-4 and x= -y-5

The solution will be worked out as –

Step 1 – plug in the value of X from one equation to the other. Such as,

3y-4 = -y-5

Step 2 – organize the equation by collecting the variable terms and constant terms on opposite sides, such as

3y+y = -5+4

Step 3 – perform the addition operation on both the side, such as

4y =-1

Step 4 – solve for y, such as

Y=-1/4=-0.25

The above are just a few of the tricks and strategies that can be adopted while dealing with algebraic expressions. If these basic rules are followed and performed, surely the equation can be solved without any hassle in the quickest time and least steps.

However, to attain expertise in arriving at the solution, practicing the equation solution is essential. The more you solve, the more you will learn. Try solving different expressions with different combinations of operations. This will help in learning and building up the Math concept better.