2. Equations and the Pythagorean Theorem (A)
  3. Lesson 10: Solving Two-Step Equations Using Algebra

Answers and Solutions


    1. To verify, we substitute \(k=20\) into \(2k+6\) and evaluate.\[\begin{align*} 2k+6 &= 2(20)+6 \\ &= 40+6 \\ &= 46 \end{align*}\]Since \(46 \neq 34\), we conclude that \(k=20\) is not the solution to the equation \(2k+6=34\).
    2. To verify, we substitute \(y=27\) into \(\dfrac{y}{3}-2\) and evaluate.\[\begin{align*} \dfrac{y}{3} - 2 &= \dfrac{27}{3} - 2 \\ &= 9 - 2 \\ &= 7 \end{align*}\]Since the result is \(7\), we conclude that \(y=27\) is the solution to the equation \(\dfrac{y}{3}-2=7\).
    3. To verify, we substitute \(g=5\) into \(3+4g\) and evaluate.\[\begin{align*} 3+4g &= 3+4(5) \\ &= 3+20 \\ &= 23 \end{align*}\]Since the result is \(23\), we conclude that \(g=5\) is the solution to the equation \(23=3+4g\).
    4. To verify, we substitute \(p=24\) into \(\dfrac{p}{3}-1\) and evaluate.\[\begin{align*} \dfrac{p}{3}-1 &= \dfrac{24}{3}-1 \\ &= 8-1 \\ &= 7 \end{align*}\]Since \(7 \neq 9\), we conclude that \(p=24\) is not the solution to the equation \(9=\dfrac{p}{3}-1\).
    1. \(4n\)\(+5\)\(=31\)
      First we would reverse the addition, and then reverse the multiplication in the second step.
    2. \(0.6g\)\(-4\)\(=3.2\)
      First we would reverse the subtraction, and then reverse the multiplication in the second step.
    3. \(16=\)\(\dfrac{y}{2}\)\(-5\)
      First we would reverse the subtraction, and then reverse the division in the second step.
  3. Notice that there are two parts to this equation: \(4\) and \(3y\).
    From order of operations, we know that the expression \(4+3y\) is multiplying \(3\) and \(y\) together and then adding the result to \(4\). Following our observations from Question 2), we should undo the addition first.
    Since subtracting \(3y\) from each side does not simplify our equation, we subtract \(4\) from both sides.
    Step 1: Subtract \(4\) from both sides.\[\begin{align*} 4+3y&=43 \\ 4+3y \class{hl2}{-4} &= 43 \class{hl2}{-4} \\ 3y &= 39 \end{align*}\]Step 2: Divide both sides by \(3\).\[\begin{align*} 3y &= 39 \\ \dfrac{3y}{\class{hl1}{3}} &= \dfrac{39}{\class{hl1}{3}} \\ y &= 13 \end{align*}\]Verifying, we have that \(4+3(13)=43\).
    Therefore, \(y=13\) is the solution to the equation.
  4. Solve each equation using algebra.
    1.  \[\begin{align*} 7x-10 &=11 \\ 7x-10 \class{hl2}{+10} &=11 \class{hl2}{+10} \\ 7x &= 21 \\ \dfrac{7x}{\class{hl1}{7}} &= \dfrac{21}{\class{hl1}{7}} \\ x &= 3 \end{align*}\]Verifying, we have that \(7(3)-10=11\).
      Therefore, \(x=3\) is the solution to the equation.
    2.  \[\begin{align*} \dfrac{k}{4}+9&=6 \\ \dfrac{k}{4}+9 \class{hl2}{-9} &= 6 \class{hl2}{-9} \\ \dfrac{k}{4} &= -3 \\ \dfrac{k}{4} \class{hl1}{(4)} &= -3 \class{hl1}{(4)} \\ k &= -12 \end{align*}\]Verifying, we have that \(\dfrac{-12}{4} + 9 = 6\).
      Therefore, \(k=-12\) is the solution to the equation.
    3.  \[\begin{align*} 7&=2d-3 \\ 7 \class{hl2}{+3} &=2d-3 \class{hl2}{+3} \\ 10 &= 2d \\ \dfrac{10}{\class{hl1}{2}} &= \dfrac{2d}{\class{hl1}{2}} \\ 5 &= d \end{align*}\]
      Verifying, we have that \(7=2(5)-3\).
      Therefore, \(d=5\) is the solution to the equation.
    4.  \[\begin{align*} 6x+7&=49 \\ 6x+7 \class{hl2}{-7} &=49 \class{hl2}{-7} \\ 6x &= 42 \\ \dfrac{6x}{\class{hl1}{6}} &= \dfrac{42}{\class{hl1}{6}} \\ x &= 7 \end{align*}\]Verifying, we have that \(6(7)+7=49\).
      Therefore, \(x=7\) is the solution to the equation.
    5.  \[\begin{align*} \dfrac{r}{5}+8&=33 \\ \dfrac{r}{5}+8 \class{hl2}{-8}&=33 \class{hl2}{-8} \\ \dfrac{r}{5} &= 25 \\ \dfrac{r}{5} \class{hl1}{(5)} &= 25\class{hl1}{(5)} \\ r &= 125 \end{align*}\]Verifying, we have that \(\dfrac{125}{5}+8=33\).
      Therefore, \(r=125\) is the solution to the equation.
    6.  \[\begin{align*} 12+\dfrac{x}{5} &= 16 \\ 12+\dfrac{x}{5} \class{hl2}{-12} &= 16 \class{hl2}{-12} \\ \dfrac{x}{5} &= 4 \\ \dfrac{x}{5} \class{hl1}{(5)} &= 4 \class{hl1}{(5)} \\ x &= 20 \end{align*}\]Verifying, we have that \(12 + \dfrac{20}{5} = 16\).
      Therefore, \(x=20\) is the solution to the equation.
    1. Answers will vary. 
      Since we know that \(3(9)-6=21\), the equation \(3k-6=21\) has the solution \(k=9\).
    2. Answers will vary.
      Since we know that \(\dfrac{9}{3} + 5 = 8\), the equation \(\dfrac{k}{3}+5=8\) has the solution \(k=9\).
    1. Let \(W\) represent the price of the watch. 
      Then the equation \(W=2r+50\) represents the price of the watch in terms of \(r\).
    2. If the price of the watch is \($208\), then \(W=208\).
      We can substitute this into the equation from part a) and solve for \(r\).\[\begin{align*} 208 &= 2r+50 \\ 208 \class{hl2}{-50} &= 2r+50 \class{hl2}{-50} \\ 158 &= 2r \\ \dfrac{158}{\class{hl1}{2}} &= \dfrac{2r}{\class{hl1}{2}} \\ 79 &= r \end{align*}\]Therefore, the price of the ring is \($79\).
  7. Solution 1
    Recall that we can multiply both sides of an equation by the same number and the resulting expressions will still be equal.  Consider multiplying both sides of the equation \(2x+5=15\) by \(3\).\[3(2x+5) = 3(15)\]Recall the distributive property which tells us we need to simplify each term on the left side. We multiply the left side by \(3\), and get that\[6x + 15 = 45\]Therefore, the value of \(6x+15\) is \(45\).
    Solution 2
    Alternatively, we can solve for \(x\) in the equation \(2x+5=15\) and determine that \(x=5\).
    Then, we can substitute \(x=5\) into the expression \(6x+15\) and evaluate. The result is that \(6x+15=45\).
    1. If \(y\) is the smallest of the three integers, then the next two even integers would be \(y+2\) and \(y+4\).
    2. The sum of the three integers \(y\), \(y+2\), and \(y+4\) is:\[y + y+2 + y+4 = 3y + 6\]
    3. Given that the sum of the three integers is \(66\), we have\[3y + 6 = 66\]We can solve this equation to determine the value of \(y\).\[\begin{align*} 3y + 6 \class{hl2}{-6} &= 66 \class{hl2}{-6} \\ 3y &= 60 \\ \dfrac{3y}{\class{hl1}{3}} &= \dfrac{60}{\class{hl1}{3}} \\ y &= 20 \end{align*}\]Then \(y=20\) is the solution to the equation.
      Therefore, the three consecutive even integers are \(20\), \(22\), and \(24\).
  9. Let \(d\) represent the distance, in metres, from the line \(M\) to the line \(L\).
    Therefore, the total length of piece \(W\) to the left of the cut is \(d\).
    Since piece \(X\) is \(3\) m from the line \(M\), then the length of piece \(X\) to the left of \(L\) is \((d-3)\) m, because \(3\) of the \(d\) metres to the left of \(L\) are empty.
    Similarly, the lengths of pieces \(Y\) and \(Z\) to the left of the line \(L\) are \((d-2)\) m and \((d-1.5)\) m.
    Therefore, the total length of lumber, in metres, to the left of the line \(L\) is\[d + (d-3) + (d-2) +(d-1.5) = 4d-6.5\]Since the total length of lumber on each side of the cut is equal, then this total length is\[\dfrac{5+3+5+4}{2} = 8.5\]Combining these two pieces of information, we have that\[\begin{align*} 4d-6.5&=8.5 \\ 4d &= 15 \\ d &= 3.75 \end{align*}\]Solving for \(d\) gives us that \(d=3.75\) m.
    Therefore, the length of the part of piece \(W\) to the left of the line \(L\) is \(3.75\) m.