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A child's Bank contains 70 coins consisting of nickels and dimes that have a total value of \$5.55. How many of each kind of coin are there?

Jun 19, 2019

#1
+102937
+3

Let x be the number of nickels.....so....70 - x  = the number of dimes

So...

number of nickels * the value of each  + number of dimes * the value of each  =total value

So we have

x * 5  +(70 - x) * 10  = 555      simplify

5x + 700 - 10x  = 555

-5x + 700  = 555       subtract 700 from both sides

-5x  = -145       divide by  -5

x = 29    nickels

And

70 - 29  =  41 dimes

Jun 19, 2019
#2
+4322
+2

We have n+d=70

5n+10d=555

Set up matrices:$$\begin{bmatrix} 1&1 \\ 5&10 \end{bmatrix}$$$$\begin{bmatrix} 70\\ 555 \end{bmatrix}$$. Take the determinant of the first matrix, and that gives $$ad-bc=10-5=5.$$ Finding the inverse of the matrix or $$A^{-1}$$, we get $$\frac{1}{5}\begin{bmatrix} 10&-1 \\ -5&1 \end{bmatrix}$$$$\begin{bmatrix} 70\\ 555 \end{bmatrix}$$, and dividing each term, gives and gets us $$\begin{bmatrix} 2 & \frac{-1}{5} \\ -1 & \frac{1}{5} \end{bmatrix}$$$$\begin{bmatrix} 70\\555 \end{bmatrix}$$, and multiplying both of them together, take the first row of the first matrix and multiply it by the first column of the second matrix, then multiply the second row of the first matrix with the second column of the second matrix to get an answer of $$\begin{bmatrix} 29\\41 \end{bmatrix}.$$ And testing the values gives us, twenty-nine(29) nickels and forty-one(41) dimes.

Systems of Equations are much easier!

-tertre

Jun 19, 2019