sddfd
Hayden Rodrigue
Mrs. Finn
Algebra 2
26 September 2014
CSI: Solving Equations For this project, all the information was set in front of us, and we had to translate that information into several mathematical equations. They started as word problems you had to change into an algebraic equation, or numbers.
An example of this is when the word problem states “The sum of a certain quantity ‘P’ together with it’s two third, and its half, becomes 234. What is the quantity of P?”. You must take that and turn it into a solvable equation. First of all, we know that sum is related to addition, therefore no subtraction or multiplication will be used. So, P together with it’s two third and it’s half, equals 234. This translates to P + 2/3P + ½ P = 234. An another example of translating an equation could be “Y times it’s half, minus 30 equals 100.” To translate this, start with the term times, which means multiplication. It would also include subtraction, hinted to by the “minus 30”. Then begin to write out the equation. Y * 1/2Y - 30 = 100.
Now to solve an equation, follow PEMDAS, or do everything in order. You begin with solving things in parentheses. If our equation is (-3+4) + 6x = 24x - 12, then you’d begin with (-3+4), which would equal 1. The equation would then be 1 + 6x = 24x -12.dsdssssssssssssssssssssssssssssssssssssssFollowing parentheses is exponents. Being there no exponents in this equation, you would move onto the next step, and that would be multiplication or division. Do these in the order they appear. There isn’t any multiplication or division in this problem not involving variables, so we’ll wait until the end. Next is addition or subtraction. Once again, do these in the order they appear. Whatever you do to one side, you must do to the other. First we’ll start by adding 12 to both sides resulting in 13 + 6x = 24x. Then subtract 6x from both sides. You’ll find it comes out to 13 = 16x. Finally, divide both sides by 16 and your answer will be 0.8125.
Mrs. Finn
Algebra 2
26 September 2014
CSI: Solving Equations For this project, all the information was set in front of us, and we had to translate that information into several mathematical equations. They started as word problems you had to change into an algebraic equation, or numbers.
An example of this is when the word problem states “The sum of a certain quantity ‘P’ together with it’s two third, and its half, becomes 234. What is the quantity of P?”. You must take that and turn it into a solvable equation. First of all, we know that sum is related to addition, therefore no subtraction or multiplication will be used. So, P together with it’s two third and it’s half, equals 234. This translates to P + 2/3P + ½ P = 234. An another example of translating an equation could be “Y times it’s half, minus 30 equals 100.” To translate this, start with the term times, which means multiplication. It would also include subtraction, hinted to by the “minus 30”. Then begin to write out the equation. Y * 1/2Y - 30 = 100.
Now to solve an equation, follow PEMDAS, or do everything in order. You begin with solving things in parentheses. If our equation is (-3+4) + 6x = 24x - 12, then you’d begin with (-3+4), which would equal 1. The equation would then be 1 + 6x = 24x -12.dsdssssssssssssssssssssssssssssssssssssssFollowing parentheses is exponents. Being there no exponents in this equation, you would move onto the next step, and that would be multiplication or division. Do these in the order they appear. There isn’t any multiplication or division in this problem not involving variables, so we’ll wait until the end. Next is addition or subtraction. Once again, do these in the order they appear. Whatever you do to one side, you must do to the other. First we’ll start by adding 12 to both sides resulting in 13 + 6x = 24x. Then subtract 6x from both sides. You’ll find it comes out to 13 = 16x. Finally, divide both sides by 16 and your answer will be 0.8125.