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Student Page 3.1.4 Solving Equations — Hands and Minds On
A4US

When set in context, it is easier to decide how to solve an equation. For example, when trying to determine how many toppings you can get on a large original pizza for $25, you might have thought about the solution in one of the following ways:

  1. An original crust large cheese pizza costs $10.99. To figure out the number of toppings I can get, I see that I have \(\$25 - \$10.99 = \$14.01\) to spend on toppings. Each topping costs $1.50. (Parts a, b, and c are different possible ways to proceed from here.)

    1. I can find the number of toppings I can afford by dividing $14.01 by $1.50. This gives me 9.34. The pizza restaurant will only allow me to buy whole numbers of toppings so I can get 9 toppings on a $25 large original crust pizza.

    2. If I start with $14.01 and subtract 1.50 until I can't subtract it anymore, I can do that 9 times, so I can afford 9 toppings on my pizza if I have $25 to spend.

    3. Two toppings cost $3. $3 × 4 = $12 with $2.01 left over for one more topping. So I can afford 8 toppings + 1 topping = 9 toppings for $25.

Notice that it helps to keep the context in mind. The context helps you make sense of the operations you are using. This allows you to be very flexible with the ways you solve the problem because you know what each of the parts represent.

Here is another context for solving equations: Let a pawn represent an unknown value of \(x\text{.}\) Let small cubes represent 1. Think about each side of an equation as sitting on a balance scale. Because one side equals the other side of the equation, the scale is balanced. In order to keep a scale balanced, it is necessary to add the same thing to both sides, remove the same thing from both sides, or divide both sides into the same number of groups then remove parts that equal each other from both sides, leaving only one group on each side of the scale. Following are examples that we will solve using this context. (The problems with negative signs and the problems with subtraction symbols require 2 different colors of cubes (pawns), one color to represent positive one (\(+x\)), and one color to represent negative one (\(-x\)).)

\(x + 1 = 5\) \(2x = 6\)
\(2x + 1 = 7\) \(2x + 1 = 3x\)
\(2x + 7 = 3x\) \(2x + 7 = 3x + 5\)
\(2 \left( x + 2 \right) = 3x + 2\) \(1 + -1 = x + -x\)
\(x - 1 = 5\) \(2x -1 = 3\)
\(3 \left( x - 1 \right) = x + 3\) \(-2x + 6 = x\)

Now resolve the above problems, this time thinking of \(x\) as some unknown number of bananas. The constants (those numbers not multiplied by \(x\)) are known numbers of bananas. (Omit the problems with subtraction symbols and the problems with negative signs.)