Kenneth S. answered • 02/04/16
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you need to ask your teacher; if you're behind the rest of the class, get one-on-one tutoring.
What needs to be done, and these concepts should have been explained to the whole class so that the understand and believe, is
1. distribute
2. collect 'like' terms
3. eventually get all of the y stuff onto one side of the equation, and the constant (number only) on the other side.
to do this, you'll TRANSPOSE a term (it goes to OPPOSITE SIDE, & takes OPPOSITE SIGN
4. finally, divide out the coefficient of the variable y, getting the answer.
Every step of this procedure requires that you understand what the words mean (that's VERY important in math--that is, what are the definitions of things.
I will now illustrate how to apply these concepts to your specific problem. Your job is to study this until there is no longer any confusion.
-9y - 3 = 10 - 5y -13 -4y
-9y - 3 = -3 -9y <-- combined the y stuff on the right & combined those two constants, also
Whoa!! The teacher threw you a curve. The left side is IDENTICAL to the right side in this equation!!!
Well, that's OK...it just means that this equation provides no single value that y must be; y could be ANYTHING.
So the solution is : ALL REAL NUMBERS
Nevertheless, what I said above is the general rule for solving equations of this type. Usually you'll get one answer.
Kenneth S.
Transpose is DEFINED adding the opposite of a term to each side of an equation.
It can be PROVED that the result of this is ALWAYS that the chosen terms disappears from one side and consequently REAPPEARS on the opposite side, WITH THE OPPOSITE SIGN. Q.E.D.
Transposing a term was common terminology until the 'reform' associated with the Sputnik overreaction in USA. From that time on, people have been forced to write down - term = - term under the original equation to show the addition of opposites explicitly. I consider that this is something that a person can do in his or her head, at least when they're in Algebra II or above.
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02/05/16
Mark M.
02/04/16