Step 1

Let x be the amount invested at 7%

y be the amount invested at 11%

Total amount invested = 20000

Thus, x+y=20000 ...(1)

Interest received at 7% =0.07x

Interest received at 11%=0.11y

Total interest received =1880

Thus, 0.07x+0.11y=1880 ...(2)

step 2

Hence we have two system of equations

x+y=20000

0.07x+0.11y=1880

we solve it by matrices

It can be expressed as

\(\begin{bmatrix}1 & 1 \\0.07 & 0.11 \end{bmatrix}\begin{bmatrix}x \\y \end{bmatrix}=\begin{bmatrix}20000 \\1880 \end{bmatrix}\)

Augmented Matrix for Ax=b is

\(\begin{bmatrix}1 & 1 &20000 \\0.07 & 0.11 &1880 \end{bmatrix}\)

\(R_2 \rightarrow R_2-0.07R_1 \begin{bmatrix}1 & 1&20000 \\0.07 & 0.11&1880 \end{bmatrix}\)

Which can be expressed as

x+y=20000

0.04y=480

\(\Rightarrow y=12000\) And, \(x=20000-12000=8000\) Hence \(\begin{bmatrix}x \\y \end{bmatrix}=\begin{bmatrix}8000 \\12000 \end{bmatrix}\)

Step 3

ANSWER:The student invested 8,000 at 7% interest and 12,000 at 11% interest

Let x be the amount invested at 7%

y be the amount invested at 11%

Total amount invested = 20000

Thus, x+y=20000 ...(1)

Interest received at 7% =0.07x

Interest received at 11%=0.11y

Total interest received =1880

Thus, 0.07x+0.11y=1880 ...(2)

step 2

Hence we have two system of equations

x+y=20000

0.07x+0.11y=1880

we solve it by matrices

It can be expressed as

\(\begin{bmatrix}1 & 1 \\0.07 & 0.11 \end{bmatrix}\begin{bmatrix}x \\y \end{bmatrix}=\begin{bmatrix}20000 \\1880 \end{bmatrix}\)

Augmented Matrix for Ax=b is

\(\begin{bmatrix}1 & 1 &20000 \\0.07 & 0.11 &1880 \end{bmatrix}\)

\(R_2 \rightarrow R_2-0.07R_1 \begin{bmatrix}1 & 1&20000 \\0.07 & 0.11&1880 \end{bmatrix}\)

Which can be expressed as

x+y=20000

0.04y=480

\(\Rightarrow y=12000\) And, \(x=20000-12000=8000\) Hence \(\begin{bmatrix}x \\y \end{bmatrix}=\begin{bmatrix}8000 \\12000 \end{bmatrix}\)

Step 3

ANSWER:The student invested 8,000 at 7% interest and 12,000 at 11% interest