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 \(4\).