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%2F20240727%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240727T032646Z&X-Amz-Expires=604800&X-Amz-SignedHeaders=host&X-Amz-Signature=e92ebce18f7f751b430621efaddcc5a8e296b8b7644416e86d1a7db7005fb085)
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%2F20240727%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240727T032646Z&X-Amz-Expires=604800&X-Amz-SignedHeaders=host&X-Amz-Signature=8971b4790e601719537231c538e4b3a26ec279eb2d7ee27c06b2a4f5886dedbe)
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%2F20240727%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240727T032646Z&X-Amz-Expires=604800&X-Amz-SignedHeaders=host&X-Amz-Signature=2339123a2c35d0cf366dcffb7e6e8f5c160f4fb9cb52c4d496c060fa1b31a600)
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%2F20240727%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240727T032646Z&X-Amz-Expires=604800&X-Amz-SignedHeaders=host&X-Amz-Signature=8bd3643a5ed16f06ee66c6981da4dc572672272b88b45444845235cdb0f04811)
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%2F20240727%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240727T032646Z&X-Amz-Expires=604800&X-Amz-SignedHeaders=host&X-Amz-Signature=852bd60988feb4f9d8fbe76b15c417314d81335eac075dfd14cb699b0de2da1e)
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%2F20240727%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240727T032646Z&X-Amz-Expires=604800&X-Amz-SignedHeaders=host&X-Amz-Signature=58ec9af573b13d22171228c141e9fc8623d0f88eecea33d887abe82514578181)
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%2F20240727%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240727T032646Z&X-Amz-Expires=604800&X-Amz-SignedHeaders=host&X-Amz-Signature=9ac8c14f965c32fe3c8aca5417e42971f8c09eab145ac0f4b282965cce60b67f)
Here is a concise way to write the steps above:
\(\begin{align} 3(x+2) &= 18 \\ x + 2 &= 6 & \text{after multiplying each side by } \tfrac13 \\ x &= 4 & \text{after subtracting 2 from each side} \\ \end{align} \)