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Question
- which shows an expression equivalent to 6x + 8.25 - 4.5x? a 1.5x - 8.25 b 10.5x + 8.25 c 1.5x + 8.25 d 10.5x - 8.25 assessment practice 13. select all expressions equivalent to -\frac{2}{3}x + 2. -2 - \frac{2}{3}x 2 - \frac{2}{3}x -1 - \frac{2}{3}x + 1 -\frac{1}{3}x - 4 + 2 -\frac{2}{3}x - 3 + 5
Question 11
Step1: Combine like terms (x - terms)
We have the expression \(6x + 8.25 - 4.5x\). First, combine the \(x\)-terms: \(6x-4.5x=(6 - 4.5)x\)
\(6-4.5 = 1.5\), so \(6x-4.5x=1.5x\)
Step2: Rewrite the expression
After combining the \(x\)-terms, the expression becomes \(1.5x+8.25\) (since the constant term \(8.25\) remains unchanged)
- \(2-\frac{2}{3}x\) can be rewritten as \(-\frac{2}{3}x + 2\) (commutative property).
- \(-\frac{2}{3}x-3 + 5\) simplifies to \(-\frac{2}{3}x+2\) (since \(-3 + 5=2\)).
- Other expressions either have incorrect constant terms or incorrect \(x\)-terms.
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C. \(1.5x + 8.25\)
Question 13
We need to check each expression to see if it is equivalent to \(-\frac{2}{3}x + 2\)
- For the expression \(-2-\frac{2}{3}x\):
The constant term here is \(- 2\) which is not equal to \(2\) in the original expression. So it is not equivalent.
- For the expression \(2-\frac{2}{3}x\):
We can rewrite it as \(-\frac{2}{3}x + 2\) (by commutative property of addition), which is the same as the original expression. So it is equivalent.
- For the expression \(-1-\frac{2}{3}x + 1\):
Simplify the constant terms: \(-1 + 1=0\), so the expression becomes \(-\frac{2}{3}x+0=-\frac{2}{3}x\), which is not equal to \(-\frac{2}{3}x + 2\) (the constant term is \(0\) instead of \(2\)). So it is not equivalent.
- For the expression \(-\frac{1}{3}x-4 + 2\):
Simplify the constant terms: \(-4 + 2=-2\), and the \(x\)-term is \(-\frac{1}{3}x\) which is not equal to \(-\frac{2}{3}x\) in the original expression. So it is not equivalent.
- For the expression \(-\frac{2}{3}x-3 + 5\):
Simplify the constant terms: \(-3 + 5 = 2\), so the expression becomes \(-\frac{2}{3}x+2\), which is the same as the original expression. So it is equivalent.