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Mariana [72]
3 years ago
15

4/3x - 4 = 5 + x Please help with this problem

Mathematics
1 answer:
stiks02 [169]3 years ago
6 0

\text {Hello! To solve for x you must...}

\text {Simplify the Equation (KCC):}

\text {Keep: 4/3x}

\text {Change: - into +}

\text {Change: 4 into -4}

\text {Switch:}

\text {5+x changes into x+5}

\text {Your new problem should be 4/3x-4=x+5}

\text {The Next Step is to Subtract x:}

\text {4/3x-4-x=x+5-x}

\text {1/3-4=5}

\text {Next you would want to add 4}

\text {Q: Why would we add 4 instead of subtracting 4?}

\text {A: Because we need 4 to equal to 0. We will add 4 to cross out 4. So -4+4 will equal 0.}

\text {1/3-4+4=5+4}

\text {1/3x=9}

\text {Your Final Step will be to Multiply 3 on both sides:}

\text {Q: Why do we multiply 3?}

\text {A: Because if we multiply 1 it won't really equal to anything. So we multiply 3 to get x}\text {by itself.}

\text 3*(1/3x)=(3)*(9)

\text {Note: 3 makes the fraction cancel out. So now x is by itself.}

\text {Your Final Answer is...}

\fbox {x=27}

\text {Best of Luck!}

\text {-LimitedX}

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2005= 18,000
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1) make 2005 year "0", and 2010 year "5"
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3) a= the y- value of y intercept ( what does y equal when x is zero) in this case a population number so, a= 18,000; b= the unknown rate of change; x= time and since we're finding  b we can plug in 5 for the time in years; y= the value of y in terms of the x-value, so when x is 5, y is 45,000
4) you end up with something like this
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8) take the fifth root of both sides 
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