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Topic: **The importance of problem solving****Question:**
First problem: z+4/6=3/2-2z+2/12 second problem: 5/-x-6=2/x third problem: 3a-2/2a+2=3/a-1 Does anyone know how to solve with correct answer? Help! No one is getting the right answer to these problems. Please help solve and show me how you did it....thanks ...You have all been helpful but answers are still coming up wrong...the answer to #1 is 2 The answer to#2 is -12/7 The answer to #3 is -1/3 , 4 Need to know how to solve these 3 problems

June 19, 2019 / By Donalda

The reason people are getting the wrong answers is that you are not properly stating the problems. If you do not use paretheses toindicate exactly what the equation, it will be interpreted several different ways. I have tried to show the proper writing of the expression in order to get the answers you say are correct. z+4/6=3/2-2z+2/12 (z+4)/6 = 3/2 - (2z +2)/12 2(z+4)/12 = 18/12 -(2z+2)/12 = (18 -2z-2)/12 = (16-2z)/12 2(z+4) = 16 -2z [denominators = so numerators =] 2z+8 =16 -2z 4z=8 z=2 5/-x-6 = 2/x 5/(-x-6) = 2/x 2(-x-6) = 5 x -2x -12 =5x -7x =12 x = -12/7 3a-2/2a+2=3/a-1 (3a-2)/(2a+2) = 3/(a-1) (3a-2)(a-1) = 3(2a+2) 3a^2 -5a+2 = 6a + 6 3a^2-11a -4 = 0 (3a+1)(a-4) = 0 3a+1 = 0 --> a= -1/3 a-4 = 0 --> a = 4 I hope this shows you the importance of the use of parentheses in clarifying exactly what you mean.

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Seamus had 32 money first of all. explanation Seamus = 32 Davey = sixty 4 (double 32) provide 24 money Seamus now has = 8 Davey now has = 40 Davey has 5 situations the quantity situation Solved

The answer to number 1 is not 2. If you plug in 2 for z you won't have both the left and the right sides equal. 1) z + 4/6 = 3/2 -2z + 2/12 3z = 3/2 + 2/12 - 4/6 3z = 18/12 + 2/12 - 8/12 3z = 12/12 = 1 z = 1/3 * previous responder is right, if for example you have a z+4 as the numerator and 6 as the denominator you need to write: (z+4)/6. Recheck the question and make sure you have copied it correctly.

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Since x is rational, x = a/b for some integers with b nonzero. Since x is nonzero, a is nonzero. Hence we have: 1/x = 1/(a/b) = b/a Which makes sense since a,b are nonzero. Now we've shown how to write 1/x as an integer fraction with nonzero denominator, hence 1/x is rational. I hope this helps!

If x is a rational selection, then there are integers p and q with x = p/q. (in certainty this purely says that rational numbers could be written as fractions.) If we even have x ? 0, then we get p ? 0. (Follows from 0/q = 0 for each integer p.) as a result we are able to divide q by p?0, and the top result's q/p, yet another rational selection. Now x * (q/p) = (p/q) * (q/p) = pq / qp = one million. So q/p is merely yet in any different case of writing one million/x. as a result one million/x, which we are able to write as q/p, is a rational selection.

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