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Is each pair of expressions equivalent? If they are not, change the second expression so that they are equivalent. Check your work on your calculator by comparing table values when you enter the equivalent expressions into \(\mathrm{Y}_{1}\) and \(\mathrm{Y}_{2}\). a. \(3-3(x+4) \quad 3 x-9\) (a) b. \(5+2(x-2) \quad 2 x+1\) c. \(5 x-3 \quad 2+5(x-1)\) d. \(-2 x-8 \quad-2(x-4)\)

Short Answer

Expert verified
a: Not equivalent, change second to \(-3x-9\). b: Equivalent. c: Equivalent. d: Not equivalent, change first to \(-2x+8\).

Step by step solution

01

Simplify the first expression for (a)

The first expression in part (a) is \(3 - 3(x + 4)\). We first distribute the \(-3\) over the terms inside the parenthesis: \[ 3 - 3(x + 4) = 3 - 3x - 12 \] Then, we combine like terms: \[ -3x - 9 \]
02

Compare the simplified expression with the second expression for (a)

The first simplified expression from Step 1 is \(-3x - 9\) and we are given a second expression \(3x - 9\). These are not equivalent since the coefficients of \(x\) are different.
03

Modify the second expression to make them equivalent for (a)

To make the expressions equivalent, modify the second expression \(3x - 9\) to \(-3x - 9\).So, the final matching expressions are \(-3x - 9\) for both.
04

Simplify the first expression for (b)

The first expression in part (b) is \(5 + 2(x - 2)\). We distribute the \(2\) over the terms inside the parenthesis: \[ 5 + 2(x - 2) = 5 + 2x - 4 \] Combine like terms: \[ 2x + 1 \]
05

Compare with the second expression for (b)

The simplified expression from Step 1 is \(2x + 1\), and the given second expression is also \(2x + 1\). They are already equivalent.
06

Simplify the second expression for (c)

The second expression in part (c) is \(2 + 5(x - 1)\). We distribute the \(5\) over the terms inside the parenthesis: \[ 2 + 5(x - 1) = 2 + 5x - 5 \] Combine like terms: \[ 5x - 3 \]
07

Compare with the first expression for (c)

The simplified expression from Step 1 is \(5x - 3\), and the given first expression is also \(5x - 3\). They are equivalent as given.
08

Simplify the second expression for (d)

The second expression in part (d) is \(-2(x - 4)\). We distribute the \(-2\) over the terms inside the parenthesis: \[ -2(x - 4) = -2x + 8 \]
09

Compare with the first expression for (d)

The simplified expression from Step 1 is \(-2x + 8\), and the given first expression is \(-2x - 8\). These are not equivalent since the constant term differs.
10

Modify the first expression to match the second for (d)

To make them equivalent, change the first expression from \(-2x - 8\) to \(-2x + 8\).So, the final expressions are \(-2x + 8\) for both.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equivalent Expressions
Equivalent expressions in algebra are expressions that are the same, even if they may look different at first glance. These expressions will give the same result for any values substituted into the variables. For example, you can have two expressions such as \(3 + 2\) and \(5\). Even though they appear different, they both equal 5, thus they are equivalent.

Checking equivalence can involve simplifying both expressions and seeing if they match. In our exercise, this step was evident when simplifying the expressions to see if they share the same form. Equivalent expressions are handy because they allow us to verify our work and confirm that transformations or simplifications maintain the original value or equation.
Simplifying Expressions
Simplifying expressions involves combining and reducing expressions to their cleanest form. This might include applying arithmetic operations, getting rid of parentheses through the distributive property, and combining like terms.

For instance, in the original solutions, expressions like \(3 - 3(x + 4)\) are simplified using distribution and combination of terms, resulting in \(-3x - 9\). Simplifying expressions is crucial as it often makes calculations easier and helps in comparing expressions. Start simplifying by handling parentheses, using the distributive property, and then reduce by joining like terms. This process makes the expression more manageable and easier to understand.
Distributive Property
The distributive property is an essential algebra concept that allows you to remove parentheses when you see an expression like \(a(b + c)\). It implies that you multiply \(a\) both by \(b\) and \(c\), resulting in \(ab + ac\).

In the exercise solutions, the distributed property was utilized to break down expressions such as \(5 + 2(x - 2)\) into \(5 + 2x - 4\). This step is pivotal because it helps to transition from a longer form with parentheses to a more straightforward, cleaner expression you can work with easily. It showcases how multiplication distributes over addition or subtraction,
making it a powerful tool for simplifying expressions in algebra.
Like Terms
In algebraic expressions, like terms are the terms that have identical variable parts but might differ in their coefficients. They can be easily combined by adding or subtracting their coefficients. The rule is simple: constants with constants, \(x\) terms with \(x\) terms, and so on.

For instance, in expressions like \(-3x - 12 + 3\), we identify like terms such as \(-12\) and \(3\), which are both constants. These terms are combined to give \(-9\), making the expression \(-3x - 9\). Recognizing like terms is critical because combining them is one of the primary tasks in simplifying expressions. When you reduce expressions in this way, you create simpler forms that are easier to understand and equate with other expressions.

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Most popular questions from this chapter

The equation \(3 x+2 y=6\) is in standard form. a. Find \(x\) when \(y\) is zero. Write your answer in the form \((x, y)\). What is the significance of this point? (a) b. Find \(y\) when \(x\) is zero. Write your answer in the form \((x, y)\). What is the significance of this point? (a) c. On graph paper, plot the points you found in \(10 \mathrm{a}\) and \(\mathrm{b}\) and draw the line through these points. (a) d. Find the slope of the line you drew in \(10 \mathrm{c}\) and write a linear equation in intercept form. e. On your calculator, graph the equation you wrote in \(10 \mathrm{~d}\). Compare this graph to the one you drew on paper. Is the intercept equation equivalent to the standard-form equation? Explain why or why not. f. Symbolically show that the equation \(3 x+2 y=6\) is equivalent to your equation from \(10 \mathrm{~d}\).

A line has the equation \(y=4-4.2 x\). a. Find the \(y\)-coordinate of the point on this line whose \(x\)-coordinate is 2 . b. Use the point you found in 11 a to write an equation in point-slope form. 2\. Find the \(x\)-coordinate of the point whose \(y\)-coordinate is \(6.1\). d. Use the point you found in \(11 \mathrm{c}\) to write a different point-slope equation. e. Show that the point-slope equations you wrote in \(11 \mathrm{~b}\) and \(\mathrm{d}\) are equivalent to the original equation in intercept form. Explain your procedure. Is the point \((4,-12)\) on the line? How about \((-3,16.6)\) ? Explain how you can determine whether a given point is on a line.

APPLICATION The table below shows fat grams and calories for some breakfast sandwiches. \begin{tabular}{|l|c|c|} \hline Breakfast sandwich & Total fat (g) \(\boldsymbol{x}\) & Calories \(\boldsymbol{y}\) \\ \hline Arby's Bacon 'n Egg Croissant & 26 & 410 \\ \hline Burger King Croissanwich with Sausage, Egg \& Cheese & 39 & 520 \\ \hline Carl's Jr. Sunrise Sandwich & 21 & 356 \\ \hline Hardee's Country Steak Biscuit & 41 & 620 \\ \hline Jack in the Box Sourdough Breakfast Sandwich & 26 & 445 \\ \hline McDonald's Sausage McMuffin with Egg & 28 & 450 \\ \hline Sonic Sausage, Egg \& Cheese Toaster & 36 & 570 \\ \hline Subway Ham \& Egg Breakfast Deli Sandwich & 13 & 310 \\ \hline \end{tabular} (www.arbys.com, www.burgerking.com, www.carlsjr.com, www. hardeesrestaurants.com, www.jackinthebox.com, www.mcdonalds.com, www.sonicdrivein.com, www.subway.com). [Data set: FFFAT, FFCAL] \(\mathrm{~ a . ~ M a ? k e ? ~ a ~ s c a ? t t e r ~ p l o o t o f o ~ t h e ? ~ d a ? t a ? . ~ D e s s c r i b e ? ~ a n n y ~ p a t t e}\) b. Select two points and find the equation of the line that passes through these two points in point-slope form. Graph the equation on the scatter plot. c. According to your model, how many calories would you expect in a Hardee's Country Steak Biscuit with 41 grams of fat? d. Does the actual data point representing the Hardee's Country Steak Biscuit lie above, on, or below the line you graphed in \(9 \mathrm{~b}\) ? Explain what the point's location means.e. Check each breakfast sandwich to find if its data point falls above, on, or below your line. f. Based on your results for \(9 \mathrm{~d}\) and e, how well does your line fit the data? g. If a sandwich has 0 grams of fat, how many calories does your equation predict? Does this answer make sense? Why or why not?

Your friend walks steadily away from you at a constant rate such that her distance at \(2 \mathrm{~s}\) is \(3.4 \mathrm{~m}\) and her distance at \(4.5 \mathrm{~s}\) is \(4.4 \mathrm{~m}\). Let \(x\) represent time in seconds, and let \(y\) represent distance in meters. a. What is the slope of the line that models this situation? (hi) b. What is the \(y\)-intercept of this line? Explain how you found it. c. Write a linear equation in intercept form that models your friend's walk.

Show how to solve the equation \(3.8=0.2(z+6.2)-5.4\) by using an undoing process to write an expression for \(z\). Check your answer by substituting it into the original equation.

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