Document Type:Essay

Subject Area:Mathematics

Document 1

If we add 10 to both sides of the equation, we will have; 3+10=3+10. Both sides remain equal. Consequently, this implies that if 3x +7=16 (x=3). If any number is added to both side of this equation, x=2 will still remain the solution to the equation. This can be demonstrated as; 3x+ 7+8=16+8. This is done by dividing both sides of an equation with a given number that multiplies the unknown. When this is done, the balance of the equation is maintained (Liao & Xu, 2014). For example, if A=B, then A/C=B/C where C is a non-zero number. If 6x=12, simplifying the equation gives you the value of x (unknown) to be 2. That is, x=2. Order of Operations These are steps that I will follow in simplifying an equation to its simplest form.

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Either the PEMDAS (Parenthesis, Exponents, Multiplication and Division, and Addition and Subtraction) or BODMAS (Bracket, Order, Division, Multiplication, Addition, and Subtraction) can be used in the same way (Pierce, 2018). Given an equation, I will first solve what is within the bracket or the parenthesis. Having tackled what is inside the bracket, I will go to order (Indices or exponent). Anything to power of anything comes next in the simplification process. That is -3*10÷-3=10. If this order of operation is not used, one may end up getting the wrong solution. Discussion of “RSTUV” Method RSTUV method is a tool that helps in demystification of a word problem in a way that it is readily translated into an equation whose solution can be solved.

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This makes it a helpful and reliable tool. RSTUV entails the following; “Read” the problem and decide what is being asked, “select” a letter to represent the unknown, “translate” the problem into an equation, use the mathematical rules of operations to solve the resulting equation, and “verify” the answer to the equation (Kaput, 2017). Consider the last (ones digit) of the difference. Determine whether the number is even or odd. 213966916 One’s digit is even. Thus, we now know that the ones digit in the difference is even. We also have learnt that a number is even if the ones/unit digit is either 0, 2, 4, 6, or 8. That makes it a contradiction. What is the implication of this? This means that the original assumption made of the existence of the finitely many primes is false.

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