Aimathic – Evaluate algebraic expressions

Evaluate algebraic expressions

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Difficulty: 1 – 5

5118856

At a produce stand, apples cost $\$2.40$ per pound and pears cost $\$3.10$ per pound. The total price is represented by $P=2.40a+3.10b$, where $a$ is the number of pounds of apples and $b$ is the number of pounds of pears. Find the price for each purchase: a) $2$ pounds of apples and $1$ pound of pears b) $1.5$ pounds of apples and no pears c) $0.5$ pound of apples and $2.5$ pounds of pears

Hints

- Substitute $0$ for the amount of a fruit that is not purchased. - Multiply each weight by its unit price. - Add the two costs.

Solution

1. For $a=2$ and $b=1$, $P=2.40\times 2+3.10\times 1=4.80+3.10=7.90$. 2. For $a=1.5$ and $b=0$, $P=2.40\times 1.5+3.10\times 0=3.60$. 3. For $a=0.5$ and $b=2.5$, $P=2.40\times 0.5+3.10\times 2.5=1.20+7.75=8.95$.

Answer

a) $\$7.90$ b) $\$3.60$ c) $\$8.95$

5121846

Let $a=-1.2$ and $b=0.5$. Evaluate each expression. a) $a+b$ b) $a-b$ c) $2ab$ d) $b^2$ e) $|a|-b$

Hints

- Substitute the values before calculating. - Track signs carefully in products and differences. - Absolute value gives the distance from $0$. - Squaring means multiplying a number by itself.

Solution

1. $a+b=-1.2+0.5=-0.7$. 2. $a-b=-1.2-0.5=-1.7$. 3. $2ab=2\times(-1.2)\times 0.5=-1.2$. 4. $b^2=(0.5)^2=0.25$. 5. $|a|-b=|-1.2|-0.5=1.2-0.5=0.7$.

Answer

a) $-0.7$ b) $-1.7$ c) $-1.2$ d) $0.25$ e) $0.7$

5124366

Complete the table by evaluating each expression for the given value of $x$. <table> <thead> <tr> <th>$x$</th> <th>$2x-5$</th> <th>$x^2+10$</th> <th>$12-3x$</th> </tr> </thead> <tbody> <tr> <td>$8$</td> <td></td> <td></td> <td></td> </tr> <tr> <td>$-4$</td> <td></td> <td></td> <td></td> </tr> <tr> <td>$0.5$</td> <td></td> <td></td> <td></td> </tr> </tbody> </table>

Hints

- Use parentheses when substituting a negative number. - A negative number squared is positive. - Follow the order of operations. - Line up decimal places when adding or subtracting.

Solution

1. For $x=8$: $2\times 8-5=11$, $8^2+10=74$, and $12-3\times 8=-12$. 2. For $x=-4$: $2\times(-4)-5=-13$, $(-4)^2+10=26$, and $12-3\times(-4)=24$. 3. For $x=0.5$: $2\times 0.5-5=-4$, $(0.5)^2+10=10.25$, and $12-3\times 0.5=10.5$.

Answer

5124396

Evaluate each expression for $z=4$ and $z=7$. a) $5z-9$ b) $z(z+2)$ c) $30-3z$

Hints

- Substitute each value for $z$. - Follow the order of operations. - Evaluate inside parentheses before multiplying.

Solution

1. For $z=4$: $5\times 4-9=11$, $4\times(4+2)=24$, and $30-3\times 4=18$. 2. For $z=7$: $5\times 7-9=26$, $7\times(7+2)=63$, and $30-3\times 7=9$.

Answer

a) $11$ when $z=4$; $26$ when $z=7$ b) $24$ when $z=4$; $63$ when $z=7$ c) $18$ when $z=4$; $9$ when $z=7$

5124406

Evaluate each expression for $b=1.5$ and $b=4$. a) $8b+2$ b) $b^2+10$ c) $2(10-b)$

Hints

- Substitute each value for $b$. - Square the value of $b$ before adding. - Evaluate inside parentheses first.

Solution

1. For $b=1.5$: $8\times 1.5+2=14$, $(1.5)^2+10=12.25$, and $2\times(10-1.5)=17$. 2. For $b=4$: $8\times 4+2=34$, $4^2+10=26$, and $2\times(10-4)=12$.

Answer

a) $14$ when $b=1.5$; $34$ when $b=4$ b) $12.25$ when $b=1.5$; $26$ when $b=4$ c) $17$ when $b=1.5$; $12$ when $b=4$

5124416

Evaluate each expression for $x=6$ and $y=2$. a) $xy+x$ b) $x\div y+5y$ c) $(x-y)(x+y)$

Hints

- Substitute both values before calculating. - Multiply and divide before adding. - Evaluate each set of parentheses first.

Solution

1. $xy+x=6\times 2+6=18$. 2. $x\div y+5y=6\div 2+5\times 2=3+10=13$. 3. $(x-y)(x+y)=(6-2)\times(6+2)=4\times 8=32$.

Answer

a) $18$ b) $13$ c) $32$

5124426

Substitute $b=0$, $b=4$, $b=10$, and $b=-3$ into $3b+14$. Find each value.

Hints

- Replace $b$ with each given number. - Multiply before adding. - Use parentheses when substituting a negative number.

Solution

1. For $b=0$, $3\times 0+14=14$. 2. For $b=4$, $3\times 4+14=26$. 3. For $b=10$, $3\times 10+14=44$. 4. For $b=-3$, $3\times(-3)+14=5$.

Answer

For $b=0$, the value is $14$. For $b=4$, the value is $26$. For $b=10$, the value is $44$. For $b=-3$, the value is $5$.

5124436

Evaluate $T(a)=25-4a$ for each value. a) $a=0$ b) $a=5$ c) $a=7$ d) $a=-2$

Hints

- Substitute each value for $a$. - Multiply before subtracting. - Use parentheses when substituting a negative number.

Solution

1. $T(0)=25-4\times 0=25$. 2. $T(5)=25-4\times 5=5$. 3. $T(7)=25-4\times 7=-3$. 4. $T(-2)=25-4\times(-2)=25+8=33$.

Answer

a) $25$ b) $5$ c) $-3$ d) $33$

5124446

Evaluate $T(x,y)=2x-3y$ for each pair. a) $x=5$, $y=2$ b) $x=3$, $y=-1$ c) $x=-2$, $y=-4$

Hints

- Substitute both values into the expression. - Use parentheses around negative values. - Track the subtraction sign in front of $3y$.

Solution

1. $T(5,2)=2\times 5-3\times 2=10-6=4$. 2. $T(3,-1)=2\times 3-3\times(-1)=6+3=9$. 3. $T(-2,-4)=2\times(-2)-3\times(-4)=-4+12=8$.

Answer

a) $4$ b) $9$ c) $8$

5124466

Evaluate $T(x,y)=3x^2+2y$ for each pair. a) $x=4$, $y=5$ b) $x=2$, $y=12$ c) $x=0.5$, $y=2$

Hints

- Substitute both values before calculating. - Evaluate the exponent before multiplying. - Add the two terms last.

Solution

1. $T(4,5)=3\times 4^2+2\times 5=48+10=58$. 2. $T(2,12)=3\times 2^2+2\times 12=12+24=36$. 3. $T(0.5,2)=3\times(0.5)^2+2\times 2=0.75+4=4.75$.

Answer

a) $58$ b) $36$ c) $4.75$

5124546

Which expressions have a value of $11$ when $x=4$? (1) $4x-5$ (2) $x^2+3$ (3) $15-x$ (4) $2(x+1)$

Hints

- Substitute $4$ for $x$ in each expression. - Evaluate one expression at a time. - Follow the order of operations.

Solution

1. Expression (1): $4\times 4-5=16-5=11$. 2. Expression (2): $4^2+3=19$. 3. Expression (3): $15-4=11$. 4. Expression (4): $2\times(4+1)=10$. 5. Expressions (1) and (3) have a value of $11$.

Answer

Expressions (1) and (3)

5124566

The expressions are $T_1=12-2x$ and $T_2=\frac{x+10}{3}$. Evaluate both expressions when $x=2$, then find $T_1-T_2$.

Hints

- Evaluate each expression separately. - Keep the two results labeled. - Subtract the value of $T_2$ from the value of $T_1$.

Solution

1. $T_1=12-2\times 2=8$. 2. $T_2=\frac{2+10}{3}=4$. 3. Therefore, $T_1-T_2=8-4=4$.

Answer

$T_1-T_2=4$

5124876

A rectangle has a fixed width of $6\,\text{cm}$ and a variable length $a$. a) Write an expression for its perimeter $P$ in terms of $a$. b) Find the perimeter when $a=4\,\text{cm}$ and when $a=12.5\,\text{cm}$.

Hints

- Add the lengths of all four sides of the rectangle. - Substitute the fixed width into the perimeter formula. - Evaluate the expression separately for each value of $a$.

Solution

1. The perimeter of a rectangle is $P=2a+2b$, where $b$ is the width. 2. Substitute $b=6$: $P=2a+2\times6=2a+12$. 3. When $a=4$, $P=2\times4+12=20\,\text{cm}$. 4. When $a=12.5$, $P=2\times12.5+12=37\,\text{cm}$.

Answer

a) $P=2a+12$ b) $20\,\text{cm}$ and $37\,\text{cm}$

5125276

Substitute $x = 5$ into each equation. Decide whether $5$ is a solution, and justify your answer. a) $4x - 7 = 13$ b) $3(x + 2) = 21$ c) $10 - 2x = 0$ d) $\frac{1}{5}x + 9 = 11$

Hints

- Replace every occurrence of $x$ with $5$. - Evaluate the left side using the order of operations. - A value is a solution only when both sides have the same value. - State the numerical comparison that supports each decision.

Solution

1. For a), the left side is $4 \times 5 - 7 = 13$, which equals the right side. Therefore, $5$ is a solution. 2. For b), the left side is $3(5 + 2) = 21$, which equals the right side. Therefore, $5$ is a solution. 3. For c), the left side is $10 - 2 \times 5 = 0$, which equals the right side. Therefore, $5$ is a solution. 4. For d), the left side is $\frac{1}{5} \times 5 + 9 = 10$, which does not equal $11$. Therefore, $5$ is not a solution.

Answer

a) Yes, $5$ is a solution. b) Yes, $5$ is a solution. c) Yes, $5$ is a solution. d) No, $5$ is not a solution because the left side equals $10$.

5222296

A hiking route has two sections. The first section is $a$ miles long, and the second section is $b$ miles long. Write an expression for the total length, then evaluate it for each pair of values. 1) $a=12$, $b=15$ 2) $a=4.7$, $b=3.8$ 3) $a=5\frac{3}{4}$, $b=6\frac{1}{2}$

Hints

- Add the lengths of sections traveled one after the other. - Line up decimal points when adding decimals. - Use a common denominator when adding fractions. - You may work with mixed numbers or convert them to another form.

Solution

1. The total length is the sum of the two sections: $a+b$. 2. For $a=12$ and $b=15$, $12+15=27$ miles. 3. For $a=4.7$ and $b=3.8$, $4.7+3.8=8.5$ miles. 4. For $a=5\frac{3}{4}$ and $b=6\frac{1}{2}$, $5\frac{3}{4}+6\frac{2}{4}=12\frac{1}{4}$ miles.

Answer

Expression: $a+b$ 1) $27$ miles 2) $8.5$ miles 3) $12\frac{1}{4}$ miles, or $12.25$ miles

5222306

A student needs $3$ wooden strips of length $x$ inches and $2$ wooden strips of length $y$ inches for a model. Write an expression for the total length of wood, then evaluate it for each pair of values. 1) $x=5$, $y=4$ 2) $x=2.5$, $y=1.8$ 3) $x=4\frac{1}{3}$, $y=1\frac{1}{2}$

Hints

- Multiply each length by the number of strips of that length. - Follow the order of operations. - A whole-number factor may cancel with a fraction's denominator. - Convert mixed numbers to improper fractions when helpful.

Solution

1. The total length is $3x+2y$. 2. For $x=5$ and $y=4$, $3\times 5+2\times 4=23$ inches. 3. For $x=2.5$ and $y=1.8$, $3\times 2.5+2\times 1.8=7.5+3.6=11.1$ inches. 4. For $x=4\frac{1}{3}=\frac{13}{3}$ and $y=1\frac{1}{2}=\frac{3}{2}$, $3\times\frac{13}{3}+2\times\frac{3}{2}=13+3=16$ inches.

Answer

Expression: $3x+2y$ 1) $23$ inches 2) $11.1$ inches 3) $16$ inches

5222486

At a school snack stand, juice drinks sell for $\$1.50$ and pretzels sell for $\$2.20$. a) Write an expression for the total revenue from selling $s$ juice drinks and $b$ pretzels. b) Find the revenue from selling $80$ juice drinks and $55$ pretzels. c) Suppose the prices change to $\$x$ for a juice drink and $\$y$ for a pretzel. Write an expression for the revenue from the same quantities sold in part b).

Hints

- Multiply each item's price by the number sold. - Add the revenue from the two types of items. - Match each quantity with the correct price. - In part c), the quantities stay fixed while the prices become variables.

Solution

1. The revenue expression is $1.50s+2.20b$. 2. Substitute $s=80$ and $b=55$: $1.50\times 80+2.20\times 55=120+121=241$. 3. With the quantities fixed at $80$ and $55$, the new revenue expression is $80x+55y$.

Answer

a) $1.50s+2.20b$ b) $\$241$ c) $80x+55y$ dollars

5222556

A hiking trail has a total length of $L$ miles. A group has already hiked $d$ miles along the trail. a) Write an expression for the distance remaining. b) Find the remaining distance when $L=18.5$ and $d=7.9$. c) In this context, why can $d$ not be greater than $L$?

Hints

- Use subtraction to find an amount remaining. - Think of removing a known length from a total length. - Interpret what a negative result would mean in this context.

Solution

1. Subtract the distance already traveled from the total length: $L-d$. 2. Substitute the values: $18.5-7.9=10.6$ miles. 3. The distance traveled is part of the trail's total length. If $d>L$, the expression $L-d$ would be negative, which cannot represent a remaining distance on this trail.

Answer

a) $L-d$ b) $10.6$ miles c) The distance already hiked cannot exceed the trail's total length; otherwise, the remaining distance would be negative.

5222646

Consider the expressions $T_1=4x-2$ and $T_2=4(x-2)$. a) Evaluate both expressions when $x=5$. b) Describe $T_2$ in words and explain why the two values are different.

Hints

- Substitute $5$ for $x$ and follow the order of operations. - What operation must be completed first inside parentheses? - Compare the first operation performed in each expression.

Solution

1. For $T_1$, substitute $x=5$: $4\times5-2=20-2=18$. 2. For $T_2$, substitute $x=5$: $4\times(5-2)=4\times3=12$. 3. The expression $T_2$ means subtract $2$ from $x$, then multiply the result by $4$. 4. The values differ because the parentheses make the subtraction happen before multiplication in $T_2$. In $T_1$, multiplication happens before subtraction.

Answer

a) $T_1=18$ and $T_2=12$ b) For $T_2$, subtract $2$ from $x$, then multiply by $4$. The parentheses change the order of operations.

5223436

Evaluate $T(x,y,z)=4xz-5y^2+12$ when $x=2.5$, $y=3$, and $z=6$.

Hints

- Substitute each value carefully. - Evaluate the exponent before multiplying. - Add and subtract after finding the two products.

Solution

1. Substitute the values: $4\times 2.5\times 6-5\times 3^2+12$. 2. The first product is $4\times 2.5\times 6=60$. 3. The second product is $5\times 3^2=5\times 9=45$. 4. Therefore, $60-45+12=27$.

Answer

$27$

5224136

A teacher uses $G=B+np$ to estimate the total cost $G$ of a class trip. Here, $B$ is the bus cost, $n$ is the number of students, and $p$ is the lodging cost per student. Find $G$ for each case. 1) $B=450$, $n=25$, $p=120$ 2) $B=520$, $n=22$, $p=145$ 3) $B=380$, $n=28$, $p=95$

Hints

- Substitute each value for its matching variable. - Multiply $n$ and $p$ before adding $B$. - Interpret the result as a total cost.

Solution

1. $G=450+25\times 120=450+3000=3450$. 2. $G=520+22\times 145=520+3190=3710$. 3. $G=380+28\times 95=380+2660=3040$.

Answer

1) $\$3450$ 2) $\$3710$ 3) $\$3040$

5224146

Evaluate $T=5a-2b$ for each pair. Pay close attention to signs. 1) $a=8$, $b=12$ 2) $a=-4$, $b=5$ 3) $a=2.5$, $b=-3$

Hints

- Use parentheses around negative values. - Multiply before subtracting. - Subtracting a negative quantity is equivalent to adding.

Solution

1. $T=5\times 8-2\times 12=40-24=16$. 2. $T=5\times(-4)-2\times 5=-20-10=-30$. 3. $T=5\times 2.5-2\times(-3)=12.5+6=18.5$.

Answer

1) $16$ 2) $-30$ 3) $18.5$

5224236

A beverage supplier delivers cases of bottled water to a school. Each case has $n$ rows with $f$ bottles in each row. The supplier delivers $k$ cases. a) Write an expression for the total number of bottles $G$. b) Evaluate $G$ for each set of values: i. $k=12$, $n=3$, $f=4$ ii. $k=25$, $n=4$, $f=5$ iii. $k=40$, $n=2$, $f=10$

Hints

- First find the number of bottles in one case. - Multiply that amount by the number of cases. - All three variables are connected by multiplication.

Solution

1. One case contains $nf$ bottles. With $k$ cases, $G=knf$. 2. For b-i, $12\times 3\times 4=144$. 3. For b-ii, $25\times 4\times 5=500$. 4. For b-iii, $40\times 2\times 10=800$.

Answer

a) $G=knf$ b) i. $144$ bottles; ii. $500$ bottles; iii. $800$ bottles

5226396

Evaluate $x = a + b + c$ for each set of values. Use convenient groupings when possible. a) $a = -3.8$, $b = 5\frac{1}{2}$, $c = -1.2$ b) $a = \frac{3}{5}$, $b = -1.4$, $c = 2$ c) $a = -12$, $b = 7\frac{3}{4}$, $c = -2\frac{1}{4}$

Hints

- Can you reorder or group the numbers to make the arithmetic easier? - Decide whether fractions or decimals are more convenient in each part. - Pay close attention to signs. - Look for numbers that combine to make a whole number.

Solution

1. For part a, group the decimals: $(-3.8) + (-1.2) = -5$. Then $-5 + 5.5 = 0.5$. 2. For part b, $\frac{3}{5} = 0.6$. Then $0.6 - 1.4 + 2 = -0.8 + 2 = 1.2$. 3. For part c, combine the mixed numbers first: $7\frac{3}{4} - 2\frac{1}{4} = 5\frac{1}{2}$. Then $-12 + 5.5 = -6.5$.

Answer

a) $x = 0.5$ b) $x = 1.2$ c) $x = -6.5$

5229596

Test whether $x-(y-z)=x-y+z$ is true for each set of values. Complete the table by evaluating both sides. <table> <tr><td>$x$</td><td>$20$</td><td>$4$</td><td>$0.7$</td><td>$\frac{5}{6}$</td></tr> <tr><td>$y$</td><td>$8$</td><td>$9$</td><td>$1.2$</td><td>$\frac{1}{3}$</td></tr> <tr><td>$z$</td><td>$5$</td><td>$2$</td><td>$0.5$</td><td>$\frac{1}{2}$</td></tr> <tr><td>Left side</td><td></td><td></td><td></td><td></td></tr> <tr><td>Right side</td><td></td><td></td><td></td><td></td></tr> </table>

Hints

- Evaluate the expression inside parentheses first. - Substitute the same values into both sides. - Use common denominators for the fractional values.

Solution

1. For $(x,y,z)=(20,8,5)$, the left side is $20-(8-5)=17$, and the right side is $20-8+5=17$. 2. For $(x,y,z)=(4,9,2)$, the left side is $4-(9-2)=-3$, and the right side is $4-9+2=-3$. 3. For $(x,y,z)=(0.7,1.2,0.5)$, the left side is $0.7-(1.2-0.5)=0$, and the right side is $0.7-1.2+0.5=0$. 4. For $(x,y,z)=\left(\frac{5}{6},\frac{1}{3},\frac{1}{2}\right)$, both sides equal $1$. 5. The two sides are equal for every listed set of values.

Answer

<table> <tr><td>$x$</td><td>$20$</td><td>$4$</td><td>$0.7$</td><td>$\frac{5}{6}$</td></tr> <tr><td>$y$</td><td>$8$</td><td>$9$</td><td>$1.2$</td><td>$\frac{1}{3}$</td></tr> <tr><td>$z$</td><td>$5$</td><td>$2$</td><td>$0.5$</td><td>$\frac{1}{2}$</td></tr> <tr><td>Left side</td><td>$17$</td><td>$-3$</td><td>$0$</td><td>$1$</td></tr> <tr><td>Right side</td><td>$17$</td><td>$-3$</td><td>$0$</td><td>$1$</td></tr> </table> The equation is true for all four sets of values.

5238176

Evaluate each expression for the given value. If an expression is undefined, state that and briefly explain why. a) $\frac{2x - 10}{x + 5}$ for $x = 5$ b) $\frac{0}{y^2 + 4}$ for $y = 3$ c) $\frac{z - 3}{z^2 - 9}$ for $z = 3$

Hints

- A fraction with numerator zero equals zero only when its denominator is nonzero. - Check the denominator after substituting. - Division by zero is undefined. - How is a zero numerator different from a zero denominator?

Solution

1. For part a, substitute $x = 5$: $\frac{2 \times 5 - 10}{5 + 5} = \frac{0}{10} = 0$. 2. For part b, substitute $y = 3$: $\frac{0}{3^2 + 4} = \frac{0}{13} = 0$. 3. For part c, substitute $z = 3$: the numerator is $3 - 3 = 0$, and the denominator is $3^2 - 9 = 0$. The result would require division by zero, so the expression is undefined.

Answer

a) $0$. b) $0$. c) Undefined because the denominator equals $0$.

5238436

For a school event, $n$ cases of drinks are purchased. Each case contains $k$ bottles, and each bottle holds $v$ fluid ounces. a) Write an expression for the total volume $V$, in fluid ounces. b) The drinks are poured into cups that each hold $8$ fluid ounces. Write an expression for the number of cups $B$ that can be filled. Assume $nkv$ is divisible by $8$. c) Find $V$ and $B$ when $n=20$, $k=12$, and $v=20$.

Hints

- Multiply to find the total number of bottles. - Multiply the number of bottles by the volume in each bottle. - Divide the total volume by $8$ to find the number of cups.

Solution

1. There are $nk$ bottles, so the total volume is $V=nkv$ fluid ounces. 2. The number of $8$-fluid-ounce cups is $B=\frac{nkv}{8}$. 3. For the given values, $V=20\times 12\times 20=4800$ fluid ounces. 4. Then $B=4800\div 8=600$ cups.

Answer

a) $V=nkv$ fluid ounces b) $B=\frac{nkv}{8}$ cups c) $V=4800$ fluid ounces and $B=600$ cups

5239336

A container originally holds $x$ gallons of juice. After $y$ gallons are used, the remaining juice is divided equally among $n$ days. a) Write an expression for the average number of cups of juice available each day. Use $1$ gallon $=16$ cups. b) Find the daily amount when $x=8$, $y=2$, and $n=24$.

Hints

- Subtract to find the amount of juice remaining. - Convert gallons to cups. - Divide the number of cups by the number of days.

Solution

1. The remaining amount is $x-y$ gallons. 2. Converting to cups gives $16(x-y)$ cups. 3. Dividing equally among $n$ days gives $\frac{16(x-y)}{n}$ cups per day. 4. For the given values, $\frac{16\times(8-2)}{24}=\frac{96}{24}=4$ cups per day.

Answer

a) $\frac{16(x-y)}{n}$ cups per day b) $4$ cups per day

5239436

A juice company mixes $3$ liters of juice that is $100\%$ fruit juice with $2$ liters of a drink that is $40\%$ fruit juice. Use the formula $G=\frac{V_1p_1+V_2p_2}{V_1+V_2}$ to find the percent of fruit juice in the mixture.

Hints

- Match each quantity with its variable. - Evaluate the numerator and denominator separately. - Divide the total fruit-juice amount by the total volume.

Solution

1. Substitute $V_1=3$, $p_1=100$, $V_2=2$, and $p_2=40$. 2. The numerator is $3\times 100+2\times 40=380$. 3. The total volume is $3+2=5$ liters. 4. Therefore, $G=\frac{380}{5}=76$.

Answer

$76\%$

5279216

A school library has $x$ books at the beginning of a month. During the month, it adds $y$ new books and removes $z$ damaged books. a) Write an expression for the number of books at the end of the month. b) Evaluate the expression for each set of values: 1) $x=1250$, $y=48$, $z=15$ 2) $x=945$, $y=112$, $z=37$

Hints

- Identify which change increases the number of books and which decreases it. - Begin with the starting number. - Substitute each value for its matching variable.

Solution

1. Add the new books and subtract the removed books: $x+y-z$. 2. For the first set, $1250+48-15=1298-15=1283$. 3. For the second set, $945+112-37=1057-37=1020$.

Answer

a) $x+y-z$ b) 1) $1283$ books 2) $1020$ books

5279236

An online-store gift card has a value of $\$G$. A game costs $\$p$. a) Write an expression for the remaining balance $R$. b) Find $R$ when $G=25.00$ and $p=12.99$. c) What does $p=G$ mean for the gift-card balance? d) Can $p$ be greater than $G$ when the purchase must be paid for only with this card? Explain.

Hints

- Subtract the purchase price from the starting balance. - Consider what happens when the two amounts are equal. - A prepaid card cannot have a negative balance.

Solution

1. Subtract the game price from the card value: $R=G-p$. 2. For the given values, $R=25.00-12.99=12.01$, so the balance is $\$12.01$. 3. If $p=G$, then $R=0$; the purchase uses the entire balance. 4. If the card cannot be combined with another payment method or overdrawn, then $p\le G$.

Answer

a) $R=G-p$ b) $\$12.01$ c) The balance is $\$0.00$. d) No. Paying only with the card requires $p\le G$.

5114306

A unit fraction can be split into two smaller unit fractions using $\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(n+1)}$. For example, when $n=4$, $\frac{1}{4}=\frac{1}{5}+\frac{1}{20}$. a) Use the rule with $n=5$ to split $\frac{1}{5}$ into two different unit fractions. b) Replace one $\frac{1}{5}$ in $\frac{2}{5}=\frac{1}{5}+\frac{1}{5}$ with your result from part a). Write $\frac{2}{5}$ as a sum of three different unit fractions.

Hints

- Substitute $5$ for $n$ in both expressions in the rule. - Replace one term in a sum with an equivalent sum. - Check that all three unit fractions are different.

Solution

1. Substitute $n=5$. Then $n+1=6$ and $n(n+1)=5\times6=30$, so $\frac{1}{5}=\frac{1}{6}+\frac{1}{30}$. 2. Replace one copy of $\frac{1}{5}$: $\frac{2}{5}=\frac{1}{5}+\frac{1}{6}+\frac{1}{30}$.

Answer

a) $\frac{1}{5}=\frac{1}{6}+\frac{1}{30}$ b) $\frac{2}{5}=\frac{1}{5}+\frac{1}{6}+\frac{1}{30}$

5114446

The following decompositions show a pattern for fractions with numerator $2$ and an odd denominator: $\frac{2}{3}=\frac{1}{2}+\frac{1}{6}$ $\frac{2}{5}=\frac{1}{3}+\frac{1}{15}$ $\frac{2}{7}=\frac{1}{4}+\frac{1}{28}$ a) Describe the pattern for the denominator of the first unit fraction. b) Use the pattern to decompose $\frac{2}{9}$ and $\frac{2}{13}$ into two unit fractions.

Hints

- Compare $3$ with $2$, $5$ with $3$, and $7$ with $4$. - Use the pattern to find the first unit fraction. - Subtract the first unit fraction from the original fraction to find the second.

Solution

1. If the original odd denominator is $n$, the first denominator is $\frac{n+1}{2}$. 2. For $\frac{2}{9}$, the first denominator is $\frac{9+1}{2}=5$. The remainder is $\frac{2}{9}-\frac{1}{5}=\frac{10}{45}-\frac{9}{45}=\frac{1}{45}$. Thus, $\frac{2}{9}=\frac{1}{5}+\frac{1}{45}$. 3. For $\frac{2}{13}$, the first denominator is $\frac{13+1}{2}=7$. The remainder is $\frac{2}{13}-\frac{1}{7}=\frac{14}{91}-\frac{13}{91}=\frac{1}{91}$. Thus, $\frac{2}{13}=\frac{1}{7}+\frac{1}{91}$.

Answer

a) Add $1$ to the original denominator and divide by $2$. b) $\frac{2}{9}=\frac{1}{5}+\frac{1}{45}$ and $\frac{2}{13}=\frac{1}{7}+\frac{1}{91}$.

5116986

For a fraction $\frac{a}{b}$, where $a$ and $b$ are positive whole numbers, describe how the value changes in each case. Give a numerical example. a) The numerator is multiplied by $4$, while the denominator stays the same. b) The denominator is multiplied by $4$, while the numerator stays the same. c) The numerator and denominator are both divided by $5$, assuming both are divisible by $5$.

Hints

- Interpret a fraction as division. - Change only one part at a time in parts a) and b). - Dividing the numerator and denominator by the same nonzero number creates an equivalent fraction.

Solution

1. For a), $\frac{4a}{b}=4\times\frac{a}{b}$, so the value is multiplied by $4$. Example: $\frac{2}{3}$ becomes $\frac{8}{3}$. 2. For b), $\frac{a}{4b}=\frac{1}{4}\times\frac{a}{b}$, so the value is divided by $4$. Example: $\frac{2}{3}$ becomes $\frac{2}{12}=\frac{1}{6}$. 3. For c), dividing both parts by $5$ creates an equivalent fraction, so the value stays the same. Example: $\frac{10}{15}=\frac{2}{3}$.

Answer

a) The value is multiplied by $4$. b) The value is divided by $4$. c) The value stays the same.

5124886

A composite figure is made from a square with side length $x$ and an adjacent rectangle. The rectangle shares one side of length $x$ with the square and has length $8\,\text{cm}$. a) Write and simplify an expression for the perimeter $P$ of the composite figure. b) Find the perimeter when $x=5\,\text{cm}$ and when $x=2.4\,\text{cm}$.

Hints

- Trace only the outside boundary of the composite figure. - Do not include the shared side in the perimeter. - Combine equal variable-length segments before evaluating.

Solution

1. The shared side is inside the composite figure, so it is not part of the perimeter. 2. The outside boundary has four segments of length $x$ and two segments of length $8\,\text{cm}$. Thus $P=4x+16$. 3. When $x=5$, $P=4\times5+16=36\,\text{cm}$. 4. When $x=2.4$, $P=4\times2.4+16=25.6\,\text{cm}$.

Answer

a) $P=4x+16$ b) $36\,\text{cm}$ and $25.6\,\text{cm}$

5124946

A rectangular lot has length $L$ meters. Its width is $10\,\text{m}$ less than its length. a) Write and simplify an expression for the perimeter $P$ using only $L$. b) Write an expression for the area $A$. c) Find the perimeter and area when $L=25\,\text{m}$.

Hints

- Express the width in terms of $L$. - Use the formulas for rectangle perimeter and area. - Keep the expression $L-10$ grouped when multiplying.

Solution

1. The width is $L-10$. 2. The perimeter is $P=2L+2(L-10)=4L-20$. 3. The area is $A=L(L-10)$. 4. When $L=25$, $P=4\times25-20=80\,\text{m}$, and $A=25\times(25-10)=375\,\text{m}^2$.

Answer

a) $P=4L-20$ b) $A=L(L-10)$ c) $P=80\,\text{m}$; $A=375\,\text{m}^2$

5133876

Consider two ways to write the product of a fraction and a whole number, where $y\neq0$: Expression $A$: $\frac{x}{y}\times z$ Expression $B$: $\frac{x\times z}{y}$ a) Evaluate both expressions for $x=5$, $y=6$, and $z=3$. b) Explain why the two expressions are equivalent for all allowed values of the variables.

Hints

- Substitute the given values into each expression separately. - A whole number can be written as a fraction with denominator $1$. - For part b), multiply the fractions symbolically instead of relying only on the numerical example.

Solution

1. For a), expression $A$ is $\frac{5}{6}\times3=\frac{15}{6}=\frac{5}{2}=2.5$. 2. Expression $B$ is $\frac{5\times3}{6}=\frac{15}{6}=\frac{5}{2}=2.5$. 3. For b), write $z=\frac{z}{1}$. Then $\frac{x}{y}\times\frac{z}{1}=\frac{xz}{y}$, which is exactly expression $B$. 4. Therefore, the expressions are equivalent whenever $y\neq0$.

Answer

a) Both expressions equal $\frac{5}{2}=2.5$. b) They are equivalent because $\frac{x}{y}\times z=\frac{xz}{y}$ for $y\neq0$.

5224246

A parking garage has $e$ levels. Each level has two sections: Section A has $p$ parking spaces, and Section B has $q$ parking spaces. a) Write a formula for the total number of parking spaces $Z$. b) Find $Z$ when $e=5$, $p=42$, and $q=58$. c) A renovation removes $4$ spaces from Section B on every level. Write a new formula $Z_{\text{new}}$, then evaluate it using the values from part b).

Hints

- First find the number of spaces on one level. - Use parentheses so the spaces on a level are added before multiplying by the number of levels. - Removing spaces changes the expression inside the parentheses.

Solution

1. Each level has $p+q$ spaces, so $Z=e(p+q)$. 2. Substitute the values: $Z=5\times(42+58)=5\times 100=500$. 3. After the renovation, Section B has $q-4$ spaces per level. Thus, $Z_{\text{new}}=e(p+q-4)$. 4. Using the given values, $Z_{\text{new}}=5\times(42+58-4)=5\times 96=480$.

Answer

a) $Z=e(p+q)$ b) $500$ c) $Z_{\text{new}}=e(p+q-4)$; $480$ spaces

5230916

Check whether $(x+4)(y-3)=xy-3x+4y-12$ for each pair. Calculate the left side and right side separately. a) $x=6$, $y=5$ b) $x=2$, $y=3$ c) $x=0$, $y=10$

Hints

- Substitute each pair into both sides. - Evaluate the two sides separately. - Follow the order of operations. - Compare the two numerical results in each case.

Solution

1. For a), the left side is $(6+4)\times(5-3)=20$, and the right side is $6\times 5-3\times 6+4\times 5-12=20$. 2. For b), the left side is $(2+4)\times(3-3)=0$, and the right side is $2\times 3-3\times 2+4\times 3-12=0$. 3. For c), the left side is $(0+4)\times(10-3)=28$, and the right side is $0\times 10-3\times 0+4\times 10-12=28$. 4. The two sides match in all three cases.

Answer

a) $20=20$ b) $0=0$ c) $28=28$ The equality holds for all three tested pairs.

5230926

Check whether $(a+b)(a-b)=a^2-b^2$ for each pair. Show the value of each side. a) $a=7$, $b=3$ b) $a=1.5$, $b=0.5$ c) $a=12$, $b=12$

Hints

- Evaluate the parentheses before multiplying. - Square each value separately on the right side. - Compare the two numerical results for each pair.

Solution

1. For part a), $(7+3)\times(7-3)=40$ and $7^2-3^2=40$. 2. For part b), $(1.5+0.5)\times(1.5-0.5)=2$ and $(1.5)^2-(0.5)^2=2$. 3. For part c), $(12+12)\times(12-12)=0$ and $12^2-12^2=0$. 4. The two sides are equal for all three tested pairs.

Answer

a) $40=40$ b) $2=2$ c) $0=0$

5245286

Hikers use the estimate $t=\frac{s}{3}+\frac{h}{2000}$, where $t$ is hiking time in hours, $s$ is horizontal distance in miles, and $h$ is total elevation gain in feet. 1) Find the estimated time for a $9$-mile hike with $1200$ feet of elevation gain. 2) Trail A is $9$ miles long with $1800$ feet of elevation gain. Trail B is $7.5$ miles long with $3000$ feet of elevation gain. Which trail has the shorter estimated time? Show the calculations.

Hints

- Substitute both the distance and elevation gain. - Evaluate the two fractions before adding. - Compare the two estimated times.

Solution

1. For the first hike, $t=\frac{9}{3}+\frac{1200}{2000}=3+0.6=3.6$ hours. 2. For Trail A, $t_A=\frac{9}{3}+\frac{1800}{2000}=3+0.9=3.9$ hours. 3. For Trail B, $t_B=\frac{7.5}{3}+\frac{3000}{2000}=2.5+1.5=4.0$ hours. 4. Since $3.9<4.0$, Trail A has the shorter estimated time.

Answer

1) $3.6$ hours 2) Trail A; its estimated time is $3.9$ hours, compared with $4.0$ hours for Trail B.

5317216

**Electricity Cost** A utility plan charges $15$ cents for each kilowatt-hour ($\text{kWh}$) of electricity used, plus a fixed monthly service charge of $\$12.00$. The expression tree shows the steps used to find the monthly bill. a) Find the monthly bill for each amount of electricity use: 1) $400\,\text{kWh}$ 2) $800\,\text{kWh}$ 3) $1500\,\text{kWh}$ b) A monthly bill is exactly $\$102.00$. Work backward through the expression tree to find the electricity use. c) Write one algebraic expression for the monthly bill, using $x$ for the number of kilowatt-hours used.

Hints

- Follow the expression tree from the usage value toward the final bill. - For part b, reverse each operation in the opposite order. - In part c, replace the usage value with $x$.

Solution

1. The tree represents $(x \times 15) \div 100 + 12$, because the usage charge is calculated in cents and then converted to dollars. 2. a) For $400\,\text{kWh}$: $(400 \times 15) \div 100 + 12 = 60 + 12 = 72$, so the bill is $\$72.00$. For $800\,\text{kWh}$: $(800 \times 15) \div 100 + 12 = 120 + 12 = 132$, so the bill is $\$132.00$. For $1500\,\text{kWh}$: $(1500 \times 15) \div 100 + 12 = 225 + 12 = 237$, so the bill is $\$237.00$. 3. b) Work backward from $102$: $102 - 12 = 90$, $90 \times 100 = 9000$, and $9000 \div 15 = 600$. The usage was $600\,\text{kWh}$. 4. c) The expression is $(x \times 15) \div 100 + 12$.

Answer

a) 1) $\$72.00$ 2) $\$132.00$ 3) $\$237.00$ b) $600\,\text{kWh}$ c) $(x \times 15) \div 100 + 12$

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