Question

Please help me on this

On the page below

Answer 1

`\textsf{a)}\quad \begin{cases}x = z - 6\\y = 3z\\x+y+z = 89\end{cases}`

`\text{b)} \quad \left[\begin{array}c1&0&-1&-6\\0&1&-3&0\\1&1&1&89\end{array}\right]`

`\textsf{c)} \quad x=13,\;\;y=57,\;\;z=19`

`\textsf{d)}\quad \textsf{Amanda has \$13, Henry has \$57, and Scott has \$19.}`

Explanation:

Part a

Define the variables and write a system of equations.

Definition of the variables:

• Let x represent the amount of money Amanda has.
• Let y represent the amount of money Henry has.
• Let z represent the amount of money Scott has.

Now, based on the given information, we can create a system of equations.

Given Amanda has $6 less than Scott:

`x = z - 6`

Given Henry has 3 times what Scott has:

`y = 3z`

Given the total money they have is $89:

`x+y+z = 89`

So, the system of equations is:

`\begin{cases}x = z - 6\\y = 3z\\x+y+z = 89\end{cases}`

`\hrulefill`

Part b

Rewrite the system of equations so that it is in the form:

`\begin{cases}a_1x + b_1y + c_1z = d_1\\a_2x + b_2y + c_2z = d_2\\a_3x + b_3y + c_3z = d_3\end{cases}`

Therefore:

`\begin{cases}x -z= - 6\\y -3x=0\\x+y+z = 89\end{cases}`

We can represent the system of equations as an augmented matrix.

An augmented matrix is formed by combining the matrices with the coefficient terms and the constant terms of a system of linear equations.

`\textbf{M}=\left[\begin{array}ca_1&b_1&c_1&d_1\\a_2&b_2&c_2&d_2\\a_3&b_3&c_3&d_3\end{array}\right]`

Therefore, the augmented matrix of the given system of equations is:

`\textbf{M}=\left[\begin{array}ccc1&0&-1&-6\\0&1&-3&0\\1&1&1&89\end{array}\right]`

`\hrulefill`

Part c

To solve the augmented matrix, use elementary row operations to convert it into the identity matrix:

`\left[\begin{array}ccc1&0&0&p\\0&1&0&q\\0&0&1&r\end{array}\right]`

where the solution to the system will be x = p, y = q, and z = r.

Elementary row operations are:

• Interchange two rows.
• Multiply a row by a constant.
• Add a multiple of a row to another row.

Subtract the first row from the third row:

`\left[\begin{array}ccc1&0&-1&-6\\0&1&-3&0\\1&1&1&89\end{array}\right]R_3-R_1\rightarrow R_3\left[\begin{array}c1&0&-1&-6\\0&1&-3&0\\0&1&2&95\end{array}\right]`

Subtract the second row from the third row to eliminate the 1 below it:

`\left[\begin{array}c1&0&-1&-6\\0&1&-3&0\\0&1&2&95\end{array}\right]R_3-R_2\rightarrow R_3\left[\begin{array}ccc1&0&-1&-6\\0&1&-3&0\\0&0&5&95\end{array}\right]`

Divide the third row by 5 to make the leading coefficient in the third row equal to 1:

`\left[\begin{array}ccc1&0&-1&-6\\0&1&-3&0\\0&0&5&95\end{array}\right](1)/(5)R_3\rightarrow R_3\left[\begin{array}c1&0&-1&-6\\0&1&-3&0\\0&0&1&19\end{array}\right]`

Add the third row to the first row to eliminate the non-zero entry in the first row:

`\left[\begin{array}c1&0&-1&-6\\0&1&-3&0\\0&0&1&19\end{array}\right]R_1+R_3\rightarrow R_1\left[\begin{array}c1&0&0&13\\0&1&-3&0\\0&0&1&19\end{array}\right]`

Add 3 times the third row to the second row to eliminate the non-zero entry in the second row:

`\left[\begin{array}ccc1&0&0&13\\0&1&-3&0\\0&0&1&19\end{array}\right]R_2+3R_3\rightarrow R_2\left[\begin{array}ccc1&0&0&13\\0&1&0&57\\0&0&1&19\end{array}\right]`

Now, the matrix is in row-echelon form, we can read off the solutions for x, y, and z:

• 
  `x=13`
• 
  `y=57`
• 
  `z=19`

`\hrulefill`

Part d

Amanda has $13, Henry has $57, and Scott has $19.