Examkin

Linear Equations in One Variable

67 soru

Tüm türler67Çoktan seçmeli35Sayısal32
Soru 21Soru →

An online store charges a flat shipping fee of 8.508.50 dollars for orders under 100100 dollars. A customer purchases 55 identical shirts and uses a coupon for 12.0012.00 dollars off the total price of the shirts. If the total charge for the order, including shipping, is 86.5086.50 dollars, what is the price, in dollars, of one shirt?

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Cevap: 18.0018.00

Cevap

The price of one shirt is 18.0018.00 dollars.
To find the price of one shirt, we set up a linear equation. Let ss represent the price of one shirt. The cost of 55 shirts after applying the 12.0012.00 dollar coupon is 5s−12.005s - 12.00. Since this amount is under 100100 dollars, the flat shipping fee of 8.508.50 dollars is added to the total. This gives the equation 5s−12.00+8.50=86.505s - 12.00 + 8.50 = 86.50. Combining the constants on the left side results in 5s−3.50=86.505s - 3.50 = 86.50. Adding 3.503.50 to both sides yields 5s=90.005s = 90.00. Dividing both sides by 55 gives s=18.00s = 18.00. Thus, the price of one shirt is 18.0018.00 dollars.

Adım Adım Çözüm

1
Represent the cost of the shirts using a variable.
Let ss represent the price of one shirt. The cost of 55 shirts with the 12.0012.00 dollar coupon discount is 5s−12.005s - 12.00 dollars.
This establishes the variable for the unknown value we need to find.
2
Set up the full linear equation by including the shipping fee and total charge.
Since 5s−12.005s - 12.00 is under 100100 dollars (which we verify at the end), a flat shipping fee of 8.508.50 dollars applies. The equation is 5s−12.00+8.50=86.505s - 12.00 + 8.50 = 86.50.
This relates all parts of the word problem into a single solvable mathematical statement.
3
Simplify the equation by combining the constant values on the left side.
5s−3.50=86.505s - 3.50 = 86.50
Combining like terms simplifies the algebraic expression.
4
Isolate the variable term by adding 3.503.50 to both sides.
5s=90.005s = 90.00
This moves all constant terms to one side of the equation.
5
Solve for the variable by dividing both sides of the equation by 55.
s=18.00s = 18.00
Dividing by the coefficient of the variable isolates ss to find the final value.

Anahtar Kavram

Linear Equations in One Variable
Soru 22Soru →

If 3(2n−5)=4n+93(2n - 5) = 4n + 9, what is the value of nn?

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Cevap: 12

Cevap

12
To solve the equation 3(2n−5)=4n+93(2n - 5) = 4n + 9, we first distribute the 3 to the terms inside the parentheses to get 6n−15=4n+96n - 15 = 4n + 9. Next, we subtract 4n4n from both sides to group the variable terms, giving 2n−15=92n - 15 = 9. We then add 15 to both sides to isolate the variable term, resulting in 2n=242n = 24. Finally, dividing both sides by 2 gives the solution n=12n = 12.

Adım Adım Çözüm

1
Distribute 3 to the terms inside the parentheses.
6n−15=4n+96n - 15 = 4n + 9
To simplify the left side of the equation and remove parentheses.
2
Subtract 4n4n from both sides of the equation.
2n−15=92n - 15 = 9
To group the variable terms on one side of the equation.
3
Add 15 to both sides of the equation.
2n=242n = 24
To isolate the term with the variable nn.
4
Divide both sides of the equation by 2.
n=12n = 12
To solve for nn.

Anahtar Kavram

Solving a linear equation in one variable by applying the distributive property and isolating the variable.
Soru 23Soru →
In the equation below, aa and bb are positive constants.
23(32x−a)−34(8−bx)=52(x−3)−12\frac{2}{3} \left( \frac{3}{2}x - a \right) - \frac{3}{4} \left( 8 - bx \right) = \frac{5}{2}(x - 3) - \frac{1}{2}
If the equation has infinitely many solutions, what is the value of aba^b?
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Cevap: 9

Cevap

The correct answer is 9.
The correct answer is 9. Expanding the left side of the equation yields (1+34b)x−(6+23a)\left(1 + \frac{3}{4}b\right)x - \left(6 + \frac{2}{3}a\right), and simplifying the right side yields 52x−8\frac{5}{2}x - 8. For the linear equation to have infinitely many solutions, the coefficient of xx on the left, 1+34b1 + \frac{3}{4}b, must equal the coefficient of xx on the right, 52\frac{5}{2}, which gives b=2b = 2. Similarly, the constant term on the left, −(6+23a)-\left(6 + \frac{2}{3}a\right), must equal the constant term on the right, −8-8, which simplifies to 6+23a=86 + \frac{2}{3}a = 8 and gives a=3a = 3. Evaluating aba^b with these values yields 32=93^2 = 9.

Adım Adım Çözüm

1
Expand both sides of the equation to collect like terms.
(1+34b)x−(6+23a)=52x−8\left( 1 + \frac{3}{4}b \right)x - \left( 6 + \frac{2}{3}a \right) = \frac{5}{2}x - 8
Expanding allows us to compare the coefficient of xx and the constant term on each side of the equation.
2
Equate the coefficients of xx on both sides of the equation.
1+34b=52  ⟹  b=21 + \frac{3}{4}b = \frac{5}{2} \implies b = 2
For a linear equation to have infinitely many solutions, the coefficient of xx must be identical on both sides.
3
Equate the constant terms on both sides of the equation.
−(6+23a)=−8  ⟹  a=3-\left( 6 + \frac{2}{3}a \right) = -8 \implies a = 3
For a linear equation to have infinitely many solutions, the constant terms must also be identical on both sides.
4
Calculate the value of aba^b using the solved values of aa and bb.
32=93^2 = 9
The question asks for the value of the expression aba^b where a=3a = 3 and b=2b = 2.

Anahtar Kavram

A linear equation in one variable of the form Ax+B=Cx+DAx + B = Cx + D has infinitely many solutions if and only if A=CA = C and B=DB = D.

Alternatif Yöntem

Since the equation must hold for all values of xx if it has infinitely many solutions, you can substitute convenient values for xx to solve for aa and bb directly. Substituting x=0x = 0 simplifies the equation to −23a−6=−8-\frac{2}{3}a - 6 = -8, which quickly yields a=3a = 3. Then, substituting x=2x = 2 and a=3a = 3 simplifies the equation to −34(8−2b)=−3-\frac{3}{4}(8 - 2b) = -3, which yields b=2b = 2. Calculating aba^b gives 32=93^2 = 9.
Tahmini Süre:3m 0s
Soru 24Soru →

If 53(3x−6)−12(4x+8)=2\frac{5}{3}(3x - 6) - \frac{1}{2}(4x + 8) = 2, what is the value of 3x−53x - 5?

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Cevap: 11

Cevap

11
Distributing the fraction 53\frac{5}{3} to (3x−6)(3x - 6) yields 5x−105x - 10. Distributing the term −12-\frac{1}{2} to (4x+8)(4x + 8) yields −2x−4-2x - 4. Combining these results gives (5x−2x)+(−10−4)=2(5x - 2x) + (-10 - 4) = 2, which simplifies to 3x−14=23x - 14 = 2. Adding 14 to both sides yields 3x=163x = 16. Finally, subtracting 5 from both sides of this equation gives 3x−5=113x - 5 = 11.

Adım Adım Çözüm

1
Distribute 53\frac{5}{3} and −12-\frac{1}{2} to the terms within their respective parentheses.
5x−10−2x−4=25x - 10 - 2x - 4 = 2
This removes the parentheses and allows us to group terms.
2
Combine the variable terms and the constant terms on the left side of the equation.
3x−14=23x - 14 = 2
Simplifying the expression on the left side makes it easier to solve.
3
Add 14 to both sides of the equation to isolate the term 3x3x.
3x=163x = 16
Since the question asks for the value of 3x−53x - 5, isolating 3x3x allows direct evaluation.
4
Subtract 5 from both sides of the equation 3x=163x = 16.
3x−5=113x - 5 = 11
This directly yields the required value without needing to compute the fractional value of xx first.

Anahtar Kavram

Solving linear equations in one variable by distribution, combining like terms, and evaluating expressions.
Tahmini Süre:1m 30s
Soru 25Soru →

In the equation 3(2x−a)−4(x+1)=2x+53(2x - a) - 4(x + 1) = 2x + 5, aa is a constant. If the equation has infinitely many solutions, what is the value of aa?

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Cevap: −3-3

Cevap

The value of aa must be −3-3.
The value of −3-3 is correct. Expanding and simplifying the left side of the equation yields 2x−3a−4=2x+52x - 3a - 4 = 2x + 5. For a linear equation to have infinitely many solutions, the constant terms on both sides must be identical. Setting −3a−4=5-3a - 4 = 5 and solving for aa gives −3a=9-3a = 9, which simplifies to a=−3a = -3.

Adım Adım Çözüm

1
Distribute the coefficients to the terms inside the parentheses on the left side of the equation.
6x−3a−4x−4=2x+56x - 3a - 4x - 4 = 2x + 5
Distributing eliminates the parentheses, allowing like terms to be combined.
2
Combine the variable terms on the left side of the equation.
2x−3a−4=2x+52x - 3a - 4 = 2x + 5
Grouping the xx terms (6x−4x=2x6x - 4x = 2x) simplifies the structure of the equation.
3
Set up the condition for the equation to have infinitely many solutions.
−3a−4=5-3a - 4 = 5
For a linear equation to have infinitely many solutions, the coefficients of xx on both sides must be equal, and the constant terms must also be equal. The xx coefficients are already equal (2=22 = 2), so we equate the constants.
4
Isolate the constant aa by adding 44 to both sides and then dividing by −3-3.
a=−3a = -3
Adding 44 yields −3a=9-3a = 9, and dividing by −3-3 isolates the variable to find the solution.

Anahtar Kavram

Linear Equations in One Variable

Alternatif Yöntem

Instead of fully simplifying the equation, substitute x=0x = 0 directly into the original equation, since a statement with infinitely many solutions must hold true for all values of xx. Substituting x=0x = 0 gives 3(−a)−4(1)=53(-a) - 4(1) = 5, which simplifies directly to −3a−4=5-3a - 4 = 5, yielding a=−3a = -3.
Tahmini Süre:1m 15s
Soru 26Soru →

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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Cevap: 72\frac{7}{2}

Cevap

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}.

Adım Adım Çözüm

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.

Anahtar Kavram

Solving linear equations in one variable by distributing terms, combining like terms, isolating the variable, and evaluating algebraic expressions.
Soru 27Soru →

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

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Cevap: 20

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Solving one-variable linear equations using inverse operations.
Soru 28Soru →

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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Cevap: 1.25

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Conditions for a linear equation in one variable to have no solution
Soru 29Soru →

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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Cevap: 6

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Solving linear equations in one variable by applying the distributive property and combining like terms.
Soru 30Soru →

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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Cevap: 108

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Formulating and solving multi-step linear equations in one variable from real-world word problems.
Tahmini Süre:2m 30s
Soru 31Soru →

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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Cevap: 5

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Solving linear equations in one variable by distributing constants, combining like terms, and evaluating expressions.
Soru 32Soru →

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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Cevap: 2

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

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

Alternatif Yöntem

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.
Tahmini Süre:2m 30s
Soru 33Soru →

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

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Cevap: 15

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

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

Alternatif Yöntem

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.
Tahmini Süre:45s
Soru 34Soru →
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?
Cevabı ve açıklamayı göster

Cevap: 3

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Linear Equations in One Variable (Infinitely Many Solutions)
Soru 35Soru →

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

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Cevap: 13

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Solving a one-variable linear equation using inverse operations.
Soru 36Soru →

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?

Cevabı ve açıklamayı göster

Cevap: 24

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Solving linear equations in one variable using the distributive property and combining like terms.
Soru 37Soru →

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?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Determining conditions for a linear equation in one variable to have infinitely many solutions (identity).
Soru 38Soru →

If 3(x−2)=−x+143(x - 2) = -x + 14, what is the value of xx?

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Cevap: 5

Cevap

5
The correct value is 5. Distributing the 3 on the left side of the equation 3(x−2)=−x+143(x - 2) = -x + 14 yields 3x−6=−x+143x - 6 = -x + 14. Adding xx to both sides gives 4x−6=144x - 6 = 14. Adding 6 to both sides gives 4x=204x = 20. Finally, dividing both sides by 4 yields x=5x = 5.

Adım Adım Çözüm

1
Distribute the 3 to the terms inside the parentheses on the left side of the equation.
3x−6=−x+143x - 6 = -x + 14
To simplify the equation and prepare to isolate the variable, the parentheses must be expanded.
2
Add xx to both sides of the equation to group all variable terms on the left side.
4x−6=144x - 6 = 14
Grouping the variable terms helps in isolating the variable.
3
Add 6 to both sides of the equation to isolate the variable term.
4x=204x = 20
Moving the constant term to the other side isolates the variable term 4x4x.
4
Divide both sides of the equation by 4 to find the value of xx.
x=5x = 5
Dividing by the coefficient of xx gives the final value of the variable.

Anahtar Kavram

Linear Equations in One Variable
Tahmini Süre:45s
Soru 39Soru →

If 3(2n+4)=423(2n + 4) = 42, what is the value of nn?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

5
To solve the linear equation 3(2n+4)=423(2n + 4) = 42, first divide both sides of the equation by 3, which simplifies it to 2n+4=142n + 4 = 14. Next, subtract 4 from both sides of the equation to isolate the term with the variable, giving 2n=102n = 10. Finally, divide both sides of the equation by 2 to solve for the variable, which yields n=5n = 5.

Adım Adım Çözüm

1
Divide both sides of the equation by 3.
2n+4=142n + 4 = 14
To simplify the equation by eliminating the outer constant multiplier.
2
Subtract 4 from both sides of the equation.
2n=102n = 10
To isolate the variable term on one side of the equation.
3
Divide both sides of the equation by 2.
n=5n = 5
To find the value of nn.

Anahtar Kavram

Solving a linear equation in one variable using inverse operations.
Tahmini Süre:45s
Soru 40Soru →

A commercial drone delivery service charges a flat rate of 15.5015.50 dollars per delivery plus a fuel surcharge of 1.801.80 dollars per mile. During a promotional event, the service offers a 20%20\% discount off the fuel surcharge only. If the total cost for a specific delivery during this promotion was 27.0227.02 dollars, how many miles did the drone travel for this delivery?

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

8
The correct answer is 8. The promotional fuel surcharge is reduced by 20%20\% from the normal 1.801.80 dollars per mile, resulting in a rate of 1.80×(1−0.20)=1.441.80 \times (1 - 0.20) = 1.44 dollars per mile. The linear equation modeling the total cost for a delivery of mm miles during the promotion is 15.50+1.44m=27.0215.50 + 1.44m = 27.02. Subtracting 15.5015.50 from both sides gives 1.44m=11.521.44m = 11.52. Dividing both sides by 1.441.44 yields the distance m=8m = 8 miles.

Adım Adım Çözüm

1
Determine the discounted fuel surcharge per mile.
The discounted surcharge is 1.441.44 dollars per mile.
A 20%20\% discount is applied to the original 1.801.80 dollars per mile fuel surcharge: 1.80×(1−0.20)=1.441.80 \times (1 - 0.20) = 1.44.
2
Set up the linear equation representing the total promotional cost.
15.50+1.44m=27.0215.50 + 1.44m = 27.02, where mm represents the number of miles traveled.
The total cost is the sum of the flat rate of 15.5015.50 dollars and the promotional fuel surcharge of 1.441.44 dollars per mile multiplied by the number of miles.
3
Subtract the flat rate from both sides of the equation.
1.44m=11.521.44m = 11.52
To isolate the term with the variable mm, subtract the flat rate of 15.5015.50 from the total promotional cost of 27.0227.02.
4
Solve for the distance by division.
m=8m = 8
Divide both sides of the equation by the promotional rate per mile, 1.441.44, to find the number of miles.

Anahtar Kavram

Formulating and solving a multi-step linear equation in one variable with decimals and percentages to represent a real-world scenario.
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Linear Equations in One Variable Alıştırma Soruları — SAT — Sayfa 2 | Examkin