Unit 4
Expressions and More Equations
Accelerated 7
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This week your student will be working with equivalent expressions (expressions that are always equal, for any value of the variable). For example, \(2x+7+4x\) and \(6x+10-3\) are equivalent expressions. We can see that these expressions are equal when we try different values for \(x\).
| \(\displaystyle 2x+7+4x\) | \(\displaystyle 6x+10-3\) | |
|---|---|---|
| when \(x\) is 5 | \(\displaystyle 2\boldcdot5+7+4\boldcdot5\) \(\displaystyle 10 + 7 + 20\) \(\displaystyle 37\) | \(\displaystyle 6\boldcdot5+10-3\) \(\displaystyle 30+10-3\) \(\displaystyle 37\) |
| when \(x\) is -1 | \(\displaystyle 2\boldcdot\text-1+7+4\boldcdot\text-1\) \(\displaystyle \text-2 + 7 + \text-4\) \(\displaystyle 1\) | \(\displaystyle 6\boldcdot\text-1+10-3\) \(\displaystyle \text-6+10-3\) \(\displaystyle 1\) |
We can also use properties of operations to see why these expressions have to be equivalent—they are each equivalent to the expression \(6x+7\).
Here is a task to try with your student:
Match each expression with an equivalent expression from the list below. One expression in the list will be left over.
List:
Solution:
This week your student will work on solving linear equations. We can think of a balanced hanger as a metaphor for an equation. An equation says that the expressions on either side have equal value, just like a balanced hanger has equal weights on either side.
If we have a balanced hanger and add or remove the same amount of weight from each side, the result will still be in balance. For example, we could remove 2 triangles from each side of this hanger and it would still balance. We could also add a square to each side and it would still balance.
We can do this with equations as well: Adding or subtracting the same amount from both sides of an equation keeps the sides equal to each other. For example, if \(4x+20\) and \(\text-6x +10\) have equal value, we can write an equation \(4x+20=\text-6x+10\). We could add -10 to both sides of the equation or divide both sides of the equation by 2 and keep the sides equal to each other. Using these moves in systematic ways, we can find that \(x=\text-1\) is a solution to this equation.
Here is a task to try with your student:
Elena and Noah work on the equation \(\frac12 \left(x+4\right) = \text-10+2x\) together. Elena’s solution is \(x=24\) and Noah’s solution is \(x=\text-8\). Here is their work:
Elena:
\(\begin{align} \frac12 \left(x+4\right) &= \text-10+2x\\ x+4 &= \text-20+2x\\ x+24 &= 2x\\ 24&=x\\ x&=24\end{align}\)
Noah:
\(\begin{align} \frac12 \left(x+4\right) &= \text-10+2x\\ x+4 &=\text -20+4x\\ \text-3x+4 &= \text-20\\ \text-3x &= \text-24\\ x&=\text-8\end{align}\)
Do you agree with their solutions? Explain or show your reasoning.
Solution:
No, they both have errors in their solutions.
Elena multiplied both sides of the equation by 2 in her first step, but forgot to multiply the \(2x\) by the 2. We can also check Elena’s answer by replacing \(x\) with 24 in the original equation and seeing if the equation is true. \(\displaystyle \frac12 \left(x+4\right) =\text -10+2x\) \(\displaystyle \frac12 \left(24+4\right) =\text -10+2(24)\) \(\displaystyle \frac12 \left(28\right) = \text-10+48\) \(\displaystyle 14=38\) Because 14 is not equal to 38, Elena’s answer is not correct.
Noah divided both sides by -3 in his last step, but wrote -8 instead of \(8\) for \(\text-24 \div \text-3\). We can also check Noah’s answer by replacing \(x\) with -8 in the original equation and seeing if the equation is true. Noah’s answer is not correct.