Algebra

432 soru

Soru 21Soru

Consider the system of equations below.

y=3x4y = 3x - 4
x+y=8x + y = 8

What is the value of xx?

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Cevap: 33

Cevap

The value of xx is 33.
To solve the system, substitute the value of yy from the first equation into the second equation: x+(3x4)=8x + (3x - 4) = 8. Combining the xx terms gives 4x4=84x - 4 = 8. Adding 44 to both sides of the equation results in 4x=124x = 12. Dividing both sides by 44 isolates the variable, giving x=3x = 3.

Adım Adım Çözüm

1
Substitute the expression for yy from the first equation into the second equation.
x+(3x4)=8x + (3x - 4) = 8
This eliminates the variable yy to create a single equation in terms of xx.
2
Combine the like terms on the left side of the equation.
4x4=84x - 4 = 8
Grouping the xx variables simplifies the expression.
3
Add 44 to both sides of the equation to isolate the term with xx.
4x=124x = 12
Performing the inverse operation of subtraction isolates the variable term.
4
Divide both sides of the equation by 44 to solve for xx.
x=3x = 3
Performing the inverse operation of multiplication isolates the variable xx.

Anahtar Kavram

Solving a system of linear equations using the substitution method.
Soru 22Soru

If (x,y)(x, y) is the solution to the system of equations below, what is the value of xx?

y=2x+1y = 2x + 1
3x+2y=163x + 2y = 16
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Cevap: 2

Cevap

The correct answer is 2.
Substituting y=2x+1y = 2x + 1 into the second equation yields 3x+2(2x+1)=163x + 2(2x + 1) = 16. Expanding this expression gives 3x+4x+2=163x + 4x + 2 = 16, which simplifies to 7x+2=167x + 2 = 16. Subtracting 2 from both sides gives 7x=147x = 14, and dividing by 7 yields x=2x = 2.

Adım Adım Çözüm

1
Substitute the expression for yy from the first equation into the second equation.
3x+2(2x+1)=163x + 2(2x + 1) = 16
This reduces the system of two variables to a single equation in terms of xx.
2
Expand and simplify the equation by distributing and combining like terms.
7x+2=167x + 2 = 16
Simplifying the equation makes it easier to isolate the variable xx.
3
Isolate the variable xx by performing inverse operations.
x=2x = 2
Subtracting 2 and then dividing by 7 isolates xx to find its value.

Anahtar Kavram

Solving a system of linear equations using the substitution method.

Alternatif Yöntem

Alternatively, you can multiply the first equation by 2 to get 4x+2y=2-4x + 2y = 2 and subtract it from 3x+2y=163x + 2y = 16 to eliminate yy, which also yields 7x=147x = 14 and thus x=2x = 2.
Tahmini Süre:1m 0s
Soru 23Soru

In the system of linear equations shown below, what value of xx satisfies both equations?

y=2x5y = 2x - 5
3x2y=123x - 2y = 12
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Cevap: 2-2

Cevap

2-2
To find the value of xx, substitute the expression for yy from the first equation, y=2x5y = 2x - 5, into the second equation, 3x2y=123x - 2y = 12. This yields 3x2(2x5)=123x - 2(2x - 5) = 12. Distributing the 2-2 across the terms in the parentheses results in 3x4x+10=123x - 4x + 10 = 12. Simplifying the left side of the equation by combining like terms gives x+10=12-x + 10 = 12. Subtracting 1010 from both sides results in x=2-x = 2. Finally, multiplying or dividing both sides by 1-1 yields x=2x = -2.

Adım Adım Çözüm

1
Substitute the expression for yy from the first equation into the second equation
3x2(2x5)=123x - 2(2x - 5) = 12
To eliminate yy and obtain an equation containing only xx.
2
Distribute 2-2 to the terms inside the parentheses
3x4x+10=123x - 4x + 10 = 12
To remove parentheses and prepare the equation for simplifying.
3
Combine like terms and solve for xx
x+10=12-x + 10 = 12, which gives x=2-x = 2, and thus x=2x = -2
To isolate and find the value of xx.

Anahtar Kavram

Solving systems of linear equations using substitution
Soru 24Soru

A line representing the linear function ff is graphed in the xyxy-plane. The line intersects the yy-axis at (0,3)(0, 3) and also passes through the point (2,11)(2, 11). What is the slope of this line?

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Cevap: 4

Cevap

The slope of the line is 4.
The slope mm of a line passing through two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by the formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}. Substituting the given points (0,3)(0, 3) and (2,11)(2, 11) into the formula gives m=11320=82=4m = \frac{11 - 3}{2 - 0} = \frac{8}{2} = 4.

Adım Adım Çözüm

1
Identify two coordinates on the line from the given information.
The points are (0,3)(0, 3) and (2,11)(2, 11).
To find the slope of a line, we need at least two points on that line.
2
Apply the slope formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}.
m=11320m = \frac{11 - 3}{2 - 0}
The slope formula calculates the ratio of the vertical change (rise) to the horizontal change (run).
3
Perform the subtraction and division to find the slope value.
m=82=4m = \frac{8}{2} = 4
Simplifying the fraction gives the final numerical slope.

Anahtar Kavram

Calculating the slope of a linear function given two points
Soru 25Soru

If 3(k+4)=5k83(k + 4) = 5k - 8, what is the value of kk?

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Cevap: 10

Cevap

10
Distributing the 33 on the left side yields 3k+12=5k83k + 12 = 5k - 8. Subtracting 3k3k from both sides yields 12=2k812 = 2k - 8. Adding 88 to both sides yields 20=2k20 = 2k. Dividing by 22 gives the solution k=10k = 10.

Adım Adım Çözüm

1
Distribute 33 to the terms inside the parentheses
3k+12=5k83k + 12 = 5k - 8
To simplify the left side of the equation
2
Subtract 3k3k from both sides of the equation
12=2k812 = 2k - 8
To group the variable terms on one side
3
Add 88 to both sides of the equation
20=2k20 = 2k
To isolate the variable term
4
Divide both sides by 22
k=10k = 10
To solve for kk

Anahtar Kavram

Solving linear equations with variables on both sides using distributive property
Soru 26Soru
In the equation below, kk is a constant.
5x2(3k)x=9x15x - 2(3 - k)x = 9x - 1
If the equation has no solution, what is the value of kk?
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Cevap: 5

Cevap

5
To find the value of kk for which the equation has no solution, we simplify the left side of the equation: 5x2(3k)x=5x6x+2kx=x+2kx=(2k1)x5x - 2(3 - k)x = 5x - 6x + 2kx = -x + 2kx = (2k - 1)x. This yields the simplified equation (2k1)x=9x1(2k - 1)x = 9x - 1. A linear equation in one variable has no solution when the variable terms on both sides cancel each other out (meaning the coefficients of xx are equal) but the constant terms are different. Setting the coefficients of xx equal to each other gives 2k1=92k - 1 = 9. Adding 11 to both sides gives 2k=102k = 10, and dividing by 22 yields k=5k = 5. Since the constant term on the left side is 00 and the constant term on the right side is 1-1, the constants are different, confirming that the equation has no solution when kk equals 55.

Adım Adım Çözüm

1
Distribute the term 2-2 to the expressions inside the parentheses on the left side of the equation.
5x6x+2kx=9x15x - 6x + 2kx = 9x - 1
To eliminate the parentheses so that all like terms can be grouped.
2
Combine the xx terms on the left side of the equation.
x+2kx=9x1-x + 2kx = 9x - 1, which can be factored as (2k1)x=9x1(2k - 1)x = 9x - 1
Grouping the coefficients of the variable xx allows direct comparison of both sides of the linear equation.
3
Set the coefficient of xx on the left side equal to the coefficient of xx on the right side.
2k1=92k - 1 = 9
For a linear equation in one variable to have no solution, the variable terms on both sides must cancel out (meaning their coefficients must be equal), while the constant terms must remain unequal.
4
Solve the resulting equation for kk.
2k=10    k=52k = 10 \implies k = 5
Isolating kk by adding 11 to both sides and then dividing by 22 determines the specific constant value.
5
Verify that the constant terms are different when k=5k = 5.
Substitute k=5k = 5 back into the original equation to get 9x=9x19x = 9x - 1, which simplifies to 0=10 = -1.
Since 0=10 = -1 is a false statement, the equation has no solution, confirming that k=5k = 5 is correct.

Anahtar Kavram

Linear Equations in One Variable

Alternatif Yöntem

Instead of algebraic simplification, the value of kk can be found by substituting the answer choices into the equation to see which value eliminates the variable xx while leaving an untrue statement. Plugging in 55 for kk gives 5x2(35)x=9x1    5x2(2)x=9x1    9x=9x1    0=15x - 2(3 - 5)x = 9x - 1 \implies 5x - 2(-2)x = 9x - 1 \implies 9x = 9x - 1 \implies 0 = -1. Because this statement is false, the equation has no solution, verifying that 55 is the correct answer.
Tahmini Süre:2m 0s
Soru 27Soru

In the equation 34(8x12)+kx=10x9\frac{3}{4}(8x - 12) + kx = 10x - 9, kk is a constant. If the equation has infinitely many solutions, what is the value of kk?

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Cevap: 4

Cevap

The value of kk is 44.
Distributing the fraction on the left side of the equation yields 34(8x)34(12)=6x9\frac{3}{4}(8x) - \frac{3}{4}(12) = 6x - 9. Substituting this back gives 6x9+kx=10x96x - 9 + kx = 10x - 9. Factoring out xx on the left side gives (6+k)x9=10x9(6 + k)x - 9 = 10x - 9. For a linear equation to have infinitely many solutions, the coefficients of xx on both sides must be identical, and the constants must be identical. Since the constants on both sides are already 9-9, we set the coefficients equal: 6+k=106 + k = 10. Subtracting 6 from both sides gives the correct value k=4k = 4.

Adım Adım Çözüm

1
Distribute the fraction 34\frac{3}{4} to the terms inside the parentheses.
6x9+kx=10x96x - 9 + kx = 10x - 9
To simplify the expression and eliminate the parentheses.
2
Factor out xx from the terms on the left side of the equation.
(6+k)x9=10x9(6 + k)x - 9 = 10x - 9
To group the xx terms together to easily compare coefficients.
3
Set the coefficient of xx on the left side equal to the coefficient of xx on the right side.
6+k=106 + k = 10
For the equation to have infinitely many solutions, the coefficients of the variable on both sides must be equal when the constant terms are equal.
4
Solve for kk by subtracting 6 from both sides of the equation.
k=4k = 4
To isolate the constant kk.

Anahtar Kavram

A linear equation in one variable has infinitely many solutions when it can be simplified to an identity of the form Ax+B=Ax+BAx + B = Ax + B, meaning both the coefficients of xx and the constant terms on both sides of the equation are equal.
Soru 28Soru

In the xyxy-plane, the graph of a linear function ff passes through the points (0,1)(0, 1) and (2,k)(2, k). The graph of another linear function gg is perpendicular to the graph of ff and passes through the points (4,8)(-4, 8) and (2,k)(2, k). Which of the following could be the value of kk?

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Cevap: 5

Cevap

5
The slope of the line representing function ff is calculated as mf=k12m_f = \frac{k - 1}{2}, and the slope of the line representing function gg is mg=k86m_g = \frac{k - 8}{6}. Since the lines are perpendicular, the product of their slopes must be 1-1. This gives the equation k12k86=1\frac{k - 1}{2} \cdot \frac{k - 8}{6} = -1, which simplifies to (k1)(k8)=12(k - 1)(k - 8) = -12. Expanding and setting the quadratic equation to zero gives k29k+20=0k^2 - 9k + 20 = 0. Factoring this expression yields (k4)(k5)=0(k - 4)(k - 5) = 0. Thus, kk can be either 44 or 55. Since 5 is the only option listed, the option containing 5 is the correct answer.

Adım Adım Çözüm

1
Express the slope of the linear function ff in terms of kk.
mf=k120=k12m_f = \frac{k - 1}{2 - 0} = \frac{k - 1}{2}
The slope formula is m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}.
2
Express the slope of the linear function gg in terms of kk.
mg=k82(4)=k86m_g = \frac{k - 8}{2 - (-4)} = \frac{k - 8}{6}
The slope formula is applied to the points (4,8)(-4, 8) and (2,k)(2, k).
3
Apply the perpendicular lines condition to set up an equation for kk.
mfmg=1    (k12)(k86)=1m_f \cdot m_g = -1 \implies \left(\frac{k - 1}{2}\right)\left(\frac{k - 8}{6}\right) = -1
Perpendicular lines have slopes that are negative reciprocals of each other, meaning their product is 1-1.
4
Solve the quadratic equation for kk.
(k1)(k8)=12    k29k+8=12    k29k+20=0    (k4)(k5)=0(k - 1)(k - 8) = -12 \implies k^2 - 9k + 8 = -12 \implies k^2 - 9k + 20 = 0 \implies (k - 4)(k - 5) = 0
Multiply both sides by 12, expand the product, move all terms to one side, and factor the quadratic expression.
5
Identify the possible values of kk and match with the options.
k=4k = 4 or k=5k = 5. Since 5 is the only value present in the options, the correct value is 5.
Both values satisfy the condition, but only one is listed among the multiple-choice options.

Anahtar Kavram

Perpendicular lines in the coordinate plane have slopes that are negative reciprocals of each other (their product is 1-1).
Soru 29Soru

In the equation below, aa and bb are constants.

13(2a5x)34(xb)=2912x+5\frac{1}{3}(2a - 5x) - \frac{3}{4}(x - b) = -\frac{29}{12}x + 5

If the equation has infinitely many solutions for xx, what is the value of 8a+9b8a + 9b?

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Cevap: 60

Cevap

The value of 8a+9b8a + 9b is 6060.
To find the value of 8a+9b8a + 9b that makes the equation have infinitely many solutions, we first expand and simplify the left side of the equation: 23a53x34x+34b=2912x+5\frac{2}{3}a - \frac{5}{3}x - \frac{3}{4}x + \frac{3}{4}b = -\frac{29}{12}x + 5. Combining the xx terms gives -\frac{29}{12}x + \(\frac{2}{3}a + \frac{3}{4}b\) = -\frac{29}{12}x + 5. Since the coefficients of xx on both sides are equal (2912-\frac{29}{12}), the equation will have infinitely many solutions if the constant terms on both sides are also equal. This requires 23a+34b=5\frac{2}{3}a + \frac{3}{4}b = 5. Multiplying this entire equation by the least common multiple of the denominators, which is 12, yields 8a+9b=608a + 9b = 60.

Adım Adım Çözüm

1
Distribute the constants through the parentheses on the left side of the equation.
23a53x34x+34b=2912x+5\frac{2}{3}a - \frac{5}{3}x - \frac{3}{4}x + \frac{3}{4}b = -\frac{29}{12}x + 5
This separates the variable terms from the constant terms so the equation can be simplified.
2
Combine the coefficients of the xx terms on the left side using 12 as the common denominator.
2912x+23a+34b=2912x+5-\frac{29}{12}x + \frac{2}{3}a + \frac{3}{4}b = -\frac{29}{12}x + 5
Simplifying the variable terms allows us to compare the coefficients on both sides of the equation.
3
Equate the constant terms from the left and right sides of the equation.
23a+34b=5\frac{2}{3}a + \frac{3}{4}b = 5
A linear equation has infinitely many solutions when the coefficients of the variable on both sides are equal and the constant terms on both sides are also equal.
4
Multiply both sides of the equation by 12 to eliminate the fractional denominators.
8a+9b=608a + 9b = 60
Multiplying the equation by the common denominator directly evaluates the target expression 8a+9b8a + 9b.

Anahtar Kavram

For a linear equation in one variable to have infinitely many solutions, it must be reducible to an identity of the form cx+d=cx+dcx + d = cx + d, where both the variable coefficients and the constant terms on both sides of the equation are equal.
Soru 30Soru

If 202(x+3)=820 - 2(x + 3) = 8, what is the value of xx?

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Cevap: 3

Cevap

3
The correct answer is 3. Distributing 2-2 to the terms inside the parentheses gives 202x6=820 - 2x - 6 = 8. Simplifying the left side yields 142x=814 - 2x = 8. Subtracting 14 from both sides results in 2x=6-2x = -6. Dividing both sides by 2-2 gives x=3x = 3.

Adım Adım Çözüm

1
Distribute the coefficient 2-2 to all terms inside the parentheses (x+3)(x + 3).
202x6=820 - 2x - 6 = 8
Applying the distributive property requires multiplying both xx and 33 by 2-2.
2
Combine the constant terms 2020 and 6-6 on the left side of the equation.
142x=814 - 2x = 8
Simplifying the expression makes it easier to isolate the variable term.
3
Subtract 1414 from both sides of the equation to isolate the variable term 2x-2x.
2x=6-2x = -6
This moves the constant term to the opposite side of the equation.
4
Divide both sides of the equation by 2-2 to solve for xx.
x=3x = 3
Dividing by the coefficient of the variable isolates the variable completely.

Anahtar Kavram

Solving linear equations in one variable with parentheses by distributing coefficients, combining like terms, and isolating the variable.
Soru 31Soru

A factory manufactures solar panels at a constant rate. At the start of a morning shift, the factory has already manufactured 8080 solar panels. The total number of solar panels manufactured tt hours after the shift begins is modeled by a linear function. If the factory has manufactured a total of 200200 solar panels 33 hours after the shift begins, how many hours after the shift begins will the factory have manufactured a total of 440440 solar panels?

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Cevap: 9

Cevap

9
To find the time when the total number of solar panels reaches 440440, we first determine the constant rate of production. The change in solar panels over the first 33 hours is 20080=120200 - 80 = 120 panels. Dividing by the 33 hours gives a constant rate of 4040 panels per hour. The linear equation representing the total panels is N(t)=40t+80N(t) = 40t + 80. Setting N(t)=440N(t) = 440 gives 440=40t+80440 = 40t + 80, which simplifies to 360=40t360 = 40t. Solving for tt yields t=9t = 9.

Adım Adım Çözüm

1
Calculate the constant rate of production (slope) of the linear function.
The rate is 4040 solar panels per hour.
The factory starts with 8080 solar panels and reaches 200200 solar panels in 33 hours. The rate of change is the change in the number of panels divided by the change in time: 2008030=1203=40\frac{200 - 80}{3 - 0} = \frac{120}{3} = 40 panels per hour.
2
Write the linear equation representing the total number of solar panels manufactured, N(t)N(t), after tt hours.
N(t)=40t+80N(t) = 40t + 80
Since the initial quantity is 8080 and the rate of production is 4040 panels per hour, the linear function is N(t)=40t+80N(t) = 40t + 80.
3
Solve for the time tt when the total number of solar panels is 440440.
t=9t = 9 hours
Set N(t)=440N(t) = 440 in the equation: 440=40t+80440 = 40t + 80. Subtract 8080 from both sides to get 360=40t360 = 40t. Divide both sides by 4040 to find t=9t = 9.

Anahtar Kavram

Linear Functions and Graphs
Tahmini Süre:1m 30s
Soru 32Soru

In the equation 12(kx4)23(xk)=56x1\frac{1}{2}(kx - 4) - \frac{2}{3}(x - k) = \frac{5}{6}x - 1, where kk is a constant, the equation has no solution for xx. What is the value of kk?

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Cevap: 3

Cevap

The value of the constant kk for which the equation has no solution is 33.
For a linear equation of the form Ax+B=Cx+DAx + B = Cx + D to have no solution, the coefficients of the variable terms must be equal (A=CA = C) but the constant terms must be unequal (BDB \neq D). Distributing the terms on the left side of the given equation and grouping them yields (12k23)x+(23k2)=56x1\left(\frac{1}{2}k - \frac{2}{3}\right)x + \left(\frac{2}{3}k - 2\right) = \frac{5}{6}x - 1. Setting the coefficients of xx equal to each other gives 12k23=56\frac{1}{2}k - \frac{2}{3} = \frac{5}{6}. Adding 23\frac{2}{3} to both sides yields 12k=96\frac{1}{2}k = \frac{9}{6}, or 12k=32\frac{1}{2}k = \frac{3}{2}, which simplifies to k=3k = 3. Evaluating the constant terms when k=3k = 3 gives 23(3)2=0\frac{2}{3}(3) - 2 = 0 on the left side and 1-1 on the right side. Since 010 \neq -1, the equation has no solution when k=3k = 3.

Adım Adım Çözüm

1
Distribute the fractional coefficients on the left side of the equation.
12kx223x+23k=56x1\frac{1}{2}kx - 2 - \frac{2}{3}x + \frac{2}{3}k = \frac{5}{6}x - 1
To separate the variable terms from the constants for further algebraic manipulation.
2
Group the terms on the left side into a coefficient for xx and a single constant term.
(12k23)x+(23k2)=56x1\left(\frac{1}{2}k - \frac{2}{3}\right)x + \left(\frac{2}{3}k - 2\right) = \frac{5}{6}x - 1
To represent the equation in the standard linear format Ax+B=Cx+DAx + B = Cx + D.
3
Set the coefficients of xx from both sides equal to each other.
12k23=56\frac{1}{2}k - \frac{2}{3} = \frac{5}{6}
A linear equation has no solution when the variable terms on both sides cancel out, requiring their coefficients to be identical.
4
Solve the linear equation to isolate the constant kk.
k=3k = 3
Add 23\frac{2}{3} to both sides to get 12k=96\frac{1}{2}k = \frac{9}{6}, which simplifies to 12k=32\frac{1}{2}k = \frac{3}{2}. Multiplying both sides by 22 yields k=3k = 3.
5
Verify that the constant terms are not equal when substituting k=3k = 3.
The left-side constant is 00 and the right-side constant is 1-1. Since 010 \neq -1, the condition is satisfied.
To ensure the equation does not simplify to an identity with infinitely many solutions (which occurs when both the variable coefficients and the constant terms are equal).

Anahtar Kavram

Determining the conditions under which a linear equation in one variable has no solution.

Alternatif Yöntem

To avoid working with fractions, multiply every term in the equation by 66 (the least common multiple of 2,3,62, 3, 6) at the start: 3(kx4)4(xk)=5x63(kx - 4) - 4(x - k) = 5x - 6. Expand the parentheses to get 3kx124x+4k=5x63kx - 12 - 4x + 4k = 5x - 6, and group the terms: (3k4)x+(4k12)=5x6(3k - 4)x + (4k - 12) = 5x - 6. For there to be no solution, set the coefficients of xx equal to each other: 3k4=5    3k=9    k=33k - 4 = 5 \implies 3k = 9 \implies k = 3. Check the constant terms with k=3k = 3: 4(3)12=04(3) - 12 = 0, which is unequal to 6-6. This confirms k=3k = 3 is the correct answer.
Tahmini Süre:2m 0s
Soru 33Soru

If 23(x6)12(x4)=4\frac{2}{3}(x - 6) - \frac{1}{2}(x - 4) = 4, what is the value of 3x123x - 12?

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Cevap: 96

Cevap

96
The correct answer is 96. Multiplying the entire equation by the common denominator 6 clears the fractions to yield 4(x6)3(x4)=244(x - 6) - 3(x - 4) = 24. Distributing the factors gives 4x243x+12=244x - 24 - 3x + 12 = 24, which simplifies to x12=24x - 12 = 24. Adding 12 to both sides determines that x=36x = 36. Evaluating the expression 3x123x - 12 with x=36x = 36 yields 3(36)12=963(36) - 12 = 96.

Adım Adım Çözüm

1
Multiply the entire equation by 6 (the least common multiple of the denominators 3 and 2) to eliminate the fractions.
4(x6)3(x4)=244(x - 6) - 3(x - 4) = 24
Clearing the denominators simplifies the algebraic manipulation and reduces the risk of fractional arithmetic errors.
2
Distribute the coefficients to the terms inside the parentheses, taking care to distribute the negative sign for the second term.
4x243x+12=244x - 24 - 3x + 12 = 24
Applying the distributive property expands the expression so that like terms can be grouped.
3
Combine like terms on the left side of the equation.
x12=24x - 12 = 24
Grouping 4x3x4x - 3x to get xx and 24+12-24 + 12 to get 12-12 simplifies the equation to a single variable and constant.
4
Isolate the variable xx by adding 12 to both sides of the equation.
x=36x = 36
Adding the opposite of the constant term isolates the variable on one side.
5
Substitute x=36x = 36 into the requested expression 3x123x - 12 to find its value.
3(36)12=963(36) - 12 = 96
The question asks for the value of the expression 3x123x - 12, not just the value of xx.

Anahtar Kavram

Solving multi-step linear equations in one variable and evaluating algebraic expressions.
Tahmini Süre:1m 30s
Soru 34Soru

In the xyxy-plane, the graph of the linear function f(x)=px+qf(x) = px + q, where pp and qq are constants, passes through the point (2,7)(2, 7). Line LL is parallel to the graph of ff and has a yy-intercept that is 33 units below the yy-intercept of the graph of ff. If the xx-intercept of line LL is (6,0)(-6, 0), what is the value of pp?

Cevabı ve açıklamayı göster

Cevap: 0.5

Cevap

The value of pp is 0.50.5 (or the fraction 1/21/2).
The correct answer is 0.50.5 (or 1/21/2). The graph of f(x)=px+qf(x) = px + q passes through (2,7)(2, 7), which means 7=2p+q7 = 2p + q, or q=72pq = 7 - 2p. Line LL is parallel to ff, so its slope is pp, and its yy-intercept is q3q - 3. Thus, the equation of line LL is y=px+q3y = px + q - 3. Since line LL has an xx-intercept at (6,0)(-6, 0), we can substitute x=6x = -6 and y=0y = 0 into its equation, yielding 0=6p+q30 = -6p + q - 3. Substituting q=72pq = 7 - 2p into this equation gives 0=6p+(72p)30 = -6p + (7 - 2p) - 3, which simplifies to 8p+4=0-8p + 4 = 0. Solving for pp gives 8p=48p = 4, or p=0.5p = 0.5.

Adım Adım Çözüm

1
Express the relationship between pp and qq using the given point (2,7)(2, 7) on the graph of ff.
q=72pq = 7 - 2p
The point (2,7)(2, 7) must satisfy the equation f(x)=px+qf(x) = px + q.
2
Formulate the equation of line LL using the parallel slope and the shifted yy-intercept.
y=px+q3y = px + q - 3
Parallel lines have equal slopes, and the yy-intercept of LL is 33 units below the yy-intercept of ff, which is qq.
3
Substitute the xx-intercept (6,0)(-6, 0) into the equation of line LL.
6p+q3=0-6p + q - 3 = 0
The xx-intercept is a point on the line where y=0y = 0.
4
Substitute the expression for qq from Step 1 into the equation from Step 3 and solve for pp.
p=0.5p = 0.5 (or 12\frac{1}{2})
Solving the linear equation 6p+(72p)3=0-6p + (7 - 2p) - 3 = 0 simplifies to 8p+4=0-8p + 4 = 0, giving p=0.5p = 0.5.

Anahtar Kavram

Understanding linear functions, their graphs, slopes of parallel lines, and intercepts.

Alternatif Yöntem

Instead of solving for qq first, you can use the point-slope form. Line LL passes through (6,0)(-6, 0) and has slope pp, so its equation is y=p(x+6)y = p(x + 6), or y=px+6py = px + 6p. The yy-intercept of LL is 6p6p. The yy-intercept of the graph of ff is qq. We are given that the yy-intercept of LL is 33 units below the yy-intercept of ff, so 6p=q36p = q - 3. Since ff passes through (2,7)(2, 7), we have 7=2p+q7 = 2p + q, which means q=72pq = 7 - 2p. Substituting this into 6p=q36p = q - 3 gives 6p=(72p)36p = (7 - 2p) - 3, or 6p=42p6p = 4 - 2p. Adding 2p2p to both sides gives 8p=48p = 4, which results in p=0.5p = 0.5.
Tahmini Süre:2m 30s
Soru 35Soru

A rental company charges a flat fee of 3535 dollars plus 1212 dollars per hour to rent a bicycle. If a customer was charged a total of 9595 dollars, for how many hours did the customer rent the bicycle?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

5
To find the number of hours the bicycle was rented, we can set up the equation 12h+35=9512h + 35 = 95, where hh is the number of hours. Subtracting 3535 from both sides gives 12h=6012h = 60. Dividing both sides by 1212 gives h=5h = 5. Thus, the customer rented the bicycle for 55 hours.

Adım Adım Çözüm

1
Set up a linear equation representing the total cost.
12h+35=9512h + 35 = 95, where hh represents the number of hours rented.
The total cost consists of a flat fee of 3535 dollars plus an hourly charge of 1212 dollars multiplied by the number of hours hh.
2
Subtract the flat fee from both sides of the equation to isolate the variable term.
12h=6012h = 60
Subtracting 3535 from both sides simplifies the equation to find the total hourly charge portion of the cost.
3
Divide by the hourly rate to solve for the number of hours hh.
h=5h = 5
Dividing both sides by 1212 isolates hh to find the number of rental hours.

Anahtar Kavram

Linear Equations in One Variable
Soru 36Soru

In the xyxy-plane, the graph of the linear function ff has a slope of 3-3 and passes through the point (4,18)(4, 18). What is the yy-coordinate of the yy-intercept of the graph of ff?

Cevabı ve açıklamayı göster

Cevap: 30

Cevap

The correct answer is 30.
The equation of a linear function can be written in slope-intercept form as f(x)=mx+bf(x) = mx + b, where mm is the slope and bb is the yy-coordinate of the yy-intercept. Substituting the given slope m=3m = -3 yields f(x)=3x+bf(x) = -3x + b. Since the point (4,18)(4, 18) lies on the graph of ff, we substitute x=4x = 4 and f(x)=18f(x) = 18 into the equation to get 18=3(4)+b18 = -3(4) + b, which simplifies to 18=12+b18 = -12 + b. Adding 1212 to both sides of the equation yields b=30b = 30. Thus, the yy-coordinate of the yy-intercept of the graph of ff is 3030.

Adım Adım Çözüm

1
Write the general slope-intercept form of a linear equation.
f(x)=3x+bf(x) = -3x + b
The slope of the line is given as 3-3, so we substitute m=3m = -3 into f(x)=mx+bf(x) = mx + b.
2
Substitute the point (4,18)(4, 18) into the equation.
18=3(4)+b18 = -3(4) + b
Since the graph of ff passes through (4,18)(4, 18), these coordinates must satisfy the function equation.
3
Solve for the yy-intercept bb.
b=30b = 30
Simplify to 18=12+b18 = -12 + b, and then add 1212 to both sides to isolate bb.

Anahtar Kavram

Finding the equation of a line using its slope and a point.
Soru 37Soru

The table below shows some values of the linear function gg.

xxg(x)g(x)
2255
441111
661717

What is the slope of the graph of gg in the xyxy-plane?

Cevabı ve açıklamayı göster

Cevap: 33

Cevap

The slope of the graph is 33.
The slope of a linear function is the constant rate of change, calculated as the change in the function values divided by the change in the input values. Choosing the points (2,5)(2, 5) and (4,11)(4, 11), the slope is 11542=62=3\frac{11 - 5}{4 - 2} = \frac{6}{2} = 3. This same slope is obtained using any other pair of points from the table, such as (4,11)(4, 11) and (6,17)(6, 17), where the slope is 171164=62=3\frac{17 - 11}{6 - 4} = \frac{6}{2} = 3.

Adım Adım Çözüm

1
Select two points from the table to find the change in the input and output values.
Using the points (2,5)(2, 5) and (4,11)(4, 11), the change in the output g(x)g(x) is 115=611 - 5 = 6, and the change in the input xx is 42=24 - 2 = 2.
The slope is defined as the change in the vertical coordinate divided by the change in the horizontal coordinate.
2
Divide the change in g(x)g(x) by the change in xx to find the slope.
The slope is 62=3\frac{6}{2} = 3.
Dividing the vertical change by the horizontal change yields the constant rate of change of the linear function.

Anahtar Kavram

Finding the slope of a linear function from a table of values
Tahmini Süre:45s
Soru 38Soru

The graph of a linear function is shown in the xyxy-plane. The line passes through the points (2,3)(-2, -3) and (2,5)(2, 5). If the point (k,11)(k, 11) lies on the line, what is the value of kk?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

5
The slope of the line is calculated as m=5(3)2(2)=2m = \frac{5 - (-3)}{2 - (-2)} = 2. Using the point-slope form with the point (2,5)(2, 5), the equation of the line is y5=2(x2)y - 5 = 2(x - 2), which simplifies to y=2x+1y = 2x + 1. Substituting the point (k,11)(k, 11) into this equation gives 11=2k+111 = 2k + 1, which solves to k=5k = 5.

Adım Adım Çözüm

1
Calculate the slope of the line using the points (2,3)(-2, -3) and (2,5)(2, 5)
m=2m = 2
The slope formula is m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}. Substituting the given coordinates: m=5(3)2(2)=84=2m = \frac{5 - (-3)}{2 - (-2)} = \frac{8}{4} = 2.
2
Determine the equation of the line using point-slope form
y=2x+1y = 2x + 1
Using yy1=m(xx1)y - y_1 = m(x - x_1) with the point (2,5)(2, 5) and m=2m = 2, we get y5=2(x2)y - 5 = 2(x - 2), which simplifies to y=2x+1y = 2x + 1.
3
Substitute (k,11)(k, 11) into the line equation and solve for kk
k=5k = 5
Substituting x=kx = k and y=11y = 11 yields 11=2k+111 = 2k + 1. Subtracting 1 from both sides gives 10=2k10 = 2k, and dividing by 2 gives k=5k = 5.

Anahtar Kavram

Finding the linear equation from coordinates and solving for variables
Soru 39Soru

In the equation below, mm is a constant.

35(5x10)2(xm)=18\frac{3}{5}(5x - 10) - 2(x - m) = 18

If the solution to the equation is x=12x = 12, what is the value of mm?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The correct value of mm is 6.
Substituting x=12x = 12 into the equation yields 35(5(12)10)2(12m)=18\frac{3}{5}(5(12) - 10) - 2(12 - m) = 18, which simplifies to 3024+2m=1830 - 24 + 2m = 18. Combining constant terms gives 6+2m=186 + 2m = 18. Subtracting 6 from both sides gives 2m=122m = 12, and dividing by 2 yields m=6m = 6.

Adım Adım Çözüm

1
Substitute x=12x = 12 into the given equation.
35(5(12)10)2(12m)=18\frac{3}{5}(5(12) - 10) - 2(12 - m) = 18
Since x=12x = 12 is the solution, it must satisfy the equation.
2
Simplify the expression inside the first set of parentheses.
5(12)10=6010=505(12) - 10 = 60 - 10 = 50
Evaluate the terms inside the parentheses first.
3
Multiply the simplified term by the fraction 35\frac{3}{5}.
35(50)=30\frac{3}{5}(50) = 30
Multiply 50 by 3 and divide by 5.
4
Substitute 30 back into the equation and distribute 2-2 to the terms inside the second set of parentheses.
3024+2m=1830 - 24 + 2m = 18
Distributing 2-2 to 12m12 - m yields 24+2m-24 + 2m due to sign rules.
5
Combine the constant terms on the left side of the equation.
6+2m=186 + 2m = 18
3024=630 - 24 = 6.
6
Subtract 6 from both sides to isolate the term with mm.
2m=122m = 12
Isolate the variable term on one side of the equation.
7
Divide both sides by 2 to solve for mm.
m=6m = 6
Divide 12 by 2 to find the final value.

Anahtar Kavram

Solving a linear equation in one variable by substitution and isolation.
Soru 40Soru

In the xyxy-plane, line kk has the equation y=2x+10y = -2x + 10. Line mm is parallel to line kk and passes through the point (6,8)(6, 8). Line ll is perpendicular to line kk and intersects the xx-axis at the point (a,0)(a, 0), where a>10a > 10. If the region bounded by lines kk, mm, ll, and the yy-axis has an area of 110110, what is the value of aa?

Cevabı ve açıklamayı göster

Cevap: 25

Cevap

25
To find the value of aa, we determine the equations of the lines mm and ll based on their geometric relationships to line kk. Line mm is parallel to line kk (y=2x+10y = -2x + 10), so its slope is 2-2. Using the point (6,8)(6, 8), its equation is y=2x+20y = -2x + 20. Line ll is perpendicular to line kk, so its slope is 12\frac{1}{2}. It intersects the xx-axis at (a,0)(a, 0), giving the equation y=12x12ay = \frac{1}{2}x - \frac{1}{2}a. The bounded region formed by the parallel lines kk and mm, the perpendicular line ll, and the yy-axis is a trapezoid. Calculating the area of this trapezoid by dividing it into a parallelogram and a triangle yields the area formula 60+2a60 + 2a. Setting this equal to the given area of 110110 yields 60+2a=11060 + 2a = 110, which solves to a=25a = 25.

Adım Adım Çözüm

1
Determine the equation of line mm using the parallel slope and the given point.
Line mm has the equation y=2x+20y = -2x + 20.
Parallel lines have equal slopes. Since line kk has a slope of 2-2, line mm also has a slope of 2-2. Substituting the point (6,8)(6, 8) into the point-slope form gives y8=2(x6)y - 8 = -2(x - 6).
2
Determine the equation of line ll using the perpendicular slope and its xx-intercept.
Line ll has the equation y=12x12ay = \frac{1}{2}x - \frac{1}{2}a.
Perpendicular lines have negative reciprocal slopes. The negative reciprocal of 2-2 is 12\frac{1}{2}. Using the point (a,0)(a, 0) in the point-slope form gives y0=12(xa)y - 0 = \frac{1}{2}(x - a).
3
Calculate the vertices of the bounded region by finding the intersection points of the boundary lines.
The vertices of the bounded region are (0,20)(0, 20), (0,10)(0, 10), (4+0.2a,20.4a)(4 + 0.2a, 2 - 0.4a), and (8+0.2a,40.4a)(8 + 0.2a, 4 - 0.4a).
The region is bounded by the parallel lines kk and mm, the perpendicular line ll, and the yy-axis (x=0x = 0).
4
Find the area of the region as an algebraic expression in terms of aa.
The area is equal to 60+2a60 + 2a.
The region can be divided into a parallelogram with a vertical base of 1010 and width 4+0.2a4 + 0.2a, and a right triangle with a vertical base of 1010 and width 44. The sum of their areas is 10(4+0.2a)+12(10)(4)=40+2a+20=60+2a10(4 + 0.2a) + \frac{1}{2}(10)(4) = 40 + 2a + 20 = 60 + 2a.
5
Set the area expression equal to the given area of 110110 and solve for aa.
a=25a = 25
Setting the area equal to 110110 yields 60+2a=11060 + 2a = 110, which simplifies to 2a=502a = 50, or a=25a = 25.

Anahtar Kavram

Linear functions, parallel and perpendicular lines, finding line equations, and coordinate geometry area.

Alternatif Yöntem

The area can also be calculated using the geometric properties of a trapezoid. The height of the trapezoid is the perpendicular distance between the parallel lines kk and mm, which is 2010(2)2+12=25\frac{|20 - 10|}{\sqrt{(-2)^2 + 1^2}} = 2\sqrt{5}. The bases of the trapezoid are the segments of lines kk and mm from the yy-axis to their intersection points with line ll. The length of the base on line kk is 959\sqrt{5} and the length of the base on line mm is 13513\sqrt{5} (when a=25a = 25). Using the formula for the area of a trapezoid, Area=95+1352×25=115×25=110\text{Area} = \frac{9\sqrt{5} + 13\sqrt{5}}{2} \times 2\sqrt{5} = 11\sqrt{5} \times 2\sqrt{5} = 110.
Tahmini Süre:3m 0s
ÖncekiSayfa 2 / 22Sonraki