Question

Gonzalez Manufacturing borrowed ​$. Part of the money was borrowed at ​%, part at ​%, and part at ​%. The annual interest was ​$​, and the total amount borrowed at ​% and ​% was twice the amount borrowed at ​%. Use Gaussian elimination or​ Gauss-Jordan elimination to find the amount borrowed at each rate.

Answer 1

Answer:

The answer is "rate 12%, 14%, 16% and value 4000, 10000, and 7000".

Step-by-step explanation:

Let the value is x, y, and z. so,

The value of x amount will be borrowed at 12% $= \frac{12}{100} \times x= 0.12x$

The value of y amount will be borrowed at 14%$= \frac{14}{100} \times x= 0.14x$

The value of z amount will be borrowed at 16% $= \frac{16}{100} \times x= 0.16x$

If the total amount= 21000

$\to x+y+z=21000$

If the total interest= 3000

calculating the total interest:

$\to 0.12x +0.14y +0.16z = 3000$

The overall amount borrowing at 12% and 14% was half as much as the amount borrowed at 16%.

$\to x + y = 2z\\\\\to x + y - 2z = 0$

by calculating the value we have  3 equations that are:

$\to x+y+z=21000\\\\\to 0.12.x +0.14.y +0.16z = 3000\\\\\to x + y - 2z= 0$

calculate the matrix by above-given value:

$\left[\begin{array}{ccc}1&1&1\\0.12&0.14&0.16\\1&1&-2\end{array}\right]$$\left[\begin{array}{c}x&y&z\\ \end{array}\right] =\left[\begin{array}{c}21000&3000&0\\ \end{array}\right]$

by solving the above matrix we get:

$\left[\begin{array}{c}x&y&z\\ \end{array}\right] =\left[\begin{array}{c}4000&10000&7000\\ \end{array}\right]$

$\to \ 4000 \text{ lent at} 12 \% \\\to \ 10000 \text{ lent at} 14 \% \\\to \ 7000 \text{ lent at} 16\%$