Answers and Solutions

1. 1. Substituting \(x=3\) into the equation we get \(y=3+11\). Evaluating, we determine \(y=14\).
   2. Substituting \(y=7\) into the equation we get \(7=\dfrac{x}{4}\). Solving, we determine \(x=28\).
   3. Substituting \(x=4\) into the equation we get \(y=4-7\). Evaluating, we determine \(y=-3\).
2. 1. Substituting \(x=6\) into the equation we get \(y=3(6)-2\). Evaluating, we determine \(y=16\).
   2. Substituting \(x=27\) into the equation we get \(y=\dfrac{27}{3}-1\). Evaluating, we determine \(y=8\).
   3. Substituting \(y=9\) into the equation we get \(9=\dfrac{x}{7}+4\). Solving, we determine \(x=35\).
3. 1. If \(128\) people are expected to attend the bake sale, we can substitute \(p=128\) into the equation.
      We get \(B=\dfrac{128}{8}-3\). Evaluating, we determine that \(B=13\).
      Therefore, \(13\) batches of cookies would be needed.
   2. If \(7\) batches of cookies have been prepared, we can substitute \(B=7\) into the equation.
      We get \(7=\dfrac{p}{8}-3\). Solving for \(p\), we determine that \(p=80\).
      Therefore, based on the batches of cookies it is expected that \(80\) people will attend the bake sale.
4. 1. If Kris' total cost was \($16\) then we can substitute \(C=16\) into the equation.
      We get \(16=2d+7\). Solving for \(d\) we determine that \(d=4.5\).
      Therefore, Kris paid \($4.50\) for each drink.
   2. If each drink costs \($5.25\) then we can substitute \(d=5.25\) into the equation.
      We get \(C=2(5.25)+7\). Evaluating we determine that \(C=17.5\).
      Therefore, Kris' total cost would be \($17.50\).
5. 1. \(S=3n+2\)
   2. When we consider the \(10^{th}\) image, we are considering when \(n=10\).
      We can substitute this into the equation from part a) and then evaluate to determine the corresponding value of \(S\).\[\begin{align*} S &= 3(10)+2 \\ &= 30 + 2 \\ &= 32 \end{align*}\]Therefore, the \(10^{th}\) image in this sequence contains \(32\) squares.
   3. If there are \(125\) squares in an image, then \(S=125\).
      We can substitute this value into the equation from part a) and then solve for \(n\).\[\begin{align*} 125 &= 3n+2 \\ 125 \class{hl2}{-2} &= 3n+2 \class{hl2}{-2} \\ 123 &= 3n \\ \dfrac{123}{\class{hl2}{3}} &= \dfrac{3n}{\class{hl2}{3}} \\ 41 &= n \end{align*}\]Therefore, the \(41^{st}\) term in the sequence corresponds to the image with \(125\) squares.
6. 1. If \(m = 1\) then the output is \(5(1+3) = 5(4) = 20\).
      If \(m = 6\) then the output is \(5(6+3) = 5(9) = 45\).
   2. If the output of the machine is \(35\), then we must have \(5(m+3) = 35\).
      Since \(5(7) = 35\) we know that \(m + 3\) must equal \(7\). This means that \(m=4\).
   3. If the output of the machine is \(225\) then we must have \(5(m+3) = 225\).  
      Let's try and solve this equation using algebra. Since the machine first adds \(3\) and then multiplies by \(5\), to undo the math machine we need to undo these operations, starting with the last operation. To undo multiplying by \(5\), we divide by \(5\). Going back to our equation this gives the following:\[\begin{align*} 5(m+3) & = 225 \\ \frac{5(m+3)}{\class{hl2}{5}} & = \frac{225}{\class{hl2}{5}} \\ m+3 & = 45 \end{align*}\]Now we can subtract \(3\) to solve the equation:\[\begin{align*} m+3 & = 45 \\ m + 3 \class{hl2}{-3} & = 45 \class{hl2}{-3}\\ m &= 42 \end{align*}\]We can verify that \(5(42+3) = 5(45) = 225\) and so the input was \(m=42\). 
7. 1. The value of \(x\) cannot directly give us the measure of \(\angle C\), but it will allow us to determine the measure of \(\angle B\). If \(x = 10\), then \(3x+5 = 3(10)+5 = 35\) and so \(\angle B = 35^{\circ}\).
      Since \(\angle A + \angle B = 50^{\circ} + 35^{\circ} = 85^{\circ}\) and \(180 - 85 = 95\) we must have \(\angle C = 95^{\circ}\).
   2. If \(\angle C = 101^{\circ}\), then we know the measure of two angles in the triangle.
      Since \(\angle A + \angle C = 50^{\circ} + 101^{\circ} = 151^{\circ}\) and \(180 - 151 = 29\) we must have \(\angle B = 29^{\circ}\).
      But we also know that \(\angle B = (3x+5)^{\circ}\) and so we must have \(3x+5 = 29\).
      Solving this equation we get \(x = 8\).
   3. Since the sum of the three angles in any triangle is \(180^{\circ}\), the sum of \(\angle A\) and \(\angle B\) must be smaller than \(180^{\circ}\).
      This means\[\angle A + \angle B = 50^{\circ} + (3x+5)^{\circ} < 180^{\circ}\]

      Using trial and error, we find that when \(x = 41\), \(3x+5 = 3(41)+5=128\), and that when \(x = 42\), \(3x+5 = 3(42)+5 = 131\). If \(\angle B = 128^{\circ}\) then the two angles add to \(178^{\circ}\) which is small enough. If \(\angle B = 131^{\circ}\) then the two angles add to \(181^{\circ}\) which is too large.
      Any value of \(x\) larger than \(42\) will also have this problem, and so \(x = 41\) is the largest possible value of \(x\).
      It is also possible to solve the inequality above using algebra! Think about how you can use the tools you have for solving equations to determine the value \(x=41\).

8. 1. Since \(\sqrt{4} = 2\), when \(n = 4\) then the output is \(2 (2) + 14 = 18\).
      Since \(\sqrt{25} = 5\), when \(n = 25\) the output is \(2 (5) + 14 = 24\).
   2. \(2\sqrt{n} + 14\).
   3. Solve \(2\sqrt{n} + 14 = 16\). We can do this through trial and error.
      If \(n = 1\) then \(\sqrt{1} = 1\) and so \(2 \sqrt{1} + 14 = 2(1) + 14 = 16\) as needed.
   4. Solve \(2\sqrt{n} + 14 = 40\). Let's solve this equation using algebra. Since the last operation performed by the math machine is adding 14, let's first undo this operation:\[\begin{align*} 2 \sqrt{n}+14 &= 40 \\ 2 \sqrt{n} +14 \class{hl2}{-14} &= 40 \class{hl2}{-14} \\ 2 \sqrt{n}&= 26 \\ \end{align*}\]The second to last operation performed by the math machine was multiplying by \(2\) and so we next undo this operation:\[\begin{align*} 2 \sqrt{n} &= 26 \\ \frac{2 \sqrt{n}}{\class{hl2}{2}} & = \frac{26}{\class{hl2}{2}}\\ \sqrt{n} & = 13 \end{align*}\]Now we are left with determining what positive integer has a square root of \(13\). Since \(13^{2} = 169\), we have \(\sqrt{169} = 13\) and so \(n = 169\). Verifying our solution gives \(2\sqrt{169} + 14 = 2(13) + 14 = 40\) as needed.
   5. The input must have been \(\left(\dfrac{p-14}{2}\right)^{2}\).