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Algebra

282 questions

All types282Multiple choice147True / False7Numeric33Matching33Fill in the blank39Short answer23
Question 221Question →

Solve the linear equation 10−4p=−1810 - 4p = -18. Which of the following is the value of pp?

Show answer & explanation

Answer: 7

Answer

The correct value for p is 7.
The correct answer is 7 because applying inverse operations accurately isolates the variable. Subtracting 10 from both sides yields −4p=−28-4p = -28, and dividing both sides by −4-4 results in p=7p = 7.

Step-by-Step Solution

1
Subtract 10 from both sides of the equation.
−4p=−18−10-4p = -18 - 10 which simplifies to −4p=−28-4p = -28
This isolates the term containing the variable pp on one side of the equation.
2
Divide both sides by -4.
p=−28−4p = \frac{-28}{-4} which simplifies to p=7p = 7
This removes the coefficient from pp, giving the final solution.

Key Concept

Solving a one-variable linear equation requires isolating the variable by performing inverse operations equally on both sides.
Question 222Question →

Read the scenario below and fill in the missing values to complete the corresponding algebraic equation.

A network administrator calculates the total required bandwidth, BB, as seventy-five less than quadruple the sum of the base traffic, tt, and twenty.

Fill in the blanks below

B=B = (t+(t + )−) -
Show answer & explanation

Answer

The completed equation is B=4(t+20)−75B = 4(t + 20) - 75. Therefore, blank 1 is 4, blank 2 is 20, and blank 3 is 75.
The scenario requires piecing together three mathematical operations. First, 'the sum of the base traffic, t, and twenty' creates the grouped expression (t+20)(t + 20). Next, 'quadruple' means multiplying that entire sum by 4, resulting in 4(t+20)4(t + 20). Finally, 'seventy-five less than' indicates that 75 must be subtracted from that product. Putting it all together gives 4(t+20)−754(t + 20) - 75, which fits perfectly into the provided blanks as 4, 20, and 75.

Step-by-Step Solution

1
Identify the core grouped expression.
The phrase 'the sum of the base traffic, t, and twenty' translates to (t+20)(t + 20).
The word 'sum' indicates addition, and it groups the terms together.
2
Apply the multiplication factor.
The phrase 'quadruple the sum' translates to 4(t+20)4(t + 20).
The word 'quadruple' means to multiply a quantity by 4.
3
Apply the subtraction.
The phrase 'seventy-five less than' means subtracting 75 from the entire previous expression, yielding 4(t+20)−754(t + 20) - 75.
The phrase 'less than' is a turnaround phrase, meaning the value being subtracted goes at the end of the expression.
4
Map the components to the provided equation structure.
Comparing 4(t+20)−754(t + 20) - 75 to the template B=B = blank_1 (t+(t + blank_2 )−) - blank_3, we find blank_1 is 4, blank_2 is 20, and blank_3 is 75.
This completes the algebraic translation in the required format.

Key Concept

Translating verbal descriptions of arithmetic operations and their structural order into an algebraic equation.
Estimated Time:45s
Question 223Question →
When the algebraic expression
(3x−2y4)39x−5y2 \frac{(3x^{-2} y^4)^3}{9x^{-5} y^2}
is fully simplified to the form CxaybC x^a y^b (where CC, aa, and bb are constants), what is the value of bb?
Show answer & explanation

Answer: 10; ten

Answer

10
By distributing the outer exponent of 3 in the numerator using multiplication (4⋅34 \cdot 3) to get y12y^{12}, and then subtracting the denominator's exponent of 2 using the quotient rule (12−212 - 2), the final exponent on the variable yy is determined to be 10.

Step-by-Step Solution

1
Apply the power of a product and power of a power rules to the numerator.
(3x−2y4)3=33⋅(x−2)3⋅(y4)3=27x−6y12(3x^{-2}y^4)^3 = 3^3 \cdot (x^{-2})^3 \cdot (y^4)^3 = 27x^{-6}y^{12}
When raising a product to a power, the outer power applies to each individual factor inside the parentheses. Exponents are multiplied when raising a power to another power.
2
Substitute the expanded numerator back into the fractional expression.
27x−6y129x−5y2 \frac{27x^{-6}y^{12}}{9x^{-5}y^2}
This sets up the expression for applying the quotient rule between the numerator and denominator.
3
Apply the quotient rule by dividing the coefficients and subtracting the exponents of like bases.
Coefficient: 27÷9=327 \div 9 = 3
Base xx: x−6−(−5)=x−6+5=x−1x^{-6 - (-5)} = x^{-6 + 5} = x^{-1}
Base yy: y12−2=y10y^{12 - 2} = y^{10}
When dividing exponential terms with the same base, subtract the denominator's exponent from the numerator's exponent.
4
Combine the simplified terms and identify the exponent bb.
The simplified expression is 3x−1y103x^{-1}y^{10}. Comparing this to the target form CxaybCx^a y^b, the value of bb is 1010.
This isolates the specific value requested by the question stem.

Key Concept

Power of a Power and Quotient of Powers Rules
Question 224Question →

An algebraic equation is shown below.

4(x−3)−2x=5x−27 4(x - 3) - 2x = 5x - 27

Which of the following represents the correct value of xx?

Show answer & explanation

Answer: 5

Answer

5
The correct answer is derived by first distributing the 4 to get 4x−124x - 12. Combining 4x4x and −2x-2x yields 2x−122x - 12. Moving the variables to one side (5x−2x=3x5x - 2x = 3x) and the constants to the other (−12+27=15-12 + 27 = 15) results in 15=3x15 = 3x. Dividing by 3 gives the final result of x=5x = 5.

Step-by-Step Solution

1
Distribute the coefficient 4 into the parentheses.
4x−12−2x=5x−27 4x - 12 - 2x = 5x - 27
According to the distributive property, the number outside the parentheses must multiply each term inside.
2
Combine the like terms on the left side of the equation.
2x−12=5x−27 2x - 12 = 5x - 27
The terms 4x4x and −2x-2x are like terms and subtract to form 2x2x.
3
Isolate the variable terms on one side by subtracting 2x2x from both sides.
−12=3x−27 -12 = 3x - 27
Subtracting 2x2x from both sides keeps the equation balanced and groups all terms containing xx on the right side.
4
Isolate the variable xx by adding 27 to both sides.
15=3x 15 = 3x
Adding 27 cancels out the −27-27 on the right, leaving the variable term by itself.
5
Solve for xx by dividing both sides by 3.
x=5 x = 5
Dividing by the coefficient of xx provides the final value of the unknown variable.

Key Concept

Solving linear equations requires correctly applying the distributive property, combining like terms accurately, and using inverse operations equally on both sides to isolate the variable.
Question 225Question →

Calculate the precise value of kk that makes the following algebraic equation true:

5(2k−3)=3k+13 5(2k - 3) = 3k + 13
Show answer & explanation

Answer: 4

Answer

4
The correct value is found by distributing the 5, gathering all variable terms on one side and constants on the other, and finally dividing by the variable's coefficient.

Step-by-Step Solution

1
Apply the distributive property on the left side of the equation.
10k−15=3k+1310k - 15 = 3k + 13
To eliminate the parentheses and expose the variable terms for grouping.
2
Subtract 3k3k from both sides.
7k−15=137k - 15 = 13
To collect all terms containing the variable kk on one side of the equation.
3
Add 1515 to both sides.
7k=287k = 28
To isolate the variable term by moving constants to the opposite side.
4
Divide both sides by 77.
k=4k = 4
To solve for the final value of kk.

Key Concept

Solving linear equations with variables on both sides using the distributive property.
Question 226Question →

Calculate the unknown value in the given equation:

5n−15=20 5n - 15 = 20

What is the value of nn?

Show answer & explanation

Answer: 7

Answer

7
To solve for n, the constant -15 must be canceled out by adding 15 to both sides, which results in 5n = 35. Then, dividing both sides by the coefficient 5 isolates n, giving the correct value of 7.

Step-by-Step Solution

1
Add 15 to both sides of the equation.
5n−15+15=20+15 5n - 15 + 15 = 20 + 15
5n=35 5n = 35
This isolates the variable term on one side of the equation by using the inverse operation of subtraction.
2
Divide both sides of the equation by 5.
5n5=355 \frac{5n}{5} = \frac{35}{5}
n=7 n = 7
This isolates the variable n completely, yielding the final numerical value.

Key Concept

Solving one-variable linear equations requires using inverse operations to isolate the unknown variable while maintaining balance across the equals sign.
Question 227Question →

An algebraic expression is given as:

(3a4b−2)2(2a−3b5)6a2b \frac{(3a^4b^{-2})^2 (2a^{-3}b^5)}{6a^2b}

What is the fully simplified form of this expression for all non-zero values of aa and bb?

Show answer & explanation

Answer: 3a33a^3

Answer

The simplified form of the expression is 3a33a^3.
The expression is correctly simplified by first applying the power to all terms inside the parentheses (resulting in 9a8b−49a^8b^{-4}), then multiplying by the adjacent term in the numerator (resulting in 18a5b18a^5b), and finally dividing by the denominator (giving 3a33a^3).

Step-by-Step Solution

1
Simplify the squared term in the numerator using the power of a product and power of a power rules.
9a8b−49a^8b^{-4}
When a product is raised to a power, raise each factor to that power (32=93^2 = 9). For variables, multiply the inner and outer exponents: (a4)2=a8(a^4)^2 = a^8 and (b−2)2=b−4(b^{-2})^2 = b^{-4}.
2
Multiply the simplified term by the second factor in the numerator, (2a−3b5)(2a^{-3}b^5).
18a5b18a^5b
Multiply the coefficients (9×2=189 \times 2 = 18). For variables with the same base, add their exponents: a8+(−3)=a5a^{8+(-3)} = a^5 and b−4+5=b1=bb^{-4+5} = b^1 = b.
3
Divide the resulting numerator by the denominator, 6a2b6a^2b.
3a33a^3
Divide the coefficients (18÷6=318 \div 6 = 3). For variables with the same base, subtract the denominator's exponent from the numerator's exponent: a5−2=a3a^{5-2} = a^3. The terms b1b^1 and b1b^1 cancel out since b1−1=b0=1b^{1-1} = b^0 = 1.

Key Concept

Operations with Variable Exponents
Question 228Question →

Solve the following linear equation for the variable ww:

7−3(2w−4)=w+5 7 - 3(2w - 4) = w + 5

What is the value of ww?

Fill in the blanks below

The value of ww is .
Show answer & explanation

Answer

2
The equation is solved by correctly applying the distributive property to get 7−6w+12=w+57 - 6w + 12 = w + 5, combining like terms to get 19−6w=w+519 - 6w = w + 5, and then using inverse operations to isolate ww, which yields w=2w = 2.

Step-by-Step Solution

1
Distribute the -3 across the terms inside the parentheses.
7−6w+12=w+57 - 6w + 12 = w + 5
The distributive property is required to expand the expression −3(2w−4)-3(2w - 4) before combining like terms. Remember that multiplying a negative by a negative results in a positive.
2
Combine the constant terms on the left side of the equation.
19−6w=w+519 - 6w = w + 5
Adding 77 and 1212 simplifies the left side.
3
Add 6w6w to both sides of the equation.
19=7w+519 = 7w + 5
This step groups all terms containing the variable ww onto the right side.
4
Subtract 5 from both sides.
14=7w14 = 7w
This isolates the variable term 7w7w on one side of the equation.
5
Divide both sides by 7.
w=2w = 2
Dividing by the coefficient of ww solves for the final value.

Key Concept

Solving linear equations involving the distributive property with negative coefficients and variables on both sides.
Estimated Time:1m 0s
Question 229Question →

An equation is given as:

2(3h−5)=4(h+1)−6 2(3h - 5) = 4(h + 1) - 6

What is the value of hh?

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Answer: 4

Answer

The correct value of hh is 4.
The correct answer is found by properly distributing the values into both sets of parentheses, combining the constants on the right side correctly (+4 - 6 = -2), and then maintaining the balance of the equation while isolating the variable using inverse operations.

Step-by-Step Solution

1
Distribute the coefficients to the terms inside the parentheses on both sides of the equation.
6h−10=4h+4−66h - 10 = 4h + 4 - 6
To eliminate the parentheses according to the distributive property.
2
Combine the constant terms on the right side of the equation.
6h−10=4h−26h - 10 = 4h - 2
To simplify the right side by calculating +4−6+4 - 6.
3
Subtract 4h4h from both sides of the equation.
2h−10=−22h - 10 = -2
To group all variable terms containing hh onto the left side.
4
Add 10 to both sides of the equation.
2h=82h = 8
To isolate the variable term by canceling out the −10-10 on the left side.
5
Divide both sides by 2.
h=4h = 4
To solve for the final value of hh.

Key Concept

Solving One-Variable Linear Equations
Question 230Question →

The length of a rectangular solar panel is given by the expression 7a4b−27a^4 b^{-2} and its width is given by 6a2b56a^2 b^5. Which of the following expressions represents the area of the solar panel?

Show answer & explanation

Answer: 42a6b342a^6 b^3

Answer

The area is represented by the expression 42a6b342a^6 b^3.
The correct answer is 42a6b342a^6 b^3. To find the area of the rectangle, you multiply its length and width: (7a4b−2)(6a2b5)(7a^4 b^{-2})(6a^2 b^5). The coefficients are multiplied together (7×6=427 \times 6 = 42). For the variables, the product rule states that when multiplying terms with the same base, you add their exponents. Thus, for base aa, 4+2=64 + 2 = 6 to get a6a^6, and for base bb, −2+5=3-2 + 5 = 3 to get b3b^3. Combining these yields 42a6b342a^6 b^3.

Step-by-Step Solution

1
Identify the formula for the area of a rectangle and set up the expression.
Area = length \times width = (7a^4 b^{-2})(6a^2 b^5)
The area of a rectangle is found by taking the product of its length and width dimensions.
2
Multiply the numerical coefficients.
7×6=427 \times 6 = 42
Constants must be multiplied independently from the variable bases.
3
Apply the product rule for exponents to the variables with identical bases.
a^{4+2} = a^6 and b^{-2+5} = b^3
When multiplying identical bases, their exponents must be added.
4
Combine the calculated coefficient and variables into a single expression.
42a6b342a^6 b^3
This forms the final simplified product representing the area.

Key Concept

Operations with Variable Exponents
Estimated Time:1m 0s
Question 231Question →

An engineer is calculating a scaling factor for a fluid dynamics simulation. The calculation reduces to the following algebraic expression:

(3m3n−4)(8m−1n2)(2m−2n3)2 \frac{ (3m^3n^{-4})(8m^{-1}n^2) }{ (2m^{-2}n^3)^2 }

Assuming mm and nn are non-zero variables, what is the fully simplified form of this scaling factor?

Show answer & explanation

Answer: 6m6n8\frac{6m^6}{n^8}

Answer

The fully simplified expression is 6m6n8\frac{6m^6}{n^8}.
The correct answer is found by simplifying the numerator and denominator independently, then dividing. Multiplying the terms in the numerator gives 24m2n−224m^2n^{-2}. Squaring the entire term in the denominator (including the 2) gives 4m−4n64m^{-4}n^6. Dividing these expressions yields a coefficient of 66. Subtracting the exponents gives m2−(−4)=m6m^{2 - (-4)} = m^6 and n−2−6=n−8n^{-2 - 6} = n^{-8}. Finally, converting the negative exponent into a fraction results in 6m6n8\frac{6m^6}{n^8}.

Step-by-Step Solution

1
Simplify the numerator using the product rule for exponents.
(3×8)m3+(−1)n−4+2=24m2n−2(3 \times 8) m^{3 + (-1)} n^{-4 + 2} = 24m^2n^{-2}
When multiplying terms with the same base, add their exponents and multiply their numerical coefficients.
2
Simplify the denominator using the power of a product rule.
22(m−2×2)(n3×2)=4m−4n62^2(m^{-2 \times 2})(n^{3 \times 2}) = 4m^{-4}n^6
When raising a product to a power, raise the coefficient and multiply every exponent by the outside power.
3
Divide the simplified numerator by the simplified denominator using the quotient rule.
(244)m2−(−4)n−2−6=6m6n−8(\frac{24}{4}) m^{2 - (-4)} n^{-2 - 6} = 6m^6n^{-8}
When dividing terms with the same base, subtract the denominator's exponent from the numerator's exponent.
4
Rewrite the expression with only positive exponents.
6m6n8\frac{6m^6}{n^8}
A negative exponent means taking the reciprocal of the base, so n−8n^{-8} becomes 1n8\frac{1}{n^8}.

Key Concept

Operations with Variable Exponents
Question 232Question →

Determine the value of the variable yy that satisfies the given linear equation.

Fill in the blanks below

Given the equation 5(y−2)+3y=4y+26 5(y - 2) + 3y = 4y + 26 , the value of yy is .
Show answer & explanation

Answer

9
By distributing the 55 to obtain 5y−105y - 10, combining the yy terms on the left to get 8y8y, and subsequently isolating yy through basic inverse operations (8y−4y=26+108y - 4y = 26 + 10), we determine that 4y=364y = 36, making y=9y = 9.

Step-by-Step Solution

1
Distribute the 5 to the terms inside the parentheses.
5y−10+3y=4y+265y - 10 + 3y = 4y + 26
Applying the distributive property removes the parentheses, allowing like terms to be combined.
2
Combine the like terms on the left side of the equation.
8y−10=4y+268y - 10 = 4y + 26
Simplifying each side individually makes it easier to isolate the variable.
3
Subtract 4y4y from both sides of the equation.
4y−10=264y - 10 = 26
This groups all terms containing the variable yy on one side of the equation.
4
Add 10 to both sides of the equation.
4y=364y = 36
This isolates the variable term by removing the constant from the left side.
5
Divide both sides by 4.
y=9y = 9
Dividing by the coefficient of yy yields the final value.

Key Concept

Solving linear equations with variables on both sides using the distributive property
Question 233Question →

A technician is testing two different electrical circuits. The voltage in Circuit A is given by the expression 5v−125v - 12, where vv is a variable setting on the power supply. The voltage in Circuit B is given by 2(v+9)2(v + 9). Both circuits are adjusted to have the exact same voltage. Fill in the blanks with the correct numerical values.

Fill in the blanks below

The required setting vv is and the resulting voltage is .
Show answer & explanation

Answer

The setting vv is 10 and the resulting voltage is 38.
By setting the two expressions equal, expanding the parentheses, and systematically isolating vv, we find the setting must be 10. Plugging 10 back into the expression evaluates the voltage to 38 for both circuits.

Step-by-Step Solution

1
Set the expressions for Circuit A and Circuit B equal to each other.
5v−12=2(v+9)5v - 12 = 2(v + 9)
The problem states that both circuits are adjusted to have the exact same voltage.
2
Distribute the 2 on the right side of the equation.
5v−12=2v+185v - 12 = 2v + 18
The distributive property is required to eliminate parentheses and expose the variable term.
3
Subtract 2v2v from both sides of the equation.
3v−12=183v - 12 = 18
To consolidate all variable terms on one side of the equation.
4
Add 12 to both sides of the equation.
3v=303v = 30
To isolate the variable term by moving constants to the other side.
5
Divide both sides by 3.
v=10v = 10
To solve for the variable vv.
6
Substitute v=10v = 10 back into either original expression.
5(10)−12=50−12=385(10) - 12 = 50 - 12 = 38
To find the resulting voltage as requested by the second blank.

Key Concept

Solving linear equations with variables on both sides and distributive property
Question 234Question →

Match each algebraic expression on the left with its fully simplified equivalent form on the right. Assume all variables represent non-zero real numbers.

Click a left item, then click its matching right item

Items

15a4b−35a−2b2\frac{15a^4b^{-3}}{5a^{-2}b^2}
(3a−2b4)2(3a^{-2}b^4)^2
(2a3b−1)(4a−1b3)(2a^3b^{-1})(4a^{-1}b^3)
(a2b−22)−3\left(\frac{a^2b^{-2}}{2}\right)^{-3}

Matches

Show answer & explanation

Answer

The quotient matches (3a^6)/b^5, the squared term matches (9b^8)/a^4, the product matches 8a^2b^2, and the fraction raised to a negative power matches (8b^6)/a^6.
Each expression is thoroughly simplified by applying the fundamental laws of exponents: adding exponents for multiplication, subtracting exponents for division, multiplying exponents for a power to a power, and reciprocating terms to handle negative exponents.

Step-by-Step Solution

1
Simplify the first expression: (15a^4b^-3) / (5a^-2b^2).
3a^6b^-5, which becomes (3a^6)/b^5.
When dividing terms with the same base, divide the coefficients and subtract the exponents. Move negative exponents to the denominator.
2
Simplify the second expression: (3a^-2b^4)^2.
9a^-4b^8, which becomes (9b^8)/a^4.
When raising a power to a power, raise the coefficient to the outer power and multiply the variable exponents.
3
Simplify the third expression: (2a^3b^-1)(4a^-1b^3).
8a^2b^2.
When multiplying terms with the same base, multiply the coefficients and add the respective exponents.
4
Simplify the fourth expression: ((a^2b^-2)/2)^-3.
a^-6b^6 / 2^-3, which becomes (8b^6)/a^6.
Apply the outer negative exponent to all factors inside the parentheses, including the denominator, then move negative powers across the fraction bar to make them positive.

Key Concept

Exponent rules: Product of powers, quotient of powers, power of a product, and negative exponents.
Question 235Question →

In a specific algorithm, the output value is determined by the algebraic rule 2m2n−3n22m^2n - 3n^2. Which of the following is the resulting output when the algorithm processes the inputs m=−4m = -4 and n=−3n = -3?

Show answer & explanation

Answer: -123

Answer

The correct evaluated result is -123.
The correct answer is found by substituting the numerical values for the variables and strictly adhering to the order of operations. First, squaring the negative integers results in positive values, transforming the expression to 2(16)(-3) - 3(9). Multiplying the terms gives -96 and 27, respectively. Finally, subtracting 27 from -96 produces a final value of -123.

Step-by-Step Solution

1
Substitute the given values m=−4m = -4 and n=−3n = -3 into the expression 2m2n−3n22m^2n - 3n^2.
2(−4)2(−3)−3(−3)22(-4)^2(-3) - 3(-3)^2
Direct substitution of variables with their numerical values is the required first step to evaluate any algebraic expression.
2
Evaluate the exponents according to the order of operations (PEMDAS).
(−4)2=16(-4)^2 = 16 and (−3)2=9(-3)^2 = 9. The expression becomes 2(16)(−3)−3(9)2(16)(-3) - 3(9).
Exponents must be calculated before multiplication to maintain mathematical accuracy. A negative number raised to an even power results in a positive number.
3
Perform the multiplication operations from left to right.
For the first term: 2(16)=322(16) = 32, and 32(−3)=−9632(-3) = -96. For the second term: 3(9)=273(9) = 27. The expression is now −96−27-96 - 27.
Multiplication precedes subtraction in the order of operations.
4
Perform the final subtraction to find the total value.
−96−27=−123-96 - 27 = -123
Combining the remaining terms yields the final evaluated numerical result of the polynomial.

Key Concept

Evaluating algebraic expressions with negative integers using substitution and the order of operations.
Question 236Question →
The algebraic expression (2x4y3)34x5y−2\frac{(2x^4 y^3)^3}{4x^5 y^{-2}} can be simplified to the form kxaybkx^a y^b, where kk is the numerical coefficient, and aa and bb are the exponents. What is the value of a+ba + b?
Show answer & explanation

Answer: 18; 18.0

Answer

18
By applying the power rule, the numerator becomes 8x12y98x^{12}y^9. Dividing this by the denominator 4x5y−24x^5y^{-2} yields a numerical coefficient of 8/4=28/4=2. For the xx terms, applying the quotient rule gives 12−5=712 - 5 = 7. For the yy terms, 9−(−2)=119 - (-2) = 11. The simplified expression is 2x7y112x^7y^{11}. The exponents are a=7a=7 and b=11b=11, so their sum is 7+11=187 + 11 = 18.

Step-by-Step Solution

1
Apply the power of a product rule to the numerator.
(23)(x4×3)(y3×3)=8x12y9(2^3)(x^{4 \times 3})(y^{3 \times 3}) = 8x^{12}y^9
To distribute the outer exponent 3 to all factors inside the parentheses.
2
Divide the numerical coefficients.
84=2\frac{8}{4} = 2
To simplify the constants to find kk.
3
Apply the quotient rule for the variable xx.
x12−5=x7x^{12 - 5} = x^7
When dividing like bases, subtract the exponent in the denominator from the exponent in the numerator. This gives a=7a = 7.
4
Apply the quotient rule for the variable yy, accounting for the negative exponent.
y9−(−2)=y9+2=y11y^{9 - (-2)} = y^{9 + 2} = y^{11}
Subtracting a negative exponent is equivalent to adding. This gives b=11b = 11.
5
Calculate the sum of the exponents aa and bb.
7+11=187 + 11 = 18
To answer the specific question asked in the prompt.

Key Concept

Applying the power of a product rule and the quotient rule for exponents, paying special attention to subtracting negative exponents.
Question 237Question →

Consider the following equation:

4(v−2)+6=7v−2(v+5) 4(v - 2) + 6 = 7v - 2(v + 5)

What is the value of vv that satisfies this equation?

Show answer & explanation

Answer: 8

Answer

8
Expanding the terms by distributing gives 4v - 8 + 6 = 7v - 2v - 10. Simplifying both sides yields 4v - 2 = 5v - 10. By subtracting 4v from both sides and adding 10 to both sides, the variable v is isolated to yield 8.

Step-by-Step Solution

1
Distribute the constants into the parentheses on both sides of the equation.
4v - 8 + 6 = 7v - 2v - 10
This removes the parentheses and expands the expressions so like terms can be combined.
2
Combine the constant terms on the left side and the variable terms on the right side.
4v - 2 = 5v - 10
Simplifying both sides makes it easier to isolate the variable.
3
Subtract 4v from both sides to group the variable terms on the right side.
-2 = v - 10
This step groups the variables together while keeping the coefficient of v positive.
4
Add 10 to both sides to solve for v.
8 = v
This step fully isolates v, giving the final solution.

Key Concept

Solving linear equations involves distributing constants to expand expressions, combining like terms, and performing identical operations on both sides to isolate the unknown variable.
Question 238Question →

Solve for tt in the equation below:

3(2t−4)+5=5t−(t−9) 3(2t - 4) + 5 = 5t - (t - 9)

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The value of tt is .
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Answer

The correct value for tt is 8.
By properly distributing values into the parentheses, combining like terms, and isolating the variable using inverse operations, the equation simplifies to 2t=162t = 16, which yields t=8t = 8.

Step-by-Step Solution

1
Distribute the 3 into the parentheses on the left side, and distribute the negative sign into the parentheses on the right side.
6t−12+5=5t−t+96t - 12 + 5 = 5t - t + 9
To eliminate the parentheses so that all terms can be properly combined.
2
Combine the constant terms on the left side and the like variable terms on the right side.
6t−7=4t+96t - 7 = 4t + 9
To simplify both sides of the equation independently before moving terms across the equals sign.
3
Subtract 4t4t from both sides of the equation.
2t−7=92t - 7 = 9
To gather all the variable terms onto the left side.
4
Add 7 to both sides of the equation.
2t=162t = 16
To isolate the term containing the variable.
5
Divide both sides by 2.
t=8t = 8
To find the final numerical value of tt.

Key Concept

Solving One-Variable Linear Equations
Estimated Time:1m 30s
Question 239Question →
Assume xx and yy are non-zero real numbers. If the algebraic expression (3x2y4)39xky7 \frac{(3x^2y^4)^3}{9x^ky^7} simplifies completely to 3xy5 3xy^5 , what is the value of the exponent kk?
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Answer: 5; k=5; k = 5

Answer

5
The correct value is 5. By expanding the numerator using the power of a product rule, (3x2y4)3(3x^2y^4)^3 becomes 27x6y1227x^6y^{12}. Dividing this by the denominator 9xky79x^ky^7 yields (279)⋅x6−k⋅y12−7\left(\frac{27}{9}\right) \cdot x^{6-k} \cdot y^{12-7}, which simplifies to 3x6−ky53x^{6-k}y^5. Equating this result to the given simplified expression 3x1y53x^1y^5 means the exponent of xx must be 1. Solving the simple equation 6−k=16 - k = 1 gives k=5k = 5.

Step-by-Step Solution

1
Expand the numerator using the power of a product and power of a power rules.
(3x2y4)3=33⋅(x2)3⋅(y4)3=27x6y12(3x^2y^4)^3 = 3^3 \cdot (x^2)^3 \cdot (y^4)^3 = 27x^6y^{12}
The outer exponent must be applied to the numerical coefficient and multiplied by the inner exponents of each variable.
2
Substitute the expanded numerator back into the fractional expression.
27x6y129xky7 \frac{27x^6y^{12}}{9x^ky^7}
This sets up the expression for division.
3
Simplify the numerical coefficients and apply the quotient rule for exponents (xa/xb=xa−bx^a / x^b = x^{a-b}).
(279)⋅x6−k⋅y12−7=3x6−ky5\left(\frac{27}{9}\right) \cdot x^{6-k} \cdot y^{12-7} = 3x^{6-k}y^5
To combine the numerical terms and the variable terms with identical bases.
4
Set the simplified expression equal to the target expression provided in the problem.
3x6−ky5=3x1y53x^{6-k}y^5 = 3x^1y^5
The problem states that the original expression simplifies completely to 3xy53xy^5.
5
Equate the exponents of the base xx to solve for kk.
6−k=1⇒k=56 - k = 1 \Rightarrow k = 5
Since the expressions are equal and the bases are the same, their corresponding exponents must be equal.

Key Concept

Applying power and quotient rules of exponents to solve for an unknown variable exponent.
Question 240Question →

What is the value of kk that makes the following equation true?

4(k−2)−3k=5k−2(k+10) 4(k - 2) - 3k = 5k - 2(k + 10)
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Answer: 6

Answer

The correct value for the variable is 6.
By following the correct order of operations, distributing carefully, and balancing the equation, we find that k = 6 makes both sides of the equation equal.

Step-by-Step Solution

1
Distribute the values outside the parentheses on both sides of the equation.
4k−8−3k=5k−2k−20 4k - 8 - 3k = 5k - 2k - 20
Applying the distributive property removes the parentheses and allows for combining like terms.
2
Combine like terms on each side of the equation.
k−8=3k−20 k - 8 = 3k - 20
Simplifying each side individually makes it easier to isolate the variable.
3
Subtract kk from both sides to collect the variables on the right side.
−8=2k−20 -8 = 2k - 20
Balancing the equation by performing the inverse operation isolates the variable terms on one side.
4
Add 20 to both sides to isolate the variable term.
12=2k 12 = 2k
Moving the constants to the opposite side of the variable completes the isolation process.
5
Divide both sides by 2.
k=6 k = 6
Dividing by the coefficient gives the final value of the variable.

Key Concept

Solving multi-step linear equations involves distributing, combining like terms, and balancing the equation using inverse operations.
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