Lesson 8
Reasoning about Solving Equations (Part 2)
Let’s use hangers to understand two different ways of solving equations with parentheses.
8.1: Equivalent to $2(x+3)$
Select all the expressions equivalent to \(2(x+3)\).
- \(2 \boldcdot (x+3) \)
- \((x + 3)2 \)
- \(2 \boldcdot x + 2 \boldcdot 3\)
- \(2 \boldcdot x + 3 \)
- \((2 \boldcdot x) + 3\)
- \((2 + x)3\)
8.2: Either Or
-
Explain why either of these equations could represent this hanger:
\(14=2(x+3)\) or \(14=2x+6\)
- Find the weight of one circle. Be prepared to explain your reasoning.
8.3: Use Hangers to Understand Equation Solving, Again
Here are some balanced hangers. Each piece is labeled with its weight.
![Four balanced hanger diagrams.](https://staging-cms-im.s3.amazonaws.com/gBmQDveKvD42swJEN7e71y9Z?response-content-disposition=inline%3B%20filename%3D%227-7.6.Revision.Image.k8.10.png%22%3B%20filename%2A%3DUTF-8%27%277-7.6.Revision.Image.k8.10.png&response-content-type=image%2Fpng&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAXQCCIHWF37H2AMFB%2F20240703%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240703T053650Z&X-Amz-Expires=604800&X-Amz-SignedHeaders=host&X-Amz-Signature=d471ce8108793d6c0a5bbdd8a8ce2598f57f81ab5c99c30dc24fb00de82ac6e8)
For each diagram:
- Assign one of these equations to each hanger:
\(2(x+5)=16\)
\(3(y+200)=3\!,000\)
\(20.8=4(z+1.1)\)
\(\frac{20}{3}=2\left(w+\frac23\right)\)
- Explain how to figure out the weight of a piece labeled with a letter by reasoning about the diagram.
- Explain how to figure out the weight of a piece labeled with a letter by reasoning about the equation.
Summary
The balanced hanger shows 3 equal, unknown weights and 3 2-unit weights on the left and an 18-unit weight on the right.
There are 3 unknown weights plus 6 units of weight on the left. We could represent this balanced hanger with an equation and solve the equation the same way we did before.
\(\begin {align} 3x+6&=18 \\ 3x&=12 \\ x&=4 \\ \end{align}\)
![Balanced hanger. Left side, circle labeled x, square labeled 2, circle labeled x, square labeled 2, circle labeled x, square labeled 2. Right side, rectangle labeled 18.](https://staging-cms-im.s3.amazonaws.com/sSxHEZP3PCGkduHAzJQn5KXL?response-content-disposition=inline%3B%20filename%3D%227-7.6.B8.Summary.png%22%3B%20filename%2A%3DUTF-8%27%277-7.6.B8.Summary.png&response-content-type=image%2Fpng&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAXQCCIHWF37H2AMFB%2F20240703%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240703T053650Z&X-Amz-Expires=604800&X-Amz-SignedHeaders=host&X-Amz-Signature=1a7df41c7214acb05d030231b2158e9578037227cc0b1535f1c78a194be02ccf)
Since there are 3 groups of \(x+2\) on the left, we could represent this hanger with a different equation: \(3(x+2)=18\).
![Balanced hanger, three groups are indicated, each group contains 1 circle labeled x and 1 square labeled 2. Right side, rectangle labeled 18. To the side, an equation 3 ( x + 2 ) = 18.](https://staging-cms-im.s3.amazonaws.com/A1yPjy8HgHHDHwt2d3LsLrR8?response-content-disposition=inline%3B%20filename%3D%227-7.6.B8.Summary2.png%22%3B%20filename%2A%3DUTF-8%27%277-7.6.B8.Summary2.png&response-content-type=image%2Fpng&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAXQCCIHWF37H2AMFB%2F20240703%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240703T053650Z&X-Amz-Expires=604800&X-Amz-SignedHeaders=host&X-Amz-Signature=7057f89fa11d864919320ddb2faf1df71977ed44743d0ee36c4649afa75dce12)
The two sides of the hanger balance with these weights: 3 groups of \(x+2\) on one side, and 18, or 3 groups of 6, on the other side.
![Balanced hanger. to the side, an equation.](https://staging-cms-im.s3.amazonaws.com/jDpG2J4dZb1d7BDCdmPXHWnH?response-content-disposition=inline%3B%20filename%3D%227-7.6.B8.Summary3.png%22%3B%20filename%2A%3DUTF-8%27%277-7.6.B8.Summary3.png&response-content-type=image%2Fpng&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAXQCCIHWF37H2AMFB%2F20240703%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240703T053650Z&X-Amz-Expires=604800&X-Amz-SignedHeaders=host&X-Amz-Signature=cec33477611cf9d0f67d4479d67e282c6f0e8014679a50ad8c02529f74fd0837)
The two sides of the hanger will balance with \(\frac13\) of the weight on each side: \(\frac13 \boldcdot 3(x+2) = \frac13 \boldcdot 18\).
![Balanced hanger, left side, 1 circle labeled x and 1 square labeled 2, right side, rectangle labeled 6. To the side, an equation says x + 2 = 6.](https://staging-cms-im.s3.amazonaws.com/9gtQB2UasbLUsTZ7dJ41ranb?response-content-disposition=inline%3B%20filename%3D%227-7.6.B8.Summaryzzz.png%22%3B%20filename%2A%3DUTF-8%27%277-7.6.B8.Summaryzzz.png&response-content-type=image%2Fpng&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAXQCCIHWF37H2AMFB%2F20240703%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240703T053650Z&X-Amz-Expires=604800&X-Amz-SignedHeaders=host&X-Amz-Signature=294a276ac573c3a66b723d4fdc27631dc0904b51cefc7ca349cc39382358940d)
We can remove 2 units of weight from each side, and the hanger will stay balanced. This is the same as subtracting 2 from each side of the equation.
![Balanced hanger.](https://staging-cms-im.s3.amazonaws.com/HjE7PGhHSR86aug9FFvNFpvw?response-content-disposition=inline%3B%20filename%3D%227-7.6.B8.Summaryyyy.png%22%3B%20filename%2A%3DUTF-8%27%277-7.6.B8.Summaryyyy.png&response-content-type=image%2Fpng&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAXQCCIHWF37H2AMFB%2F20240703%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240703T053650Z&X-Amz-Expires=604800&X-Amz-SignedHeaders=host&X-Amz-Signature=58f76e8fbd4ae09cbfb8d5fa30a4fff389a45acc35b961eab01d0ceb24ef3513)
An equation for the new balanced hanger is \(x=4\). This gives the solution to the original equation.
![Balanced hanger, left side, circle labeled x, right side, rectangle labeled 4. To the side, an equation x = 4.](https://staging-cms-im.s3.amazonaws.com/7NffLYDYUXH9MNVhJBJRYTGY?response-content-disposition=inline%3B%20filename%3D%227-7.6.B8.Summaryasdfwer.png%22%3B%20filename%2A%3DUTF-8%27%277-7.6.B8.Summaryasdfwer.png&response-content-type=image%2Fpng&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAXQCCIHWF37H2AMFB%2F20240703%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240703T053650Z&X-Amz-Expires=604800&X-Amz-SignedHeaders=host&X-Amz-Signature=fdfac2a3ba1cd723b2679968475d29aa8cda853d4047729d831b36ad04986987)
Here is a concise way to write the steps above:
\(\begin{align} 3(x+2) &= 18 \\ x + 2 &= 6 & \text{after multiplying each side by } \tfrac13 \\ x &= 4 & \text{after subtracting 2 from each side} \\ \end{align} \)