In high school level, students are mainly asked to evaluate their answers in simplest terms. Long and complex expressions are given which are asked to simplify and evaluate into simplest rather elegant terms. A mathematics problem is never complete unless and until the answer is evaluated in asingle simplified expression. To simplify the expressions, learning crucial steps and skills are necessary.
Every student and aspiring mathematicians must know the use of operators. One just only canâ€™t start evaluating from left to right by adding, subtracting, multiplying or dividing. This may look the same, but ultimately the answers are going to differ.
Some operators get precedence over others and should be done first and stepwise. The order of use of operations comes like this: parenthesis, exponents, multiplication, division, addition and lastly subtraction. There is a nice short form for these operators. It is â€œPlease Excuse Me Dear Aunt Sallyâ€ or â€œPEMDASâ€.
This basic knowledge of operators is required not only to solve basic expressions but also helps in complex ones. Special techniques are required for evaluating almost all kinds of polynomials.
In maths, parenthesis or brackets are given so as to indicate that these are necessary parts of the expression. Whatever be the terms inside them must be evaluated irrespective of all operators that are there inside the brackets or in expression.
At first, evaluation of the brackets is needed but one should note that inside the brackets, laws of operators are still applied. Like, inside brackets you should multiply and divide first then go for addition and subtraction.
For example, let us try to simplify the expression:
Here, we need to evaluate the brackets at first. 5+2=7 and 3+ (4/2) =3+2= 5.
In the second brackets, we actually operate the division first and then get on to division according to the order of precedence.
If we would have done the operation from left to right, then theanswer would have been 7/2 that is incorrect. Suppose there are multiple parentheses, then the innermost parentheses are solved at first, then the second innermost and so on.
Once youâ€™re done with parentheses, evaluate the exponents. These can be remembered easilywhen you see a power just next to a number, and you can evaluate it & place it just in place of the exponent.
Using the same example, we evaluate 4(7)+32â€“ (5). Now in place of 32we can write 9 and replace in the same place to get 4(7)+9 â€“ (5).
Now you need to start multiplying your expressions. You must keep in mind that multiplication can be written in many ways like â€˜xâ€™ symbol, an asterisk or a dot, all denotes the symbol of multiplication. A number within parentheses also denotes multiplication (like 2(3), indicates 2×3)
In this problem, there is just an example of multiplication and the other one just need to open the parentheses. The problem is reduced to 4(7) + 9-(5), which can be written as;
4×7+9â€“5, Here 4×7=28, so, actually the expression turns out to be 28+9-5.
While you are searching for divisions in an expression, you must keep in mind that not only â€˜Ã·â€™ denotes division but also â€˜/â€™ and slashes as in (4/2) in this expression that was solved earlier.
Earlier in this expression, while solving for brackets and parentheses, we already evaluated division operations. So, we donâ€™t have to perform any division over here. You must also keep in mind that you do not have to perform every step of operations present in PEMDAS that are not there in the expression. Only find and evaluate those in the expression.
Coming to the penultimate step is anaddition. This has got zero complexity, and you can perform the operations anyway from left to right as in the expression. Now this expression reads, 28+9-5. Now we need to add up whatever is capable of doing so.
Finally, we add up 28 and 9 and thus our expression stands out to be 37 â€“ 5.
Ultimate step in an expression is subtraction as in PEMDAS. Proceeding, we solve all the subtraction problems as there in this expression. In other words, it can also be termed as an addition of negative numbers.
Performing this ultimate operation of subtraction, we need to subtract 5 from 37, and the result gives out to be 32. Finally our taken expression yields to 32.
Next we come into the steps for simplification of complex expressions