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Question 61Question →

If 3(x−2)−4(2−x)=12(6x−4)+13(x - 2) - 4(2 - x) = \frac{1}{2}(6x - 4) + 1, what is the value of 2x−32x - 3?

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Answer: 72\frac{7}{2}

Answer

The value of the expression is 72\frac{7}{2}.
The correct answer is 72\frac{7}{2}. Distributing the constants on both sides of the equation 3(x−2)−4(2−x)=12(6x−4)+13(x - 2) - 4(2 - x) = \frac{1}{2}(6x - 4) + 1 yields 3x−6−8+4x=3x−2+13x - 6 - 8 + 4x = 3x - 2 + 1. Combining like terms on both sides gives 7x−14=3x−17x - 14 = 3x - 1. Subtracting 3x3x from both sides and adding 1414 to both sides results in 4x=134x = 13, which simplifies to x=134x = \frac{13}{4}. Substituting this value into 2x−32x - 3 gives 2(134)−3=132−3=722\left(\frac{13}{4}\right) - 3 = \frac{13}{2} - 3 = \frac{7}{2}.

Step-by-Step Solution

1
Distribute the constants on both sides of the equation.
3x−6−8+4x=3x−2+13x - 6 - 8 + 4x = 3x - 2 + 1
To simplify the linear terms and constants before isolating the variable.
2
Combine like terms on both sides of the equation.
7x−14=3x−17x - 14 = 3x - 1
To reduce the equation to a simpler form with one variable term and one constant term on each side.
3
Isolate the variable term by subtracting 3x3x from both sides and adding 1414 to both sides.
4x=134x = 13
To group all variable terms on one side and constant terms on the other.
4
Solve for xx by dividing both sides by 44.
x=134x = \frac{13}{4}
To isolate the variable xx.
5
Substitute the value of xx into the expression 2x−32x - 3.
2(134)−3=132−3=722\left(\frac{13}{4}\right) - 3 = \frac{13}{2} - 3 = \frac{7}{2}
To calculate the final value requested by the question.

Key Concept

Solving linear equations in one variable by distributing terms, combining like terms, isolating the variable, and evaluating algebraic expressions.
Question 62Question →

If 35p−4=8\frac{3}{5}p - 4 = 8, what is the value of pp?

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Answer: 20

Answer

20
To solve the equation 35p−4=8\frac{3}{5}p - 4 = 8, we perform inverse operations to isolate pp. First, add 4 to both sides of the equation to get 35p=12\frac{3}{5}p = 12. Next, multiply both sides by the reciprocal of the coefficient of pp, which is 53\frac{5}{3}. This gives p=12×53=20p = 12 \times \frac{5}{3} = 20. Substituting 20 back into the original equation confirms it is the correct solution.

Step-by-Step Solution

1
Add 4 to both sides of the equation.
35p=12\frac{3}{5}p = 12
To isolate the term containing the variable pp.
2
Multiply both sides of the equation by 53\frac{5}{3}.
p=20p = 20
To solve for pp by multiplying by the reciprocal of its coefficient.

Key Concept

Solving one-variable linear equations using inverse operations.
Question 63Question →

In the equation below, kk is a constant.

13(2kx−9)−56(x+4)=112\frac{1}{3}(2kx - 9) - \frac{5}{6}(x + 4) = \frac{11}{2}

If the equation has no solution, what is the value of kk?

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Answer: 1.25

Answer

1.25 (or 5/4)
The correct answer is 1.25 (or 5/4). A linear equation in one variable of the form Ax+B=CAx + B = C has no solution if the variable terms on both sides of the equation are equal (meaning A=0A = 0) and the constant terms are unequal (B≠CB \neq C). Expanding the left side of the given equation yields 23kx−3−56x−103=112\frac{2}{3}kx - 3 - \frac{5}{6}x - \frac{10}{3} = \frac{11}{2}. Combining the constant terms gives (23k−56)x−193=112\left(\frac{2}{3}k - \frac{5}{6}\right)x - \frac{19}{3} = \frac{11}{2}. Setting the coefficient of xx to 00 yields 23k−56=0\frac{2}{3}k - \frac{5}{6} = 0. Solving for kk gives k=56×32=54k = \frac{5}{6} \times \frac{3}{2} = \frac{5}{4}, which is equivalent to 1.25. Since the remaining constant terms are unequal (−193≠112-\frac{19}{3} \neq \frac{11}{2}), the equation has no solution when k=1.25k = 1.25.

Step-by-Step Solution

1
Expand the expression on the left side of the equation
23kx−3−56x−103=112\frac{2}{3}kx - 3 - \frac{5}{6}x - \frac{10}{3} = \frac{11}{2}
Apply the distributive property to remove the parentheses.
2
Group the xx terms and combine the constants on the left side
(23k−56)x−193=112\left(\frac{2}{3}k - \frac{5}{6}\right)x - \frac{19}{3} = \frac{11}{2}
Simplify the equation by combining like terms: −3−103=−93−103=−193-3 - \frac{10}{3} = -\frac{9}{3} - \frac{10}{3} = -\frac{19}{3}.
3
Set the coefficient of the xx term equal to 00
23k−56=0\frac{2}{3}k - \frac{5}{6} = 0
For a linear equation to have no solution, the variable terms on both sides of the equation must cancel out (meaning the coefficient of the variable must be 00), while the remaining constant terms must not be equal (−193≠112-\frac{19}{3} \neq \frac{11}{2}).
4
Solve the resulting equation for kk
k=1.25k = 1.25
Add 56\frac{5}{6} to both sides to get 23k=56\frac{2}{3}k = \frac{5}{6}, then multiply by the reciprocal of 23\frac{2}{3}, which gives k=56×32=1512=54=1.25k = \frac{5}{6} \times \frac{3}{2} = \frac{15}{12} = \frac{5}{4} = 1.25.

Key Concept

Conditions for a linear equation in one variable to have no solution
Question 64Question →

A rectangle has a length of 2x+52x + 5 centimeters and a width of x−2x - 2 centimeters. If the perimeter of the rectangle is 4242 centimeters, what is the value of xx?

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Answer: 6

Answer

6
The correct answer is 6. The perimeter of a rectangle is calculated using the formula P=2l+2wP = 2l + 2w. Substituting the given expressions for length and width yields the linear equation 2(2x+5)+2(x−2)=422(2x + 5) + 2(x - 2) = 42. Distributing the 2 gives 4x+10+2x−4=424x + 10 + 2x - 4 = 42. Combining like terms on the left side simplifies this to 6x+6=426x + 6 = 42. Subtracting 6 from both sides yields 6x=366x = 36. Finally, dividing both sides by 6 gives the value x=6x = 6.

Step-by-Step Solution

1
Write the perimeter equation in terms of xx.
2(2x+5)+2(x−2)=422(2x + 5) + 2(x - 2) = 42
The perimeter of a rectangle is the sum of all its sides, represented by the formula P=2l+2wP = 2l + 2w.
2
Distribute the 2 into the parentheses.
4x+10+2x−4=424x + 10 + 2x - 4 = 42
Applying the distributive property yields 2(2x+5)=4x+102(2x + 5) = 4x + 10 and 2(x−2)=2x−42(x - 2) = 2x - 4.
3
Combine like terms on the left side.
6x+6=426x + 6 = 42
Grouping the variable terms gives 4x+2x=6x4x + 2x = 6x, and grouping the constant terms gives 10−4=610 - 4 = 6.
4
Subtract 6 from both sides of the equation.
6x=366x = 36
This isolates the variable term on the left side of the equation.
5
Divide both sides of the equation by 6.
x=6x = 6
Dividing by the coefficient of xx solves for the variable.

Key Concept

Solving linear equations in one variable by applying the distributive property and combining like terms.
Question 65Question →

A line in the xyxy-plane has the equation 3y−2x=123y - 2x = 12. What is the slope of this line?

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Answer: 23\frac{2}{3}

Answer

The slope of the line is 23\frac{2}{3}.
To find the slope, the equation 3y−2x=123y - 2x = 12 is rewritten in slope-intercept form (y=mx+by = mx + b). Adding 2x2x to both sides gives 3y=2x+123y = 2x + 12. Dividing by 33 yields y=23x+4y = \frac{2}{3}x + 4. The slope is the coefficient of xx, which is 23\frac{2}{3}.

Step-by-Step Solution

1
Isolate the term containing yy by adding 2x2x to both sides of the equation.
3y=2x+123y = 2x + 12
To express the equation in slope-intercept form (y=mx+by = mx + b).
2
Divide all terms in the equation by 33 to solve for yy.
y=23x+4y = \frac{2}{3}x + 4
To isolate yy and identify the slope, mm, which is the coefficient of xx.

Key Concept

Finding the slope of a line from its linear equation by rewriting it in slope-intercept form.
Question 66Question →

A solar energy company offers two payment plans for installing solar panels. Under Plan A, the customer pays a one-time installation fee of 1,200andamonthlymaintenancefeeof1,200 and a monthly maintenance fee of 25. Under Plan B, there is no installation fee, but the customer pays a monthly maintenance fee of 45forthefirst12months,and45 for the first 12 months, and 35 per month for each month thereafter. If a customer chooses Plan A, after how many months of service will the total cost of Plan A be exactly equal to the total cost of Plan B?

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Answer: 108

Answer

The correct answer is 108 months.
The correct answer is 108. Setting the total cost of Plan A, 1200+25m1200 + 25m, equal to the total cost of Plan B, 45(12)+35(m−12)45(12) + 35(m-12), yields the linear equation 1200+25m=120+35m1200 + 25m = 120 + 35m. Isolating mm by subtracting 25m25m and 120120 from both sides gives 10m=108010m = 1080, which simplifies to m=108m = 108.

Step-by-Step Solution

1
Set up the total cost expression for Plan A.
CostA=1200+25m\text{Cost}_A = 1200 + 25m
Plan A has a fixed setup fee of 1,200plusarecurringcostof1,200 plus a recurring cost of 25 for each of the mm months.
2
Set up the total cost expression for Plan B, assuming the number of months m>12m > 12.
CostB=45(12)+35(m−12)=540+35m−420=120+35m\text{Cost}_B = 45(12) + 35(m - 12) = 540 + 35m - 420 = 120 + 35m
Plan B charges 45permonthforthefirst12monthsand45 per month for the first 12 months and 35 per month for the remaining m−12m - 12 months.
3
Equate the two expressions and solve for mm.
1200+25m=120+35m⇒1080=10m⇒m=1081200 + 25m = 120 + 35m \Rightarrow 1080 = 10m \Rightarrow m = 108
Setting the two costs equal allows us to isolate the variable mm by subtracting 25m25m and 120120 from both sides.

Key Concept

Formulating and solving multi-step linear equations in one variable from real-world word problems.
Estimated Time:2m 30s
Question 67Question →

If 5−3(2x−4)=2(x+3)−55 - 3(2x - 4) = 2(x + 3) - 5, what is the value of 4x−34x - 3?

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Answer: 5

Answer

5
The correct answer is the value of the expression when the linear equation is solved correctly. Distributing the terms on both sides of the equation yields 5−6x+12=2x+6−55 - 6x + 12 = 2x + 6 - 5. Simplifying each side gives 17−6x=2x+117 - 6x = 2x + 1. Adding 6x6x to both sides and subtracting 11 from both sides results in 16=8x16 = 8x, which means x=2x = 2. Substituting x=2x = 2 into the expression 4x−34x - 3 gives 4(2)−3=54(2) - 3 = 5.

Step-by-Step Solution

1
Distribute the constants on both sides of the equation.
5−6x+12=2x+6−55 - 6x + 12 = 2x + 6 - 5
To eliminate the parentheses so that like terms can be combined.
2
Combine like terms on both sides of the equation.
17−6x=2x+117 - 6x = 2x + 1
To simplify the linear expression on each side.
3
Isolate the variable xx by adding 6x6x to both sides and subtracting 11 from both sides.
16=8x16 = 8x, which simplifies to x=2x = 2
To find the value of xx that satisfies the equation.
4
Substitute the value of xx into the target expression 4x−34x - 3.
4(2)−3=54(2) - 3 = 5
To find the final value requested by the question.

Key Concept

Solving linear equations in one variable by distributing constants, combining like terms, and evaluating expressions.
Question 68Question →

In the equation below, cc is a constant.

3c(2x−1)−2(x+4c)5=2x−3\frac{3c(2x - 1) - 2(x + 4c)}{5} = 2x - 3

If the equation has no solution for xx, what is the value of cc?

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Answer: 2

Answer

The correct answer is 2.
To find the value of cc that results in no solution, we first clear the fraction by multiplying both sides of the equation by 5, giving 3c(2x−1)−2(x+4c)=10x−153c(2x - 1) - 2(x + 4c) = 10x - 15. Next, we expand the terms to get 6cx−3c−2x−8c=10x−156cx - 3c - 2x - 8c = 10x - 15, and group the xx terms and constant terms: (6c−2)x−11c=10x−15(6c - 2)x - 11c = 10x - 15. For a linear equation to have no solution, the coefficients of xx on both sides must be equal while the constant terms must be different. Setting the coefficients equal gives 6c−2=106c - 2 = 10, which solves to c=2c = 2. Checking the constant terms when c=2c = 2, we get −11(2)=−22-11(2) = -22 on the left and −15-15 on the right. Since −22≠−15-22 \neq -15, the equation has no solution, confirming c=2c = 2 is correct.

Step-by-Step Solution

1
Multiply both sides of the equation by 5 to eliminate the denominator.
3c(2x−1)−2(x+4c)=10x−153c(2x - 1) - 2(x + 4c) = 10x - 15
Clearing the denominator simplifies the equation into standard polynomial terms.
2
Distribute the terms on the left side of the equation.
6cx−3c−2x−8c=10x−156cx - 3c - 2x - 8c = 10x - 15
Applying the distributive property expands the expression so terms can be grouped.
3
Group the xx terms and constant terms on the left side.
(6c−2)x−11c=10x−15(6c - 2)x - 11c = 10x - 15
Putting the equation in the standard form Ax+B=Cx+DAx + B = Cx + D allows us to easily set up the conditions for no solution.
4
Set the coefficients of xx on both sides equal to each other.
6c−2=106c - 2 = 10, which simplifies to 6c=126c = 12, and thus c=2c = 2.
For the equation to have no solution, the variable terms on both sides must cancel each other out.
5
Verify that the constant terms are not equal when c=2c = 2.
The left-side constant is −11(2)=−22-11(2) = -22, and the right-side constant is −15-15. Since −22≠−15-22 \neq -15, the equation has no solution.
If the constant terms were equal, the equation would have infinitely many solutions instead of no solution.

Key Concept

Identifying conditions for a linear equation in one variable to have no solution.

Alternative Method

Instead of clearing the fraction first, write the left side of the equation as (6c−25)x−11c5(\frac{6c - 2}{5})x - \frac{11c}{5}. For there to be no solution, the coefficient of xx on the left side, 6c−25\frac{6c - 2}{5}, must equal the coefficient of xx on the right side, which is 2. Solving 6c−25=2\frac{6c - 2}{5} = 2 gives 6c−2=10  ⟹  c=26c - 2 = 10 \implies c = 2.
Estimated Time:2m 30s
Question 69Question →

If 2(x−6)=102(x - 6) = 10, what is the value of x+4x + 4?

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Answer: 15

Answer

15
The correct answer is 15. Isolating the variable in the equation 2(x−6)=102(x - 6) = 10 can be done by first dividing both sides by 2, which gives x−6=5x - 6 = 5. Adding 6 to both sides yields x=11x = 11. Substituting this value into the expression x+4x + 4 results in 11+4=1511 + 4 = 15. Alternatively, distributing the 2 gives 2x−12=102x - 12 = 10, and adding 12 to both sides gives 2x=222x = 22, which leads to x=11x = 11 and thus x+4=15x + 4 = 15.

Step-by-Step Solution

1
Distribute the 2 to the terms inside the parentheses on the left side of the equation.
2x−12=102x - 12 = 10
To remove the parentheses and prepare to isolate the variable term.
2
Add 12 to both sides of the equation.
2x=222x = 22
To isolate the variable term 2x2x on the left side of the equation.
3
Divide both sides of the equation by 2.
x=11x = 11
To find the value of xx.
4
Substitute the value of xx into the expression x+4x + 4.
11+4=1511 + 4 = 15
To find the final value requested by the question.

Key Concept

Solving linear equations in one variable by applying the distributive property, isolating the variable, and evaluating an expression.

Alternative Method

Instead of distributing first, divide both sides of the equation 2(x−6)=102(x - 6) = 10 by 2 directly to get x−6=5x - 6 = 5. Then, add 10 to both sides of the equation x−6=5x - 6 = 5 to get the value of x+4x + 4 directly: (x−6)+10=5+10⇒x+4=15(x - 6) + 10 = 5 + 10 \Rightarrow x + 4 = 15.
Estimated Time:45s
Question 70Question →

A linear function ff has a slope of 23\frac{2}{3}. The function gg is defined by g(x)=f(x+4)−3g(x) = f(x + 4) - 3. If the graph of y=g(x)y = g(x) in the xyxy-plane has an xx-intercept at (5,0)(5, 0), what is the xx-intercept of the graph of y=f(x)y = f(x)?

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Answer: 4.5

Answer

4.5
The correct answer is 4.54.5 (or 92\frac{9}{2}). By using the function transformation relation, we find g(5)=f(9)−3=0g(5) = f(9) - 3 = 0, which yields f(9)=3f(9) = 3. With a slope of 23\frac{2}{3}, the equation of the line is f(x)=23x−3f(x) = \frac{2}{3}x - 3. Setting f(x)=0f(x) = 0 yields the xx-intercept at x=4.5x = 4.5.

Step-by-Step Solution

1
Use the definition of function gg and its given xx-intercept to find a point on the graph of ff.
f(9)=3f(9) = 3
Since the graph of y=g(x)y = g(x) has an xx-intercept at (5,0)(5, 0), we know g(5)=0g(5) = 0. Substituting x=5x = 5 into the definition g(x)=f(x+4)−3g(x) = f(x + 4) - 3 gives g(5)=f(5+4)−3=f(9)−3g(5) = f(5 + 4) - 3 = f(9) - 3. Since g(5)=0g(5) = 0, it follows that f(9)−3=0f(9) - 3 = 0, or f(9)=3f(9) = 3.
2
Determine the equation of the linear function ff using its slope and the point identified in Step 1.
f(x)=23x−3f(x) = \frac{2}{3}x - 3
The function ff is linear with a slope of 23\frac{2}{3} and passes through the point (9,3)(9, 3). Using the point-slope formula, we get f(x)−3=23(x−9)f(x) - 3 = \frac{2}{3}(x - 9), which simplifies to f(x)=23x−3f(x) = \frac{2}{3}x - 3.
3
Find the xx-intercept of the graph of y=f(x)y = f(x) by setting f(x)=0f(x) = 0.
x=4.5x = 4.5 (or 92\frac{9}{2})
To find the xx-intercept, set f(x)=0f(x) = 0. This gives 23x−3=0\frac{2}{3}x - 3 = 0. Adding 33 to both sides and multiplying by 32\frac{3}{2} yields x=92x = \frac{9}{2}, which is equal to 4.54.5.

Key Concept

Linear Functions and Graphs

Alternative Method

Alternatively, we can write the equation of ff in slope-intercept form as f(x)=23x+bf(x) = \frac{2}{3}x + b. Then the definition of g(x)g(x) becomes g(x)=23(x+4)+b−3=23x+83+b−3=23x+b−13g(x) = \frac{2}{3}(x + 4) + b - 3 = \frac{2}{3}x + \frac{8}{3} + b - 3 = \frac{2}{3}x + b - \frac{1}{3}. Since the graph of gg has an xx-intercept at (5,0)(5, 0), we substitute x=5x = 5 and g(5)=0g(5) = 0 to get 23(5)+b−13=0\frac{2}{3}(5) + b - \frac{1}{3} = 0, which simplifies to 3+b=03 + b = 0, so b=−3b = -3. This gives f(x)=23x−3f(x) = \frac{2}{3}x - 3. Finally, the xx-intercept is found by solving 23x−3=0\frac{2}{3}x - 3 = 0, resulting in x=4.5x = 4.5.
Estimated Time:2m 0s
Question 71Question →

In the xyxy-plane, line LL passes through the points (0,4)(0, 4) and (6,0)(6, 0). A second line, MM, is perpendicular to line LL and intersects line LL at a point on the line y=xy = x. If line MM is represented by the equation y=px+qy = px + q, where pp and qq are constants, what is the value of p+qp + q?

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Answer: 310\frac{3}{10}

Answer

The correct answer is the value stating three-tenths
To find the sum of the slope pp and yy-intercept qq of line MM, we first determine the equation of line LL. Line LL passes through (0,4)(0, 4) and (6,0)(6, 0), so its slope is 0−46−0=−23\frac{0 - 4}{6 - 0} = -\frac{2}{3}. The equation of line LL is therefore y=−23x+4y = -\frac{2}{3}x + 4. Setting y=xy = x gives x=−23x+4x = -\frac{2}{3}x + 4, which yields x=2.4x = 2.4. The intersection point is (2.4,2.4)(2.4, 2.4). Because line MM is perpendicular to line LL, its slope is the negative reciprocal of −23-\frac{2}{3}, which is p=1.5p = 1.5. Using point-slope form with the intersection point (2.4,2.4)(2.4, 2.4), the equation of line MM is y−2.4=1.5(x−2.4)y - 2.4 = 1.5(x - 2.4), which simplifies to y=1.5x−1.2y = 1.5x - 1.2. Thus, q=−1.2q = -1.2. The sum p+qp + q is 1.5−1.2=0.31.5 - 1.2 = 0.3, or 310\frac{3}{10}.

Step-by-Step Solution

1
Find the equation of line LL using the given points (0,4)(0, 4) and (6,0)(6, 0).
The slope of line LL is mL=0−46−0=−23m_L = \frac{0 - 4}{6 - 0} = -\frac{2}{3}. Since the yy-intercept is (0,4)(0, 4), the equation of line LL is y=−23x+4y = -\frac{2}{3}x + 4.
To find where line LL intersects another line, we first need to establish its linear equation.
2
Find the intersection point of line LL and the line y=xy = x.
Substitute y=xy = x into the equation for line LL: x=−23x+4  ⟹  53x=4  ⟹  x=2.4x = -\frac{2}{3}x + 4 \implies \frac{5}{3}x = 4 \implies x = 2.4. Thus, the intersection point is (2.4,2.4)(2.4, 2.4).
The problem states that line MM intersects line LL at a point on the line y=xy = x.
3
Determine the equation of line MM using the intersection point and the perpendicular relationship.
Since line MM is perpendicular to line LL, its slope pp is the negative reciprocal of −23-\frac{2}{3}, which is p=32=1.5p = \frac{3}{2} = 1.5. Using point-slope form at (2.4,2.4)(2.4, 2.4): y−2.4=1.5(x−2.4)  ⟹  y=1.5x−1.2y - 2.4 = 1.5(x - 2.4) \implies y = 1.5x - 1.2. This gives q=−1.2q = -1.2.
Perpendicular lines in the coordinate plane have slopes that multiply to negative one. We then use the point-slope formula to define the equation of line MM.
4
Calculate the sum of pp and qq.
p+q=1.5+(−1.2)=0.3=310p + q = 1.5 + (-1.2) = 0.3 = \frac{3}{10}.
The question asks for the value of the expression p+qp + q.

Key Concept

Perpendicular lines and linear graph intersections
Estimated Time:3m 0s
Question 72Question →

In the xyxy-plane, the graphs of the linear equations y=3x−5y = 3x - 5 and y=−x+7y = -x + 7 intersect at the point (x,y)(x, y). What is the value of xx?

Show answer & explanation

Answer: 3

Answer

3
To find the xx-value of the intersection point of the two graphs, set the two equations equal to each other because both represent the same yy-value at that point: 3x−5=−x+73x - 5 = -x + 7. Adding xx to both sides of the equation gives 4x−5=74x - 5 = 7. Adding 55 to both sides gives 4x=124x = 12. Dividing both sides by 44 yields x=3x = 3.

Step-by-Step Solution

1
Set the two expressions for yy equal to each other to find the xx-coordinate of the intersection point.
3x−5=−x+73x - 5 = -x + 7
Since both equations are solved for yy, their right-hand sides must be equal at the point of intersection.
2
Add xx to both sides of the equation to collect the variable terms on one side.
4x−5=74x - 5 = 7
Moving the variable terms together allows us to isolate xx.
3
Add 55 to both sides of the equation to isolate the term with xx.
4x=124x = 12
Grouping the constant terms on the opposite side prepares the equation for final division.
4
Divide both sides by 44 to find the value of xx.
x=3x = 3
This isolates xx and gives the final solution.

Key Concept

Solving a system of linear equations by setting the equations equal to find the point of intersection in the coordinate plane.
Question 73Question →
In the equation below, aa and bb are constants.
a(3x−5)−b(2x+2)=4x−12a(3x - 5) - b(2x + 2) = 4x - 12
If the equation has infinitely many solutions for xx, what is the value of a+ba + b?
Show answer & explanation

Answer: 3

Answer

3
To find the value of a+ba + b that results in infinitely many solutions, we first distribute the constants aa and bb to rewrite the equation as (3a−2b)x−(5a+2b)=4x−12(3a - 2b)x - (5a + 2b) = 4x - 12. For a linear equation to have infinitely many solutions, the coefficient of xx on both sides must be equal, and the constant terms on both sides must be equal. This gives the system of equations 3a−2b=43a - 2b = 4 and 5a+2b=125a + 2b = 12. Adding these two equations yields 8a=168a = 16, which gives a=2a = 2. Substituting a=2a = 2 into 3a−2b=43a - 2b = 4 yields 6−2b=46 - 2b = 4, which gives b=1b = 1. Therefore, the value of a+ba + b is 2+1=32 + 1 = 3.

Step-by-Step Solution

1
Distribute the constants aa and bb on the left side of the equation and group the terms.
(3a−2b)x−(5a+2b)=4x−12(3a - 2b)x - (5a + 2b) = 4x - 12
To write the linear equation in the standard form Ax+B=Cx+DAx + B = Cx + D so that coefficients can be compared.
2
Set up a system of equations by equating the coefficient of xx and the constant term on both sides of the equation.
3a−2b=43a - 2b = 4 and −(5a+2b)=−12-(5a + 2b) = -12 (which simplifies to 5a+2b=125a + 2b = 12)
For a linear equation in one variable to have infinitely many solutions, the coefficient of xx on both sides must be equal, and the constant terms on both sides must be equal.
3
Solve the system of equations by adding them to eliminate bb.
8a=16  ⟹  a=28a = 16 \implies a = 2
Adding 3a−2b=43a - 2b = 4 and 5a+2b=125a + 2b = 12 eliminates bb, allowing us to solve directly for aa.
4
Substitute a=2a = 2 back into 3a−2b=43a - 2b = 4 to solve for bb.
3(2)−2b=4  ⟹  6−2b=4  ⟹  −2b=−2  ⟹  b=13(2) - 2b = 4 \implies 6 - 2b = 4 \implies -2b = -2 \implies b = 1
Substituting the value of aa allows us to determine bb.
5
Calculate the value of a+ba + b.
a+b=2+1=3a + b = 2 + 1 = 3
To find the sum of the two constants as requested.

Key Concept

Linear Equations in One Variable (Infinitely Many Solutions)
Question 74Question →

If 23(x−4)=6\frac{2}{3}(x - 4) = 6, what is the value of xx?

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Answer: 13

Answer

13
To solve the linear equation 23(x−4)=6\frac{2}{3}(x - 4) = 6, multiply both sides of the equation by 32\frac{3}{2} to isolate x−4x - 4. This yields x−4=9x - 4 = 9. Adding 44 to both sides of the equation isolates xx and gives the value x=13x = 13.

Step-by-Step Solution

1
Multiply both sides of the equation 23(x−4)=6\frac{2}{3}(x - 4) = 6 by the reciprocal of the fraction, which is 32\frac{3}{2}.
x−4=9x - 4 = 9
Multiplying a fraction by its reciprocal simplifies it to 1, leaving the term in parentheses isolated on the left side.
2
Add 44 to both sides of the equation x−4=9x - 4 = 9.
x=13x = 13
Adding 44 isolates the variable xx on the left side.

Key Concept

Solving a one-variable linear equation using inverse operations.
Question 75Question →

In the xyxy-plane, line ll passes through the point (3,4)(3, 4) and has a slope of mm, where m>0m > 0. Line kk is perpendicular to line ll and passes through the point (2,2)(2, 2). If the sum of the yy-intercepts of line ll and line kk is 11, what is the value of mm?

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Answer: 22

Answer

The correct value of mm is 22.
The correct answer is the value 22. By finding the equations of both lines in slope-intercept form, we express their yy-intercepts in terms of mm: the yy-intercept of line ll is 4−3m4 - 3m and the yy-intercept of line kk is 2m+2\frac{2}{m} + 2. Setting their sum to 11 gives the equation 3m2−5m−2=03m^2 - 5m - 2 = 0. Factoring this quadratic yields the solutions m=−13m = -\frac{1}{3} and m=2m = 2. Since mm must be positive, m=2m = 2 is the only valid solution.

Step-by-Step Solution

1
Find the equation and yy-intercept of line ll.
Using the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) with the point (3,4)(3, 4) and slope mm, the equation of line ll is y−4=m(x−3)y - 4 = m(x - 3), which simplifies to y=mx+4−3my = mx + 4 - 3m. Thus, the yy-intercept of line ll is 4−3m4 - 3m.
Expressing the yy-intercept of line ll in terms of mm allows us to use it in the sum equation.
2
Find the equation and yy-intercept of line kk.
Since line kk is perpendicular to line ll, its slope is the negative reciprocal of mm, which is −1m-\frac{1}{m}. Using the point-slope form with the point (2,2)(2, 2), the equation of line kk is y−2=−1m(x−2)y - 2 = -\frac{1}{m}(x - 2), which simplifies to y=−1mx+2m+2y = -\frac{1}{m}x + \frac{2}{m} + 2. Thus, the yy-intercept of line kk is 2m+2\frac{2}{m} + 2.
Expressing the yy-intercept of line kk in terms of mm allows us to use it in the sum equation.
3
Set up the equation for the sum of the yy-intercepts and solve for mm.
The sum of the yy-intercepts is (4−3m)+(2m+2)=1(4 - 3m) + (\frac{2}{m} + 2) = 1. Simplifying this equation gives 6−3m+2m=1  ⟹  5−3m+2m=06 - 3m + \frac{2}{m} = 1 \implies 5 - 3m + \frac{2}{m} = 0. Multiplying by mm yields 3m2−5m−2=03m^2 - 5m - 2 = 0. Factoring the quadratic gives (3m+1)(m−2)=0(3m + 1)(m - 2) = 0, which has solutions m=−13m = -\frac{1}{3} and m=2m = 2. Since m>0m > 0, we have m=2m = 2.
To determine the unique positive slope that satisfies the given conditions.

Key Concept

Writing equations of perpendicular lines and finding their intercepts using slopes and points.
Question 76Question →

A food truck sells tacos and burritos. On Tuesday, the food truck sold a total of 120 tacos and burritos. The number of tacos sold, tt, was 20 more than the number of burritos sold, bb. This situation is represented by the system of equations below:

t+b=120t + b = 120
t=b+20t = b + 20

How many burritos did the food truck sell on Tuesday?

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Answer: 50

Answer

The food truck sold 50 burritos on Tuesday.
To find the number of burritos sold, bb, we substitute t=b+20t = b + 20 into the first equation t+b=120t + b = 120, giving (b+20)+b=120(b + 20) + b = 120. Combining like terms yields 2b+20=1202b + 20 = 120. Subtracting 20 from both sides results in 2b=1002b = 100. Finally, dividing both sides by 2 gives b=50b = 50. Thus, the food truck sold 50 burritos on Tuesday.

Step-by-Step Solution

1
Substitute the expression for tt from the second equation into the first equation
(b+20)+b=120(b + 20) + b = 120
To eliminate the variable tt and set up an equation with only one variable, bb.
2
Combine like terms on the left side of the equation
2b+20=1202b + 20 = 120
To simplify the equation for solving.
3
Subtract 20 from both sides of the equation
2b=1002b = 100
To isolate the variable term on one side of the equation.
4
Divide both sides of the equation by 2
b=50b = 50
To solve for the variable bb representing the number of burritos.

Key Concept

Solving systems of linear equations using the substitution method
Question 77Question →

During a chemistry experiment, the temperature of a liquid sample is decreased at a constant rate. At the start of the experiment, the temperature of the liquid is 80∘C80^\circ\text{C}. After 44 minutes, the temperature of the liquid is 68∘C68^\circ\text{C}. If T(t)T(t) represents the temperature of the liquid, in degrees Celsius, tt minutes after the experiment starts, which of the following equations defines TT?

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Answer: T(t)=−3t+80T(t) = -3t + 80

Answer

The equation T(t)=−3t+80T(t) = -3t + 80 correctly defines the temperature function.
The correct equation shows a starting value of 8080 and a rate of change of −3-3, which corresponds to the initial temperature of 80∘C80^\circ\text{C} and a decrease of 3∘C3^\circ\text{C} per minute.

Step-by-Step Solution

1
Identify the yy-intercept (initial value) of the linear function.
The initial temperature at t=0t = 0 is 80∘C80^\circ\text{C}, so the yy-intercept bb is 8080.
The initial value of a linear relationship in context corresponds to the yy-intercept.
2
Calculate the slope (constant rate of change) using the coordinate points (0,80)(0, 80) and (4,68)(4, 68).
The slope mm is 68−804−0=−124=−3\frac{68 - 80}{4 - 0} = \frac{-12}{4} = -3.
The slope is defined as the change in the dependent variable (temperature) divided by the change in the independent variable (time).
3
Substitute the slope mm and yy-intercept bb into the slope-intercept form T(t)=mt+bT(t) = mt + b.
T(t)=−3t+80T(t) = -3t + 80
Substituting the specific slope and intercept values into the general form defines the particular function.

Key Concept

Determining a linear equation from a real-world scenario by identifying the initial value and constant rate of change.
Question 78Question →

If 34(8x−12)−12(2x−6)=18\frac{3}{4}(8x - 12) - \frac{1}{2}(2x - 6) = 18, what is the value of 5x5x?

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Answer: 24

Answer

24
By distributing the fractions to the terms inside the parentheses, we get 6x−9−x+3=186x - 9 - x + 3 = 18. Combining like terms on the left side yields 5x−6=185x - 6 = 18. Adding 6 to both sides gives the value of 5x5x as 24.

Step-by-Step Solution

1
Distribute the coefficients outside the parentheses to the terms inside.
34(8x)−34(12)−12(2x)+12(6)=18\frac{3}{4}(8x) - \frac{3}{4}(12) - \frac{1}{2}(2x) + \frac{1}{2}(6) = 18, which simplifies to 6x−9−x+3=186x - 9 - x + 3 = 18.
To eliminate the parentheses and prepare to combine like terms.
2
Combine like terms on the left side of the equation.
(6x−x)+(−9+3)=18(6x - x) + (-9 + 3) = 18, which simplifies to 5x−6=185x - 6 = 18.
To simplify the linear expression on the left side.
3
Isolate the term 5x5x by adding 6 to both sides of the equation.
5x=18+65x = 18 + 6, which simplifies to 5x=245x = 24.
To find the value of 5x5x directly as requested by the question.

Key Concept

Solving linear equations in one variable using the distributive property and combining like terms.
Question 79Question →

In the equation 3(4x+b)−2(x−5)=2(5x+8)3(4x + b) - 2(x - 5) = 2(5x + 8), bb is a constant. If the equation has infinitely many solutions for xx, what is the value of bb?

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Answer: 2

Answer

The value of bb is 2.
Distributing the constants in the equation 3(4x+b)−2(x−5)=2(5x+8)3(4x + b) - 2(x - 5) = 2(5x + 8) yields 12x+3b−2x+10=10x+1612x + 3b - 2x + 10 = 10x + 16. Combining like terms on the left side simplifies the equation to 10x+3b+10=10x+1610x + 3b + 10 = 10x + 16. For a linear equation in one variable to have infinitely many solutions, both sides of the equation must be identical. Since the coefficients of xx are equal (10=1010 = 10), the constant terms must also be equal: 3b+10=163b + 10 = 16. Solving for bb yields 3b=63b = 6, which simplifies to b=2b = 2.

Step-by-Step Solution

1
Distribute the constants through the parentheses on both sides of the equation 3(4x+b)−2(x−5)=2(5x+8)3(4x + b) - 2(x - 5) = 2(5x + 8).
12x+3b−2x+10=10x+1612x + 3b - 2x + 10 = 10x + 16
To eliminate parentheses and allow grouping of like terms.
2
Combine like terms on the left side of the equation.
10x+3b+10=10x+1610x + 3b + 10 = 10x + 16
To simplify the left-hand expression into the standard linear form.
3
Equate the constant terms on both sides of the equation.
3b+10=163b + 10 = 16
A linear equation in one variable has infinitely many solutions when both sides are identical. Since the coefficients of the variable xx are both 10, the constant terms must be equal.
4
Solve for bb by isolating it.
b=2b = 2
Subtracting 10 from both sides gives 3b=63b = 6, and dividing by 3 yields the final value.

Key Concept

Determining conditions for a linear equation in one variable to have infinitely many solutions (identity).
Question 80Question →

In the xyxy-plane, line ll passes through the points (−2,−3)(-2, -3) and (2,5)(2, 5). Line kk is parallel to line ll and has a yy-intercept of (0,−1)(0, -1). If the point (a,9)(a, 9) lies on line kk, what is the value of aa?

Show answer & explanation

Answer: 5

Answer

The value of aa is 55.
To find the value of aa, we first determine the slope of line ll using the two given points, (−2,−3)(-2, -3) and (2,5)(2, 5). The slope mm is given by 5−(−3)2−(−2)=84=2\frac{5 - (-3)}{2 - (-2)} = \frac{8}{4} = 2. Since line kk is parallel to line ll, it has the same slope of 22. The yy-intercept of line kk is (0,−1)(0, -1), so the equation of line kk is y=2x−1y = 2x - 1. To find the value of aa, we substitute the point (a,9)(a, 9) into this equation: 9=2a−19 = 2a - 1. Solving for aa gives 10=2a10 = 2a, which simplifies to a=5a = 5.

Step-by-Step Solution

1
Calculate the slope of line ll using the two given points.
The slope of line ll is 22.
Using the slope formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} with the points (−2,−3)(-2, -3) and (2,5)(2, 5), we get m=5−(−3)2−(−2)=84=2m = \frac{5 - (-3)}{2 - (-2)} = \frac{8}{4} = 2.
2
Find the equation of line kk.
The equation of line kk is y=2x−1y = 2x - 1.
Line kk is parallel to line ll, so it has the same slope, m=2m = 2. Its yy-intercept is (0,−1)(0, -1), which gives the equation y=2x−1y = 2x - 1 in slope-intercept form.
3
Solve for aa by substituting the point (a,9)(a, 9) into the equation of line kk.
The value of aa is 55.
Substituting x=ax = a and y=9y = 9 into y=2x−1y = 2x - 1 gives 9=2a−19 = 2a - 1. Solving for aa yields 10=2a10 = 2a, or a=5a = 5.

Key Concept

Linear Functions and Graphs
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