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2 years ago

(5 marks)

Mathematics

1 answer:

Shtirlitz [24]2 years ago

3 0

I thought this right but sorry if it is wrong

first make the 6% in amount by 6×10,000÷100=3600

then, the convert 4 year into month by (4×12)=48month. after that divide 48months by 4 month.i.e.48÷4=12.at last multiply 12 by 3600.i.e. 12×3600=43200

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Step-by-step explanation:

The idea is to find values of a, b, and c that satisfy all three equations. Any of the methods you learned for systems of 2 equations in 2 variables will work with 3 equations in 3 variables. Often, folks find the "elimination" method to be about the easiest to do by hand.

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Add 6 times the 4th equation to the 5th equation:

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c = 5 . . . . . multiply by -1

We can now work backwards to find the other variable values. Substituting into the 4th equation, we have ...

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The solution to this set of equations is (a, b, c) = (-2, -3, 5).

_____

Comment on steps

At each stage, we made choices calculated to simplify the process. By using equations that had a variable coefficient of 1, we avoided messy fractions or multiplying by more numbers than necessary when we used the elimination process. That is, the procedure is guided by the idea of elimination, but the specific steps are ad hoc. Using these same specific steps on different equations will likely be useless.

_____

Alternate solution methods

The coefficients of these equations can be put into the form called an "augmented matrix" as follows:

$\left[\begin{array}{ccc|c}-3&-4&2&28\\1&3&-4&-31\\2&0&3&11\end{array}\right]$

Many graphing and/or scientific calculators are able to solve equations written in this form. The function used is the one that puts this matrix into "reduced row-echelon form". The result will look like ...

$\left[\begin{array}{ccc|c}1&0&0&-2\\0&1&0&-3\\0&0&1&5\end{array}\right]$

where the rightmost column is the solution for the variables in the same order they appear in the equations. The vertical line in the body of the matrix may or may not be present in a calculator view. The square matrix to the left of the vertical bar is an identity matrix (1 in each diagonal element) when there is exactly one solution.

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