Detailed Answer

a) We multiply the second equation by 2 and subtract it from the first one:

\[2x + 5y - (2x - 2y) = 11 - 4\]

\[7y = 7\]

\[y = 1\]

We have found \(y\), and we can reinject that value in the second equation:

\[2x - 2*1 = 4\]

\[x = 3\]

Therefore, the solution is \((3,1)\).

b) We solve the first equation for \(x\), and find:

\[x = \frac{3}{2}y - 2\]

We substitute this into the second equation, which gives us:

\[4(\frac{3}{2}y - 2) + y = 20 \Rightarrow 6y - 8 + y = 20 \Rightarrow 7y = 28 \Rightarrow y = 4\]

We know the value of \(y\) and can use it to find \(x\):

\[x = \frac{3}{2} * 4 - 2 = 4\]

So, the solution is \((4,4)\).

c) We multiply the first equation by 3 and add it to the second equation:

\[2x + 15y + 3[3x - 5y] = 19 + 3*1\]

\[2x + 9x + 15y - 15y = 22\]

\[11x = 22 \Rightarrow x = 2\]

Putting that value back into the second equation we get:

\[3 * 2 - 5y = 1 \Rightarrow y = 1\]

Then, the solution is \((2,1)\).

Answers d), e), f) are given by the calculator.

d) \((2.73,2.52)\)

e) \((1,10,-12)\)

f) \((6,5,-1)\)

Detailed Answer

a) According to the text, we find:

\[ (1) \ s + b = 15\]

\[ (2) \ b = 2s\]

So the standardized form is:

\[ (1) \ s + b = 15\]

\[ (2) \ -2s + b = 0\]

b)

\[ (1) \ 6n + 12p = 86\]

\[ (2) \ n = p + 10\]

So the standardized form is:

\[ (1) \ 6n + 12p = 86\]

\[ (2) \ n - p = 10\]

c)

\[ (1) \ f + w + a = 151\]

\[ (2) \ f = 2w\]

\[ (3) \ w = a - 23\]

So the standardized form is:

\[ (1) \ f + w + a = 151\]

\[ (2) \ f - 2w = 0\]

\[ (3) \ w - a = -23\]

d)

\[ (1) \ 450a + 100s + 500c = 200,000\]

\[ (2) \ 630a + 90s + 940c = 278,760\]

\[ (3) \ 540a + 110s + 610c = 239,680\]

This form is already standardized.

Detailed Answer

a) If there’s one more Maxi than there are Normal menus, it means \(x = y + 1\), which is the same as \(x - y = 1\). There are 15 friends so in total 15 menus ordered. Also, the number of orders multiplied by their respective prices has to be equal to \(81\) dollars, resulting in the final system of equations as follows:

\[ (1) \ 5x + 8y + 4z = 81\]

\[ (2) \ x + y + z = 15\]

\[ (3) \ x - y = 1\]

b) The answers can be found using a calculator.

\[ x=5, \ y=4, \ z=6 \]

c) She has to pay for her own share (\(1 \times 5\)) plus the price of 4 maxi menus (\(4 \times 8\)), for a total of \(37\) dollars.

Detailed Answer

a) The key here is to express everything in terms of hours, so in the first equation \(8 \, \text{h} \, 30 \, \text{min}\) is equal to \(8.5\) hours.

b) Using the method of substitution:

\[ (1) \ 9w + 2d = 8.5\]

\[ (2) \ 6w + 3d = 9\]

\[ (1) \ 9w + 2d = 8.5\]

\[ (2) \ d = 3 - 2w\]

\[ (1) \ 9w + 2(3 - 2w) = 8.5\]

\[ (2) \ d = 3 - 2w\]

\[ (1) \ 9w + 6 - 4w = 8.5\]

\[ (2) \ d = 3 - 2w\]

\[ w = 0.5\]

\[ d = 3 - 2w\]

\[ w = 0.5\]

\[ d = 2\]

c) We draw both lines and the solution is at the intersection:

Detailed Answer

a) We call \(x\) the length of a ball of black yarn, and \(y\) the length of a ball of gray yarn.

\[\ (1) \ \frac{2}{3}x + y = 500\]

\[\ (2) \ \frac{1}{3}x + y = 350\]

b) We read on the graph the coordinates of the intersection of the two lines: \((450, 200)\). This gives us \(x\) and \(y\).

c) Let \(l_b\) be the length of black yarn we need for this scarf, and \(l_g\) be the length of gray yarn. We can write the system:

\[(1) \ l_g = 2l_b\]

\[(2) \ l_g + l_b = 1800\]

and solve it with a calculator or by hand. After solving it we get:

\[(1) \ l_g = 1200\]

\[(2) \ l_b = 600\]

Thus, we need 600 meters of black yarn, and 1,200 meters of gray yarn.

d) To find how many balls of each we need, we can calculate:

\[(1) \ \frac{l_g}{y} = \frac{1200}{200} = 6\]

\[(2) \ \frac{l_b}{x} = \frac{600}{450} = \frac{4}{3}\]

You can’t order \(\frac{4}{3}\) of a ball of yarn, so she will have to order two balls of black yarn.

Detailed Answer

a) If we write \(d\) as the number David got and \(m\) as the number Maddie got, we can write thanks to the hints:

\[(1) \ m = d - 3\]

\[(2) \ md = 10\]

b) We replace \(m\) with \((d - 3)\) in equation \((2)\), and find \(d(d - 3) = 10\). This is equivalent to \(d^2 - 3d - 10 = 0\).

c) We can plot \(y = x^2 - 3x - 10\) with a calculator to get the following graph:

The roots are \(-2\) and \(5\), which means the equation is solved for \(d = -2\) or \(d = 5\).

d) The equation we wrote gave us the possible numbers on David’s die. The number on the face of a die cannot be \(-2\), so David must have gotten a \(5\). This means Maddie got \(5 - 3 = 2\).

Detailed Answer

a) We write \(m\), \(j\), and \(d\) as the numbers that Maddie, Jane, and David respectively got on their die.

\[(1) \ m = j + 1\]

\[(2) \ m = d - 2\]

\[(3) \ mdj = 8\]

b) Systems of equations from part (a) can be rewritten as the following:

\[(1) \ j = m - 1\]

\[(2) \ d = m + 2\]

\[(3) \ m(m - 1)(m + 2) = 8\]

Resulting in the final polynomial equation of: \(x^3 + x^2 - 2x - 8\)

c) We can plot \(y = x^3 + x^2 - 2x - 8\) with a calculator to get the following graph:

We find that there is only one root for \(x = 2\).

d) Since we know that \(m = 2\) is the solution of our polynomial equation, we can find \(j = 2 - 1 = 1\) and \(d = 2 + 2 = 4\). This means Maddie got \(2\), Jane got \(1\), and David got \(4\).

Detailed Answer

a) Let \(m_1\) and \(m_2\) be the cost for each motor at either factory. By multiplying the quantity of motors produced in each factory by \(m_1\) and \(m_2\) respectively, we can derive two equations for the production costs (one per factory):

\[56\, m_1 + 37\,m_2 = 8892\]

\[23 \,m_1 + 41\,m_2 = 6026\]

b) We can solve the system using the Apps function on the TI-84 calculator. Select PlySmlt2 and choose the appropriate size for the system (in this case, a \(2 \times 2\) system). After entering the values, the solution is:

The solution is \(m_1 = \$98\) and \(m_2 = \$92\)

c) Let us denote \(q_1\) and \(q_2\) as the additional quantities produced for motor 1 and motor 2, respectively. Using the results from \(m_1\) and \(m_2\), and the previous quantities produced, we can derive the following equations:

\[(23+q_1)\, 98 + (41+q_2)\, 92 = 10500\]

\[(41+q_2) - (23+q_1) = 15\]

Reducing this system, we get:

\[98\,q_1 + 92\,q_2 = 4474\]

\[q_2 - q_1 = -3\]

Using the same process as in part (b), we find \(q_1 = 25\) and \(q_2 = 22\).

Detailed Answer

a) We use the entries in the table for the coefficients of each appliance and their corresponding variable in that order and table row (equation). Notice the values for the last column are in kWh, which we'll have to convert to Wh to get the values of power consumption of each appliance in W (multiply by 1000). The following system is then obtained:

\[ 24\, x + 5\, y + 4\, z = 1272 \\ 9\, x + 12\, y + 16\, z = 1215 \\ 15\, x + 8\, y + 7\, z = 1128 \]

b) We can solve the system using the Apps function on the TI-84 calculator. Select PlySmlt2 and choose the appropriate size for the system (in this case, a \(3 \times 3\) system). After entering the values, the solution is:

The solution is \(x = 39\), \(y = 60\), and \(z = 9\).

c) Using the results from part (b) and the values in the table for Sarah's energy consumption habits, the following equations are obtained with \(w\) as the power consumption for the washing machine:

\[ 10\, x + 3\, y + 11\, z + 2\, w = 7005 \\ 10 \times 39 + 3 \times 60 + 11 \times 9 + 2 \times w = 7005 \]

\[ 669 + 2\, w = 7005 \\ 2\, w = 7005 - 669 \\ w = \frac{7005 - 669}{2} \\ w = 2328 \]

The power consumption of Sarah's washing machine is \(w = 528\) W.

Detailed Answer

a) Evaluating the function at the values provided in the table, we get:

\[ a(-1)^2 + b(-1) + c = f(-1) \]

\[ a(2)^2 + b(2) + c = f(2) \]

\[ a(3)^2 + b(3) + c = f(3) \]

\[ a - b + c = 0 \]

\[ 4 \cdot a + 2 \cdot b + c = 66 \]

\[ 9 \cdot a + 3 \cdot b + c = 148 \]

b)

\[ a - b + c = 0 \text{ (Eq. 1)} \]

\[ 4 \cdot a + 2 \cdot b + c = 66 \text{ (Eq. 2)} \]

\[ 9 \cdot a + 3 \cdot b + c = 148 \text{ (Eq. 3)} \]

Let us subtract Eq. 1 from Eq. 2 and divide by 3. Likewise, subtract Eq. 1 from Eq. 3 and divide by 4:

\[ a + b = 22 \text{ (Eq. 2')} \]

\[ 2 \cdot a + b = 37 \text{ (Eq. 3')} \]

To find the value of \(a\), subtract Eq. 2' from Eq. 3' to get \(a = 15\). Substituting this value into Eq. 2', we get \(b = 22 - 15 = 7\). Finally, from Eq. 1, \(c = b - a = 7 - 15 = -8\).

c) Let us use the technique of completing squares for the function \(f(x) = 15x^2 + 7x - 8\). Add and subtract \(\frac{49}{60}\) to get:

\[ f(x) = \left(15x^2 + 7x + \frac{49}{60}\right) - \frac{49}{60} - 8 \]

\[ f(x) = 15 \left(x^2 + \frac{7}{15}x + \frac{49}{900}\right) - \frac{49 + 480}{60} \]

\[ f(x) = 15 \left(x + \frac{7}{30}\right)^2 - \frac{529}{60} \]

If we evaluate \(f(x)\) at \(x = -\frac{7}{30}\), we get the minimum value possible to be:

\[ f\left(-\frac{7}{30}\right) = -\frac{529}{60} \]

Detailed Answer

(a) Given that we have two values that render the same result, namely the points \((-8, 54)\) and \((-2, 54)\), then the axis of symmetry for this quadratic function can be found at \(x = \frac{-8 - 2}{2} = -5\). Then the equation can be rewritten as:

\[ f(x) = a(x + 5)^2 + k \]

\[ f(x) = ax^2 + 10ax + (25a + k) \]

From this equation, we find \(b = 10a\) and \(c = 25a + k\). Evaluating at either \(x = -8\) or \(x = -2\) and at \(x = 1\), we obtained the following equations:

\[ f(-2) = -16a + c = 54 \]

\[ f(1) = 11a + c = 0 \]

By eliminating \(c\) we get: \(-27a = 54\) and thus \(a = -2\). Using this value of \(a\), we find \(b = -20\) and \(c = 22\).

(b) From the graph of the function, we can see that \(f(x)\) has a root at \(x = 1\), that is, \(x - 1\) is a factor of \(f(x)\). By a simple inspection we realize:

\[ f(x) = -2x^2 - 20x + 22 \]

\[ f(x) = -2(x - 1)(x + 11) \]

Similarly, for the equation of the line:

\[ y - 8x = 88 \]

\[ y = 8(x + 11) \]

Equating these expressions, we get:

\[ -2(x - 1)(x + 11) = 8(x + 11) \]

\[ -2(x - 1) = 8 \]

\[ x = -3\]

(c) From part (a), we know that \(c = 25a + k\) for the vertex form of the function, where \(k\) is the \(y\) coordinate of the vertex. Therefore:

\[ k = c - 25a = 22 + 50 = 72 \]

Detailed Answer

(a) Evaluating \( f(x) \) at the \( x \) coordinates of the points given, we get:

\[ (-3)^3 \times a + (-3)^2 \times b + (-3) \times c + d = 76 \]

\[ (-1)^3 \times a + (-1)^2 \times b + (-1) \times c + d = 32 \]

\[ 2^3 \times a + 2^2 \times b + 2 \times c + d = -49 \]

\[ 4^3 \times a + 4^2 \times b + 4 \times c + d = 27 \]

\[ -27a + 9b - 3c + d = 76 \]

\[ -a + b - c + d = 32 \]

\[ 8a + 4b + 2c + d = -49 \]

\[ 64a + 16b + 4c + d = 27\]

(b) We can solve the problem by using the Apps function on the TI-84 calculator and then select PlySmlt2. Select the appropriate size for the system (this is a \(4 \times 4\) system). Press Graph to introduce the values according to the equations in part (a), to get: \( a = 2 \), \( b = 3 \), \( c = -36 \), and \( d = -5. \)

(c) Using the formula for the slope, together with the two points given in the graph:

\[ m = \frac{27 - 76}{4 - (-3)} = \frac{-49}{7} = -7 \]

Taking the point \((4, 27)\) and the slope-point form equation of a line, we get:

\[ y - 27 = -7(x - 4) \]

\[ y - 27 = -7x + 28 \]

\[ y = -7x + 55 \]

Since the line and the function meet at \((-3, 76)\) and \((4, 27)\), the polynomial for \(f(x) - y\) can be written as:

\[ f(x) - y = (2x - e)(x + 3)(x - 4) \]

\[ 2x^2 + 3x^2 - 36x - 5 + 7x - 55 = (2x - e)(x^2 - x - 12) \]

\[ 2x^2 + 3x^2 - 29x - 60 = 2x^3 - (2 + e)x^2 + (e - 24)x - 12 \]

By comparing terms, it follows that \( e = -5 \) and thus:

\[ x = \frac{-5}{2} \text{ and } y = -7 \cdot \frac{-5}{2} + 55 = \frac{-145}{2} \]

(d) According to the problem, \( x - 5 \) is a factor of \( f(x) \). Therefore, using Horner's rule, we find that:

\[ \frac{2x^2 + 3x^2 - 36x - 5}{x - 5} = 2x^2 - 7x - 1 \]

We now find the discriminant of this quadratic formula to get:

\[ r = 7^2 - 4 \times 2 \times (-1) = 49 + 8 = 57 \]