Evaluate algebraic expressions 1
Find the value of \(x+9\) when \(x=4\).
Hints
- Find the variable in the expression.
- Replace it with the given number, then calculate.
Solution
1. Replace \(x\) with \(4\).
2. Add: \(4+9=13\).
Answer
\(13\)
Evaluate algebraic expressions
Learning goals
- Substitutes given values correctly.
- Applies grouping symbols and order of operations correctly.
- Computes the value of a numerical expression.
- Checks the result against the expression structure.
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Evaluate algebraic expressions 2
Evaluate \(12-2x\) when \(x=0\). Explain what happens to the term \(2x\).
Hints
- What happens when a factor of a product is zero?
- Perform the multiplication before subtracting.
Solution
1. Substitute \(0\) for \(x\): \(12-2\times0\).
2. The product \(2\times0\) is \(0\), leaving \(12\).
Answer
\(12\). The term \(2x\) becomes \(0\).
Evaluate algebraic expressions 3
Find \(3a+b\) when \(a=2\) and \(b=5\).
Hints
- Match each letter with its own given number.
- Multiply before adding.
Solution
1. Replace \(a\) with \(2\) and \(b\) with \(5\).
2. Evaluate \(3\times2+5=6+5=11\).
Answer
\(11\)
Evaluate algebraic expressions 4
Evaluate \(2(x+3)\) for \(x=4\). Which calculation must be done first?
Hints
- Substitute before changing the operation order.
- What do parentheses tell you to do first?
Solution
1. Substitute \(4\): \(2(4+3)\).
2. Add inside parentheses, then multiply: \(2\times7=14\).
Answer
\(14\). Add inside the parentheses first.
Evaluate algebraic expressions 5
Find the value of \(7+x\) when \(x=-2\).
Hints
- Keep the negative sign with the substituted number.
- What does adding a negative number do to the total?
Solution
1. Replace \(x\) with \(-2\).
2. Calculate \(7+(-2)=5\).
Answer
\(5\)
Evaluate algebraic expressions 6
Evaluate \(\frac{x}{4}\) when \(x=20\). What does the fraction bar tell you to do?
Hints
- Write the fraction using the given number in place of \(x\).
- Try the same calculation using a smaller value of \(x\). What operation gives the same result?
Solution
1. Replace \(x\) with \(20\).
2. The fraction bar means division, so \(20\div4=5\).
Answer
\(5\). Divide by \(4\).
Evaluate algebraic expressions 7
Evaluate \(3x+2\) for each value of \(x\).
a) \(x=1\)
b) \(x=4\)
c) \(x=10\)
Hints
- Replace \(x\) with the given value.
- Multiply before adding.
- Evaluate each part separately.
Solution
1. For \(x=1\), substitute and evaluate: \(3\times1+2=5\).
2. For \(x=4\), substitute and evaluate: \(3\times4+2=14\).
3. For \(x=10\), substitute and evaluate: \(3\times10+2=32\).
Answer
a) \(5\)
b) \(14\)
c) \(32\)
Evaluate algebraic expressions 8
Evaluate \(\frac{a}{b}\) for each pair of values.
a) \(a=8\), \(b=2\)
b) \(a=3\), \(b=4\)
c) \(a=0\), \(b=5\)
Hints
- Replace \(a\) and \(b\) with the given values.
- A fraction bar represents division.
- Zero divided by a nonzero number equals zero.
Solution
1. For \(a=8\) and \(b=2\), \(\frac{a}{b}=\frac{8}{2}=4\).
2. For \(a=3\) and \(b=4\), \(\frac{a}{b}=\frac{3}{4}\).
3. For \(a=0\) and \(b=5\), \(\frac{a}{b}=\frac{0}{5}=0\).
Answer
a) \(4\)
b) \(\frac{3}{4}\)
c) \(0\)
Evaluate algebraic expressions 9
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) \(\$8.95\)
Evaluate algebraic expressions 10
Let \(a=1.2\) and \(b=0.5\). Evaluate each expression.
a) \(a+b\)
b) \(a-b\)
c) \(2ab\)
d) \(b^2\)
e) \(a^2-b\)
Hints
- Substitute the values before calculating.
- Use parentheses when replacing a variable in a power.
- Follow the order of operations.
- Keep decimal places aligned when adding or subtracting.
Solution
1. \(a+b=1.2+0.5=1.7\).
2. \(a-b=1.2-0.5=0.7\).
3. \(2ab=2\times1.2\times0.5=1.2\).
4. \(b^2=(0.5)^2=0.25\).
5. \(a^2-b=(1.2)^2-0.5=1.44-0.5=0.94\).
Answer
a) \(1.7\)
b) \(0.7\)
c) \(1.2\)
d) \(0.25\)
e) \(0.94\)
Evaluate algebraic expressions 11
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 \(z=7\)
Evaluate algebraic expressions 12
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 \(b=4\)
Evaluate algebraic expressions 13
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) \(32\)
Evaluate algebraic expressions 14
Substitute \(b=0\), \(b=4\), \(b=10\), and \(b=2.5\) into \(3b+14\). Find each value.
Hints
- Replace \(b\) with each given number.
- Multiply before adding.
- Keep decimal places aligned when adding.
Solution
1. For \(b=0\), \(3\times0+14=14\).
2. For \(b=4\), \(3\times4+14=26\).
3. For \(b=10\), \(3\times10+14=44\).
4. For \(b=2.5\), \(3\times2.5+14=21.5\).
Answer
For \(b=0\), the value is \(14\).
For \(b=4\), the value is \(26\).
For \(b=10\), the value is \(44\).
For \(b=2.5\), the value is \(21.5\).
Evaluate algebraic expressions 15
Evaluate \(25-4a\) for each value.
a) \(a=0\)
b) \(a=5\)
c) \(a=7\)
d) \(a=2.5\)
Hints
- Substitute each value for \(a\).
- Multiply before subtracting.
- Evaluate each part separately.
Solution
1. For \(a=0\), \(25-4\times0=25\).
2. For \(a=5\), \(25-4\times5=5\).
3. For \(a=7\), \(25-4\times7=-3\).
4. For \(a=2.5\), \(25-4\times2.5=25-10=15\).
Answer
a) \(25\)
b) \(5\)
c) \(-3\)
d) \(15\)
Evaluate algebraic expressions 16
Evaluate \(2x-3y\) 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.
- Multiply before subtracting.
- Evaluate the two products separately before finding their difference.
Solution
1. For \(x=5\) and \(y=2\), \(2\times5-3\times2=10-6=4\).
2. For \(x=3\) and \(y=1\), \(2\times3-3\times1=6-3=3\).
3. For \(x=2\) and \(y=4\), \(2\times2-3\times4=4-12=-8\).
Answer
a) \(4\)
b) \(3\)
c) \(-8\)
Evaluate algebraic expressions 17
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)
Evaluate algebraic expressions 18
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 \(37\,\text{cm}\)
Evaluate algebraic expressions 19
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
Evaluate algebraic expressions 20
Evaluate each expression when \(x=3\).
a) \(3x+10\)
b) \(x^2-5\)
c) \(2x^2+x\)
d) \(\frac{1}{3}x^3\)
Hints
- Substitute \(3\) for every occurrence of \(x\).
- Evaluate exponents before multiplication, addition, or subtraction.
- Work through one expression at a time.
Solution
1. \(3\times3+10=9+10=19\).
2. \(3^2-5=9-5=4\).
3. \(2\times3^2+3=18+3=21\).
4. \(\frac{1}{3}\times3^3=\frac{1}{3}\times27=9\).
Answer
a) \(19\)
b) \(4\)
c) \(21\)
d) \(9\)
Evaluate algebraic expressions 21
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) \(\$3040\)
Evaluate algebraic expressions 22
Evaluate \(5a-2b\) for each pair.
1) \(a=8\), \(b=12\)
2) \(a=4\), \(b=5\)
3) \(a=2.5\), \(b=3\)
Hints
- Substitute both values before calculating.
- Find the two products first.
- Subtract the second product from the first.
Solution
1. \(5\times8-2\times12=40-24=16\).
2. \(5\times4-2\times5=20-10=10\).
3. \(5\times2.5-2\times3=12.5-6=6.5\).
Answer
1) \(16\)
2) \(10\)
3) \(6.5\)
Evaluate algebraic expressions 23
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
Evaluate algebraic expressions 24
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
- Look for two values that combine to make a whole number.
- Decide whether fractions or decimals are more convenient in each part.
- Reordering addends does not change their sum.
Solution
1. For part a, group the decimals: \(3.8+1.2=5\). Then \(5+5.5=10.5\).
2. For part b, \(\frac{3}{5}=0.6\). Then \(0.6+1.4+2=2+2=4\).
3. For part c, combine the mixed numbers first: \(7\frac{3}{4}+2\frac{1}{4}=10\). Then \(12+10=22\).
Answer
a) \(x=10.5\)
b) \(x=4\)
c) \(x=22\)
Evaluate algebraic expressions 25
Evaluate each expression for the given value.
a) \(\frac{2x-10}{5}\) for \(x=5\)
b) \(\frac{0}{y^2+4}\) for \(y=3\)
c) \(\frac{z+3}{6}\) for \(z=3\)
Hints
- Substitute the value before simplifying the fraction.
- A fraction with numerator zero and a nonzero denominator equals zero.
- Evaluate exponents before addition.
Solution
1. For part a, substitute \(x=5\): \(\frac{2\times5-10}{5}=\frac{0}{5}=0\).
2. For part b, substitute \(y=3\): \(\frac{0}{3^2+4}=\frac{0}{13}=0\).
3. For part c, substitute \(z=3\): \(\frac{3+3}{6}=\frac{6}{6}=1\).
Answer
a) \(0\)
b) \(0\)
c) \(1\)
Evaluate algebraic expressions 26
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
Evaluate algebraic expressions 27
A kayak rental uses the expression \(C=0.75m+1.20\) for a short practice session, where \(m\) is the number of minutes and \(C\) is the cost in dollars. Evaluate \(C\) when \(m=8.4\).
Hints
- Replace the variable with the given session length before calculating.
- Follow the multiplication before the addition and keep the money units.
Solution
1. Substitute \(m=8.4\): \(C=0.75\times 8.4+1.20\).
2. Multiply: \(0.75\times8.4=6.30\).
3. Add the fixed amount: \(C=6.30+1.20=7.50\).
Answer
\(C=\$7.50\)
Evaluate algebraic expressions 28
Eva evaluates \(3x+2\) at \(x=4\) and gets \(18\) by calculating \(3(4+2)\). Explain her mistake and give the correct value.
Hints
- Compare the grouping in Eva's work with the original expression.
- Decide which term has the coefficient \(3\).
Solution
1. Eva added parentheses that are not in the given expression, so she multiplied the sum instead of only \(x\).
2. The correct substitution is \(3\times4+2=12+2=14\).
Answer
Eva multiplied \(3\) by the whole sum. The correct value is \(14\).
Evaluate algebraic expressions 29
Use the calculation tree with \(x=5\). What is the value of the addition branch, and what is the final value?
Hints
- Start with the operation that contains the variable.
- Use its result in the operation that comes next.
Solution
1. The addition branch has value \(5+3=8\).
2. The outer multiplication gives \(2\times8=16\).
Answer
Addition branch: \(8\). Final value: \(16\).
Evaluate algebraic expressions 30
What is the value of \(2x^2-(x-2)^2\) when \(x=5\)?
a) \(21\)
b) \(29\)
c) \(41\)
d) \(91\)
Hints
- Substitute the given number for the variable.
- Evaluate parentheses and exponents before multiplication and subtraction.
- Square the entire value inside the parentheses.
Solution
1. Substitute \(5\) for \(x\): \(2\times5^2-(5-2)^2\).
2. Evaluate inside the parentheses: \((5-2)^2=3^2=9\).
3. Evaluate the first term: \(2\times 5^2=2\times 25=50\).
4. Subtract: \(50-9=41\).
Answer
c) \(41\)
Evaluate algebraic expressions 31
Compare these two expressions:
\(A=\frac{1}{n}\) and \(B=\frac{1}{n+1}+\frac{1}{n(n+1)}\).
a) Find \(A\) and \(B\) separately when \(n=2\). Show the two fractions you add to find \(B\).
b) Repeat for \(n=5\).
c) Do the expressions have the same value in both cases? Use your calculations to explain.
Hints
- Replace every occurrence of \(n\) with the same given number.
- Evaluate the denominators before adding the fractions in \(B\).
- Compare the final values only after finding each expression separately.
Solution
1. For \(n=2\), \(A=\frac{1}{2}\), while \(B=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}\).
2. For \(n=5\), \(A=\frac{1}{5}\), while \(B=\frac{1}{6}+\frac{1}{30}=\frac{1}{5}\).
3. The separately evaluated expressions agree for both given values of \(n\). This checks these two cases; it does not by itself prove the expressions agree for every \(n\).
Answer
a) \(A=\frac{1}{2}\); \(B=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}\).
b) \(A=\frac{1}{5}\); \(B=\frac{1}{6}+\frac{1}{30}=\frac{1}{5}\).
c) Yes. Each pair has the same value for the two values tested.
Evaluate algebraic expressions 32
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
- Substitute one value of \(x\) at a time.
- Evaluate the exponent before addition.
- Follow the order of operations.
- Line up decimal places when adding or subtracting.
Solution
1. For \(x=8\): \(2\times8-5=11\), \(8^2+10=74\), and \(12-3\times8=-12\).
2. For \(x=4\): \(2\times4-5=3\), \(4^2+10=26\), and \(12-3\times4=0\).
3. For \(x=0.5\): \(2\times0.5-5=-4\), \((0.5)^2+10=10.25\), and \(12-3\times0.5=10.5\).
Answer
<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>\(11\)</td> <td>\(74\)</td> <td>\(-12\)</td> </tr> <tr> <td>\(4\)</td> <td>\(3\)</td> <td>\(26\)</td> <td>\(0\)</td> </tr> <tr> <td>\(0.5\)</td> <td>\(-4\)</td> <td>\(10.25\)</td> <td>\(10.5\)</td> </tr> </tbody> </table>
Evaluate algebraic expressions 33
Evaluate \(3x^2+2y\) 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. For \(x=4\) and \(y=5\), \(3\times4^2+2\times5=48+10=58\).
2. For \(x=2\) and \(y=12\), \(3\times2^2+2\times12=12+24=36\).
3. For \(x=0.5\) and \(y=2\), \(3\times(0.5)^2+2\times2=0.75+4=4.75\).
Answer
a) \(58\)
b) \(36\)
c) \(4.75\)
Evaluate algebraic expressions 34
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.
- Check which of the two evaluated amounts is larger before finding their difference.
Solution
1. \(T_1=12-2\times2=8\).
2. \(T_2=\frac{2+10}{3}=4\).
3. Therefore, \(T_1-T_2=8-4=4\).
Answer
\(T_1=8\), \(T_2=4\), and \(T_1-T_2=4\)
Evaluate algebraic expressions 35
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 \(10\).
Evaluate algebraic expressions 36
Two students evaluate \(\frac{xz}{y}\) for \(x=5\), \(y=6\), and \(z=3\).
Student A first calculates \(\frac{x}{y}\times z\).
Student B first calculates \(\frac{x\times z}{y}\).
a) Use each method to find the value.
b) Explain why the two methods give the same result.
Hints
- Substitute the three values before calculating.
- Keep the fraction value exact until the last step.
- Compare the multiplication and division performed in the two methods.
Solution
1. Student A calculates \(\frac{5}{6}\times3=\frac{15}{6}=\frac{5}{2}=2.5\).
2. Student B calculates \(\frac{5\times3}{6}=\frac{15}{6}=\frac{5}{2}=2.5\).
3. Both methods multiply \(x\) and \(z\) and divide the product by \(y\); they only perform the operations in a different order.
Answer
a) Both methods give \(\frac{5}{2}=2.5\).
b) Both methods multiply \(x\) and \(z\) and divide by \(y\), so they evaluate the same expression.
Evaluate algebraic expressions 37
Aino 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 \(y=4\), \(3\times5+2\times4=23\), so the total is \(23\,\text{in.}\).
3. For \(x=2.5\) and \(y=1.8\), \(3\times2.5+2\times1.8=7.5+3.6=11.1\), so the total is \(11.1\,\text{in.}\).
4. For \(x=4\frac{1}{3}=\frac{13}{3}\) and \(y=1\frac{1}{2}=\frac{3}{2}\), \(3\times\frac{13}{3}+2\times\frac{3}{2}=13+3=16\), so the total is \(16\,\text{in.}\).
Answer
Expression: \(3x+2y\)
1) \(23\,\text{in.}\)
2) \(11.1\,\text{in.}\)
3) \(16\,\text{in.}\)
Evaluate algebraic expressions 38
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
Evaluate algebraic expressions 39
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.
Evaluate algebraic expressions 40
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.
Evaluate algebraic expressions 41
Evaluate \(4xz-5y^2+12\) 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: \(4\times2.5\times6-5\times3^2+12\).
2. The first product is \(4\times2.5\times6=60\).
3. The second product is \(5\times3^2=5\times9=45\).
4. Therefore, \(60-45+12=27\).
Answer
\(27\)
Evaluate algebraic expressions 42
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
Evaluate algebraic expressions 43
Evaluate each expression for the given values.
1) \(2a^2-3(a-b)\) when \(a=2\) and \(b=1.5\)
2) \(\frac{x+y}{xy}\) when \(x=\frac{1}{2}\) and \(y=\frac{1}{4}\)
Hints
- Substitute each given value before simplifying.
- Evaluate exponents and parentheses before multiplication or subtraction.
- For the fraction, simplify the numerator and denominator separately.
- How can you divide by a fraction?
Solution
1. \(2\times2^2-3(2-1.5)=8-3\times0.5=6.5\).
2. The numerator is \(\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\). The denominator is \(\frac{1}{2}\times\frac{1}{4}=\frac{1}{8}\).
3. Dividing gives \(\frac{3}{4}\div\frac{1}{8}=6\).
Answer
1) \(6.5\)
2) \(6\)
Evaluate algebraic expressions 44
Evaluate each expression for the given values.
1) \(s(t^2-0.5)+\frac{2}{s}\) when \(s=4\) and \(t=1.5\)
2) \(\frac{12}{x}-\frac{y}{0.2}\) when \(x=1.5\) and \(y=0.04\)
Hints
- Substitute the given values before evaluating.
- Evaluate powers and parentheses before multiplication.
- Simplify each quotient separately.
- Keep decimal place value aligned in the final subtraction.
Solution
1. Substitute: \(4(1.5^2-0.5)+\frac24=4(2.25-0.5)+0.5=4\times1.75+0.5=7.5\).
2. Substitute and simplify each quotient: \(\frac{12}{1.5}-\frac{0.04}{0.2}=8-0.2=7.8\).
Answer
1) \(7.5\)
2) \(7.8\)
Evaluate algebraic expressions 45
Evaluate each expression for the given values.
1) \(4(a+3)-(b-5)\times3\) when \(a=2\) and \(b=7\)
2) \(\frac{xy+8}{x-2}\) when \(x=4\) and \(y=3\)
Hints
- Substitute each value using parentheses where needed.
- Work from the innermost parentheses outward.
- Apply multiplication before the final subtraction.
- Simplify the numerator and denominator separately before dividing.
Solution
1. Substitute: \(4(2+3)-(7-5)\times3=4\times5-2\times3=20-6=14\).
2. Substitute: \(\frac{4\times3+8}{4-2}=\frac{20}{2}=10\).
Answer
1) \(14\)
2) \(10\)
Evaluate algebraic expressions 46
Evaluate each expression for the given values.
1) \(2x^2-3y\) when \(x=3\) and \(y=\frac{1}{2}\)
2) \(\frac{a+b}{a-b}\) when \(a=0.8\) and \(b=0.2\)
3) \(z^3+2z^2-5\) when \(z=2\)
Hints
- Substitute the values before applying the order of operations.
- Evaluate exponents before multiplication and subtraction.
- In the fraction, simplify the numerator and denominator separately.
- Keep fractional and decimal values exact until the final simplification.
Solution
1. \(2\times3^2-3\times\frac{1}{2}=18-1.5=16.5\).
2. \(\frac{0.8+0.2}{0.8-0.2}=\frac{1}{0.6}=\frac{5}{3}\).
3. \(2^3+2\times2^2-5=8+8-5=11\).
Answer
1) \(16.5\)
2) \(\frac{5}{3}\)
3) \(11\)
Evaluate algebraic expressions 47
Evaluate each expression for the given values.
1) \(4(p-q)^2\) when \(p=2.5\) and \(q=0.5\)
2) \(\frac{x^2-1}{x+1}\) when \(x=3\)
3) \(0.5a^2b-ab^2\) when \(a=4\) and \(b=1\)
Hints
- Substitute before simplifying.
- Evaluate parentheses before exponents, and exponents before multiplication.
- Simplify the numerator and denominator separately in part 2.
- In part 3, evaluate each product term before subtracting.
Solution
1. \(4(2.5-0.5)^2=4\times2^2=16\).
2. \(\frac{3^2-1}{3+1}=\frac84=2\).
3. \(0.5\times4^2\times1-4\times1^2=8-4=4\).
Answer
1) \(16\)
2) \(2\)
3) \(4\)
Evaluate algebraic expressions 48
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.
Evaluate algebraic expressions 49
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.
Evaluate algebraic expressions 50
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) \(0=0\)
Evaluate algebraic expressions 51
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
Evaluate algebraic expressions 52
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
Evaluate algebraic expressions 53
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
\(76\%\)
Evaluate algebraic expressions 54
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.
Evaluate algebraic expressions 55
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 \(p\le G\).
Evaluate algebraic expressions 56
Evaluate \(2a^2-3b+ab\) for \(a=3\) and \(b=4\).
Hints
- Substitute both variable values everywhere they appear.
- Evaluate the exponent before the multiplications and additions.
Solution
1. Substitute: \(2\times 3^2-3\times 4+3\times 4\).
2. Evaluate the power and products: \(2\times9-12+12\).
3. Combine the results: \(18\).
Answer
\(18\)
Evaluate algebraic expressions 57
Jordan evaluates \(5-x^2\) for \(x=2\) and writes \((5-2)^2=9\). Identify the substitution error and find the correct value.
Hints
- Mark exactly which part of the original expression has the exponent.
- Evaluate the power before the subtraction.
Solution
1. The exponent applies only to \(x\), so substitution gives \(5-(2)^2\), not \((5-2)^2\).
2. Evaluate the power: \(2^2=4\).
3. Subtract: \(5-4=1\).
Answer
Jordan incorrectly squared the entire difference. The correct value is \(1\).
Evaluate algebraic expressions 58
Two game booths use the score expressions \(A=3n+8\) and \(B=5n-4\). Evaluate both expressions when \(n=7\), then state which score is greater and by how much.
Hints
- Evaluate each rule separately using the same input.
- After finding both values, compare them and calculate the difference.
Solution
1. \(A=3\times 7+8=29\).
2. \(B=5\times 7-4=31\).
3. Since \(31>29\), score \(B\) is greater by \(2\).
Answer
\(A=29\), \(B=31\). Score \(B\) is greater by \(2\).
Evaluate algebraic expressions 59
Evaluate \(R=\frac{m^2+n^2}{m+n}\) when \(m=3\) and \(n=1\). Give the result as a decimal.
Hints
- Evaluate the numerator and denominator as separate grouped expressions.
- Complete the division only after both grouped values are known.
Solution
1. Substitute: \(R=\frac{(3)^2+(1)^2}{3+1}\).
2. The numerator is \(10\), and the denominator is \(4\).
3. Divide: \(R=2.5\).
Answer
\(R=2.5\)
Evaluate algebraic expressions 60
The calculation tree gives a monthly electricity bill in dollars. The variable \(x\) is electricity used, in kilowatt-hours (\(\text{kWh}\)). The energy charge is in cents, and the final addition is a monthly fee.
a) Find the bill for 1) \(400\,\text{kWh}\), 2) \(800\,\text{kWh}\), and 3) \(1500\,\text{kWh}\).
b) A monthly bill is \(\$102.00\). Work backward through the tree to find the electricity use.
c) Write a single expression for the bill in dollars, using \(x\).
Hints
- Follow the branches from the electricity use toward the final bill.
- To work backward, undo the last operation first.
- Keep track of whether an amount is in cents or dollars.
Solution
1. The tree calculates \((15x)\div100+12\), converting cents to dollars before adding the monthly fee.
2. For \(x=400\), the bill is \(6000\div100+12=72\) dollars. For \(x=800\), it is \(12000\div100+12=132\) dollars. For \(x=1500\), it is \(22500\div100+12=237\) dollars.
3. Work backward from \(102\): subtract \(12\), multiply by \(100\), and divide by \(15\). This gives \(600\,\text{kWh}\).
4. The expression is \(\frac{15x}{100}+12\).
Answer
a) 1) \(\$72.00\); 2) \(\$132.00\); 3) \(\$237.00\)
b) \(600\,\text{kWh}\)
c) \(\frac{15x}{100}+12\)
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