Systems of Linear Equations

32 soru

Soru 1Soru

If x+y=8x + y = 8 and 2xy=72x - y = 7, what is the value of xx?

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Cevap: 5

Cevap

5
Adding the two equations yields (x+y)+(2xy)=8+7(x + y) + (2x - y) = 8 + 7, which simplifies to 3x=153x = 15. Dividing both sides by 3 gives the correct value, x=5x = 5.

Adım Adım Çözüm

1
Add the two equations together to eliminate the variable yy.
(x+y)+(2xy)=8+7    3x=15(x + y) + (2x - y) = 8 + 7 \implies 3x = 15
Since the coefficients of yy are +1+1 and 1-1, adding the equations eliminates yy directly.
2
Solve for xx by dividing both sides of the equation by 3.
x=153=5x = \frac{15}{3} = 5
Isolating xx gives the required solution.

Anahtar Kavram

Elimination Method in Systems of Linear Equations
Soru 2Soru
Consider the system of linear equations in two variables xx and yy:
2x3y=a4x6y=b\begin{aligned} 2x - 3y &= a \\ 4x - 6y &= b \end{aligned}
where aa and bb are real constants. Which of the following statements MUST be true? Select all that apply.

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Cevap: If b=2ab = 2a, the system has infinitely many solutions.; If b2ab \neq 2a, the system has no solution.; If b=2a+1b = 2a + 1, the lines represented by the equations in the xyxy-plane are parallel and distinct.

Cevap

The correct statements are: 'If b=2ab = 2a, the system has infinitely many solutions.', 'If b2ab \neq 2a, the system has no solution.', and 'If b=2a+1b = 2a + 1, the lines represented by the equations in the xyxy-plane are parallel and distinct.'
Statements asserting that the system has infinitely many solutions when b=2ab = 2a, no solution when b2ab \neq 2a, and parallel distinct lines when b=2a+1b = 2a + 1 are all mathematically sound. Multiplying the first equation by 2 reveals that the left sides are identical (4x6y4x - 6y). Equality of the right sides (b=2ab = 2a) makes the lines identical, whereas inequality (b2ab \neq 2a) makes them parallel and distinct.

Adım Adım Çözüm

1
Analyze the coefficients of the system of equations.
The coefficients of xx and yy in the second equation (44 and 6-6) are exactly twice the coefficients of xx and yy in the first equation (22 and 3-3).
Comparing coefficient ratios determines line relationships (slopes).
2
Multiply the first equation by 2.
2(2x3y)=2(a)    4x6y=2a2(2x - 3y) = 2(a) \implies 4x - 6y = 2a.
This puts the left side of the first equation in exact alignment with the second equation (4x6y=b4x - 6y = b).
3
Evaluate the condition for infinitely many solutions.
If b=2ab = 2a, the two equations become 4x6y=2a4x - 6y = 2a and 4x6y=2a4x - 6y = 2a, which describe the exact same line, giving infinitely many solutions.
Coincident lines intersect at every point along the line.
4
Evaluate the condition for no solution.
If b2ab \neq 2a, subtracting the equations gives 0=b2a00 = b - 2a \neq 0, a contradiction. Hence, the lines are parallel and distinct, meaning no solution exists.
Distinct parallel lines never intersect.
5
Examine specific cases such as b=2a+1b = 2a + 1 and a=0,b=0a=0, b=0.
For b=2a+1b = 2a + 1, since 2a+12a2a + 1 \neq 2a, b2ab \neq 2a, confirming parallel distinct lines. For a=0,b=0a=0, b=0, b=2(0)=0b = 2(0) = 0, which yields infinitely many solutions rather than a unique solution.
Verifies specific option claims.

Anahtar Kavram

Systems of Linear Equations (Solvability and Geometric Interpretation)
Soru 3Soru

If the pair (x,y)(x, y) satisfies the system of linear equations below, what is the value of xyx - y?

3x5y=142x+7y=1\begin{aligned} 3x - 5y &= 14 \\ 2x + 7y &= -1 \end{aligned}
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Cevap: 4

Cevap

The value of xyx - y is 44.
Solving the system of linear equations by elimination gives x=3x = 3 and y=1y = -1. Substituting these values into the target expression xyx - y yields 3(1)=3+1=43 - (-1) = 3 + 1 = 4.

Adım Adım Çözüm

1
Eliminate one variable using the elimination method.
Multiply the first equation by 22 and the second equation by 33:
6x10y=286x - 10y = 28
6x+21y=36x + 21y = -3
Creating matching coefficients for xx allows elimination of xx by subtraction.
2
Subtract the first modified equation from the second modified equation to solve for yy.
(6x+21y)(6x10y)=328(6x + 21y) - (6x - 10y) = -3 - 28
31y=31    y=131y = -31 \implies y = -1
Subtracting eliminates xx, isolating yy.
3
Substitute y=1y = -1 back into one of the original equations to solve for xx.
3x5(1)=14    3x+5=14    3x=9    x=33x - 5(-1) = 14 \implies 3x + 5 = 14 \implies 3x = 9 \implies x = 3
Plugging in yy allows direct solution for xx.
4
Evaluate the requested expression xyx - y.
xy=3(1)=3+1=4x - y = 3 - (-1) = 3 + 1 = 4
Subtracting a negative value is equivalent to adding its positive value.

Anahtar Kavram

Solving systems of linear equations using elimination and evaluating algebraic expressions.
Tahmini Süre:1m 30s
Soru 4Soru

A chemistry laboratory prepares a 100 mL100\text{ mL} mixture using three solutions: Solution XX (10%10\% acid by volume), Solution YY (30%30\% acid by volume), and Solution ZZ (50%50\% acid by volume). The resulting mixture is 37%37\% acid by volume. If the volume of Solution YY used is 5 mL5\text{ mL} more than twice the volume of Solution XX, what is the volume, in mL\text{mL}, of Solution ZZ used in the mixture?

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Cevap: 5050

Cevap

50 mL50\text{ mL}
Setting up the system of equations gives x+y+z=100x + y + z = 100, x+3y+5z=370x + 3y + 5z = 370, and y=2x+5y = 2x + 5. Substituting y=2x+5y = 2x + 5 into the first two equations yields 3x+z=953x + z = 95 and 7x+5z=3557x + 5z = 355. Solving for xx gives x=15x = 15, which leads to z=953(15)=50z = 95 - 3(15) = 50. Therefore, 50 mL50\text{ mL} of Solution Z was used.

Adım Adım Çözüm

1
Define variables and set up the system of linear equations
Let xx, yy, and zz be the volumes in mL\text{mL} of Solutions XX, YY, and ZZ, respectively.
1) Total volume: x+y+z=100x + y + z = 100
2) Total acid volume: 0.10x+0.30y+0.50z=0.37(100)    x+3y+5z=3700.10x + 0.30y + 0.50z = 0.37(100) \implies x + 3y + 5z = 370
3) Relationship between YY and XX: y=2x+5y = 2x + 5
Translate the word problem statements into mathematical equations.
2
Substitute y=2x+5y = 2x + 5 into equations (1) and (2) to reduce to a two-variable system
From equation (1):
x+(2x+5)+z=100    3x+z=95    z=953xx + (2x + 5) + z = 100 \implies 3x + z = 95 \implies z = 95 - 3x

From equation (2):
x+3(2x+5)+5z=370    7x+15+5z=370    7x+5z=355x + 3(2x + 5) + 5z = 370 \implies 7x + 15 + 5z = 370 \implies 7x + 5z = 355
Eliminating yy simplifies the system to two equations in xx and zz.
3
Substitute z=953xz = 95 - 3x into 7x+5z=3557x + 5z = 355 and solve for xx
7x+5(953x)=355    7x+47515x=355    8x=120    x=157x + 5(95 - 3x) = 355 \implies 7x + 475 - 15x = 355 \implies -8x = -120 \implies x = 15
Solves for the unknown volume of Solution X.
4
Calculate zz using z=953xz = 95 - 3x
z=953(15)=9545=50z = 95 - 3(15) = 95 - 45 = 50
Finds the requested volume of Solution Z.

Anahtar Kavram

Setting up and solving 3x3 systems of linear equations using substitution or elimination
Tahmini Süre:2m 0s
Soru 5Soru
Consider the following system of linear equations in variables xx and yy, where kk is a real constant:
kx+4y=8x+ky=k+2\begin{aligned} kx + 4y &= 8 \\ x + ky &= k + 2 \end{aligned}
Which of the following statements are true? Select all that apply.

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Cevap: If k=2k = 2, the system has infinitely many solutions.; If k=2k = -2, the system has no solutions.; If k=0k = 0, the unique solution to the system is (x,y)=(2,2)(x, y) = (2, 2).

Cevap

The correct statements are that setting k=2k = 2 results in infinitely many solutions, setting k=2k = -2 results in no solutions, and setting k=0k = 0 yields the unique solution (2,2)(2, 2).
The system has a coefficient matrix determinant of k24k^2 - 4. Setting k=2k = 2 produces identical equations (x+2y=4x + 2y = 4), giving infinitely many solutions. Setting k=2k = -2 produces parallel equations with different constants (x+2y=4-x + 2y = 4 vs x+2y=0-x + 2y = 0), giving no solutions. Setting k=0k = 0 reduces the system directly to y=2y = 2 and x=2x = 2, confirming the unique point (2,2)(2, 2).

Adım Adım Çözüm

1
Analyze the determinant of the coefficient matrix to identify conditions for unique vs. non-unique solutions.
The coefficient matrix determinant is Δ=kk41=k24=(k2)(k+2)\Delta = k\cdot k - 4\cdot 1 = k^2 - 4 = (k - 2)(k + 2).
If Δ0\Delta \neq 0 (i.e., k±2k \neq \pm 2), the system has a unique solution. If Δ=0\Delta = 0 (i.e., k=2k = 2 or k=2k = -2), the lines are either identical or parallel.
2
Test k=2k = 2 in the system.
Equation 1 becomes 2x+4y=8    x+2y=42x + 4y = 8 \implies x + 2y = 4. Equation 2 becomes x+2y=4x + 2y = 4.
Since the equations are identical, the lines coincide, resulting in infinitely many solutions.
3
Test k=2k = -2 in the system.
Equation 1 becomes 2x+4y=8    x+2y=4-2x + 4y = 8 \implies -x + 2y = 4. Equation 2 becomes x2y=0    x+2y=0x - 2y = 0 \implies -x + 2y = 0.
The slopes are equal (1/21/2) but the y-intercepts differ (22 vs 00), meaning the lines are parallel and distinct, yielding zero solutions.
4
Test k=0k = 0 and k=1k = 1 to evaluate remaining choices.
For k=0k = 0, 4y=8    y=24y = 8 \implies y = 2 and x+0=2    x=2x + 0 = 2 \implies x = 2, giving (2,2)(2, 2). For k=1k = 1, Equation 2 is directly x+y=3x + y = 3.
This verifies that the statement for k=0k = 0 is correct, while the statement for k=1k = 1 claiming x+y=4x + y = 4 is false.

Anahtar Kavram

Parametric Linear Systems and Conditions for Solvability
Soru 6Soru

A logistics warehouse uses two sizes of shipping crates: small crates and large crates. A shipment containing 33 small crates and 55 large crates has a total weight of 4747 pounds. A second shipment containing 66 small crates and 22 large crates has a total weight of 3838 pounds. What is the total weight, in pounds, of a shipment containing 22 small crates and 33 large crates?

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Cevap: 29

Cevap

29 pounds
By setting up the linear system 3s+5l=473s + 5l = 47 and 6s+2l=386s + 2l = 38, using elimination gives s=4s = 4 pounds for a small crate and l=7l = 7 pounds for a large crate. Substituting these values into the target expression 2s+3l2s + 3l yields 2(4)+3(7)=292(4) + 3(7) = 29 pounds.

Adım Adım Çözüm

1
Set up a system of linear equations from the given context.
Let ss be the weight of a small crate and ll be the weight of a large crate. The equations are: (1) 3s+5l=473s + 5l = 47 and (2) 6s+2l=386s + 2l = 38.
Translate the physical constraints of each shipment into algebraic equations.
2
Solve for the variable ll by eliminating ss.
Multiply Equation (1) by 22 to get 6s+10l=946s + 10l = 94. Subtract Equation (2) (6s+2l=386s + 2l = 38) from this new equation: (6s+10l)(6s+2l)=9438    8l=56    l=7(6s + 10l) - (6s + 2l) = 94 - 38 \implies 8l = 56 \implies l = 7.
Align the coefficients of ss so that elimination via subtraction yields a single-variable linear equation for ll.
3
Substitute l=7l = 7 back into Equation (2) to solve for ss.
6s+2(7)=38    6s+14=38    6s=24    s=46s + 2(7) = 38 \implies 6s + 14 = 38 \implies 6s = 24 \implies s = 4.
Determine the individual weight of a small crate.
4
Calculate the requested total weight for 22 small crates and 33 large crates.
2s+3l=2(4)+3(7)=8+21=292s + 3l = 2(4) + 3(7) = 8 + 21 = 29.
Substitute the individual values of ss and ll into the target expression.

Anahtar Kavram

Solving 2x2 Systems of Linear Equations by Elimination and Linear Combination Evaluation
Tahmini Süre:1m 30s
Soru 7Soru

A chemist mixes xx ounces of a 40% acid solution with yy ounces of a 70% acid solution to produce a 20-ounce mixture that is 52% acid. What is the value of xyx - y?

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Cevap: 4

Cevap

4
Setting up the linear system x+y=20x + y = 20 and 0.40x+0.70y=10.40.40x + 0.70y = 10.4 leads to x=12x = 12 and y=8y = 8. Subtracting yy from xx yields 128=412 - 8 = 4.

Adım Adım Çözüm

1
Set up the system of linear equations based on total solution volume and pure acid content.
Total volume equation: x+y=20x + y = 20. Pure acid equation: 0.40x+0.70y=0.52(20)=10.40.40x + 0.70y = 0.52(20) = 10.4.
The sum of the component volumes equals the total volume, and the sum of the pure acid contents equals the total acid content.
2
Multiply the acid equation by 10 to eliminate decimals.
4x+7y=1044x + 7y = 104.
Clearing decimals simplifies the subsequent elimination calculation.
3
Solve for yy using the elimination method.
Multiply x+y=20x + y = 20 by 4 to get 4x+4y=804x + 4y = 80. Subtract this from 4x+7y=1044x + 7y = 104: (4x+7y)(4x+4y)=10480    3y=24    y=8(4x + 7y) - (4x + 4y) = 104 - 80 \implies 3y = 24 \implies y = 8.
Eliminating the variable xx isolates yy.
4
Solve for xx and compute xyx - y.
x=208=12x = 20 - 8 = 12. Therefore, xy=128=4x - y = 12 - 8 = 4.
Substitute y=8y = 8 back into the first equation and calculate the requested expression.

Anahtar Kavram

Systems of Linear Equations in Mixture Problems
Soru 8Soru
Consider the following system of linear equations:
2x+3y=124x+6y=24\begin{aligned} 2x + 3y &= 12 \\ 4x + 6y &= 24 \end{aligned}

Which of the following statements about this system must be true? Select all that apply.

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Cevap: The system has infinitely many solutions.; The graphs of the two equations represent the exact same line in the xyxy-plane.; The ordered pair (3,2)(3, 2) is a solution to the system.

Cevap

The correct statements are that the system has infinitely many solutions, the graphs represent the exact same line, and (3,2)(3, 2) is a solution to the system.
Dividing 4x+6y=244x + 6y = 24 by 22 produces 2x+3y=122x + 3y = 12, showing that both equations represent the exact same line. Therefore, the system has infinitely many solutions. Substituting x=3x = 3 and y=2y = 2 yields 2(3)+3(2)=122(3) + 3(2) = 12, confirming that (3,2)(3, 2) is one of the infinitely many valid solutions.

Adım Adım Çözüm

1
Analyze the relationship between the two linear equations
Dividing the second equation 4x+6y=244x + 6y = 24 by 22 gives 2x+3y=122x + 3y = 12, which is identical to the first equation.
Comparing coefficients and constants determines whether equations in a system are dependent, independent, or inconsistent.
2
Determine the number of solutions and geometric structure
Because the equations are mathematically equivalent, they describe the same line in the coordinate plane and have infinitely many intersection points.
Identical linear equations form a dependent system with infinitely many solutions.
3
Test the ordered pair (3,2)(3, 2)
Evaluating 2(3)+3(2)=6+6=122(3) + 3(2) = 6 + 6 = 12 confirms that (3,2)(3, 2) lies on the line.
Any point satisfying one equation satisfies the entire system of equivalent equations.
4
Determine the axis intercepts
Setting y=0y=0 gives 2x=12x=62x=12 \Rightarrow x=6 (the xx-intercept is (6,0)(6,0)); setting x=0x=0 gives 3y=12y=43y=12 \Rightarrow y=4 (the yy-intercept is (0,4)(0,4)).
Checking coordinates of axis intersections prevents mislabeling xx- and yy-intercepts.

Anahtar Kavram

Dependent Systems of Linear Equations
Soru 9Soru
Consider the following system of linear equations:
x+2y=103xy=9\begin{aligned} x + 2y &= 10 \\ 3x - y &= 9 \end{aligned}
Which of the following statements are true? Select all that apply.

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Cevap: The value of xx is 44.; The value of x+yx + y is 77.; The value of 2x+y2x + y is 1111.

Cevap

The true statements are those asserting that x=4x = 4, that x+y=7x + y = 7, and that 2x+y=112x + y = 11.
Solving the system of linear equations yields the unique solution x=4x = 4 and y=3y = 3. Substituting these values into the given choices demonstrates that the statements claiming x=4x = 4, x+y=7x + y = 7, and 2x+y=112x + y = 11 are all mathematically correct.

Adım Adım Çözüm

1
Express yy in terms of xx using the second equation.
y=3x9y = 3x - 9
Isolating yy allows for simple substitution into the first equation.
2
Substitute y=3x9y = 3x - 9 into the first equation x+2y=10x + 2y = 10.
x+2(3x9)=10    x+6x18=10    7x=28    x=4x + 2(3x - 9) = 10 \implies x + 6x - 18 = 10 \implies 7x = 28 \implies x = 4
This yields a single linear equation in terms of xx.
3
Calculate yy using x=4x = 4.
y=3(4)9=3y = 3(4) - 9 = 3
Substituting x=4x = 4 back gives the unique solution for yy.
4
Evaluate the given statements with (x,y)=(4,3)(x, y) = (4, 3).
x=4x = 4 is true; x+y=4+3=7x + y = 4 + 3 = 7 is true; y=4y = 4 is false (y=3y = 3); 2x+y=2(4)+3=112x + y = 2(4) + 3 = 11 is true; xy=43=1x - y = 4 - 3 = 1 is false.
Testing each condition determines which options are valid.

Anahtar Kavram

Solving a 2x2 system of linear equations using substitution to evaluate linear expressions.
Tahmini Süre:1m 0s
Soru 10Soru

An investor divides a total of $15,000\$15,000 between two accounts, Account P and Account Q. Account P earns an annual simple interest rate of 6%6\%, while Account Q earns an annual simple interest rate of 8%8\%. If the total interest earned from both accounts combined after one year is $1,020\$1,020, how much more money was invested in Account P than in Account Q?

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Cevap: $3,000\$3,000

Cevap

$3,000\$3,000
Setting up the system of linear equations p+q=15,000p + q = 15,000 and 0.06p+0.08q=1,0200.06p + 0.08q = 1,020 yields p=9,000p = 9,000 and q=6,000q = 6,000. Subtracting the amount in Account Q from Account P gives 9,0006,000=3,0009,000 - 6,000 = 3,000. Thus, $3,000\$3,000 is the correct answer.

Adım Adım Çözüm

1
Define variables and set up the system of linear equations
Let pp be the amount invested in Account P and qq be the amount invested in Account Q. The equations are p+q=15,000p + q = 15,000 and 0.06p+0.08q=1,0200.06p + 0.08q = 1,020.
The first equation represents the total investment amount, and the second equation represents the total annual interest earned.
2
Simplify the interest equation and solve for one variable using elimination
Multiply 0.06p+0.08q=1,0200.06p + 0.08q = 1,020 by 100 to obtain 6p+8q=102,0006p + 8q = 102,000, which simplifies to 3p+4q=51,0003p + 4q = 51,000. Multiplying the first equation by 3 yields 3p+3q=45,0003p + 3q = 45,000. Subtracting this from 3p+4q=51,0003p + 4q = 51,000 gives q=6,000q = 6,000.
Eliminating pp allows for direct calculation of the amount invested in Account Q.
3
Calculate the amount invested in Account P
p=15,0006,000=9,000p = 15,000 - 6,000 = 9,000.
Substitute the value of qq back into the total investment equation.
4
Determine the requested difference
pq=9,0006,000=3,000p - q = 9,000 - 6,000 = 3,000.
The question specifically asks how much more money was invested in Account P than in Account Q.

Anahtar Kavram

Solving Systems of 2x2 Linear Equations for Word Problems
Soru 11Soru

A logistics company packages cargo using three types of containers: small, medium, and large.

- A shipment of 33 small, 22 medium, and 11 large container has a total weight of 130130 kilograms.
- A shipment of 11 small, 44 medium, and 22 large containers has a total weight of 185185 kilograms.
- A shipment of 22 small, 11 medium, and 33 large containers has a total weight of 160160 kilograms.

What is the weight, in kilograms, of one large container?

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Cevap: 35

Cevap

The weight of one large container is 35 kilograms.
Representing the weights of small, medium, and large containers as variables SS, MM, and LL yields the 3x3 system of linear equations:
1) 3S+2M+L=1303S + 2M + L = 130
2) S+4M+2L=185S + 4M + 2L = 185
3) 2S+M+3L=1602S + M + 3L = 160

Solving for SS in equation (2) gives S=1854M2LS = 185 - 4M - 2L. Substituting this into equations (1) and (3) reduces the system to:
- 2M+L=852M + L = 85
- 7M+L=2107M + L = 210

Subtracting the first equation from the second yields 5M=1255M = 125, so M=25M = 25. Substituting M=25M = 25 into 2M+L=852M + L = 85 gives 50+L=8550 + L = 85, which simplifies to L=35L = 35.

Adım Adım Çözüm

1
Set up a system of three linear equations based on the shipment descriptions
3S+2M+L=1303S + 2M + L = 130, S+4M+2L=185S + 4M + 2L = 185, and 2S+M+3L=1602S + M + 3L = 160
Translate the physical constraints of the three shipments into mathematical relationships
2
Isolate variable SS in the second equation and substitute into the first and third equations
Two equations in two variables: 2M+L=852M + L = 85 and 7M+L=2107M + L = 210
Reduce the 3x3 system to a 2x2 system to eliminate variable SS
3
Subtract the two simplified equations to solve for MM
5M=125    M=255M = 125 \implies M = 25
Eliminate variable LL to obtain the value of MM
4
Substitute M=25M = 25 back into 2M+L=852M + L = 85 to solve for LL
L=35L = 35
Find the requested value for the weight of one large container

Anahtar Kavram

Solving 3x3 Systems of Linear Equations using Substitution and Elimination
Soru 12Soru
Consider the following system of linear equations in xx, yy, and zz, where cc is a real constant:
32xy+2z=5\frac{3}{2}x - y + 2z = 5
x+13yz=2x + \frac{1}{3}y - z = 2
6xy+z=c6x - y + z = c

If the system has at least one solution (x,y,z)(x, y, z), what is the value of 7x23y7x - \frac{2}{3}y?

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Cevap: 18

Cevap

18
To find the value of 7x23y7x - \frac{2}{3}y without individual values for x,y,x, y, and zz, we express 7x23y7x - \frac{2}{3}y as a linear combination m(Eq. 1)+n(Eq. 2)m(\text{Eq. 1}) + n(\text{Eq. 2}). Matching the zz-coefficients requires 2mn=0    n=2m2m - n = 0 \implies n = 2m. Matching the yy-coefficients yields m+13(2m)=23    m=2-m + \frac{1}{3}(2m) = -\frac{2}{3} \implies m = 2, which gives n=4n = 4. Verifying the xx-coefficient gives 2(32)+4(1)=72\left(\frac{3}{2}\right) + 4(1) = 7. Applying these multipliers to the right-hand sides gives 2(5)+4(2)=10+8=182(5) + 4(2) = 10 + 8 = 18.

Adım Adım Çözüm

1
Identify the target expression 7x23y7x - \frac{2}{3}y as a linear combination of the first two equations
Express m(32xy+2z)+n(x+13yz)=7x23y+0zm\left(\frac{3}{2}x - y + 2z\right) + n\left(x + \frac{1}{3}y - z\right) = 7x - \frac{2}{3}y + 0z
Because the system is dependent when consistent, individual variable values cannot be uniquely determined, but specific linear combinations independent of zz can be evaluated.
2
Set up a system of equations for the scalar multipliers mm and nn
Equating coefficients of zz: 2mn=0    n=2m2m - n = 0 \implies n = 2m. Equating coefficients of yy: m+13n=23-m + \frac{1}{3}n = -\frac{2}{3}.
Eliminating zz requires the net coefficient of zz to equal 0.
3
Solve for mm and nn
Substitute n=2mn = 2m into the yy-coefficient equation: m+23m=13m=23    m=2-m + \frac{2}{3}m = -\frac{1}{3}m = -\frac{2}{3} \implies m = 2, which gives n=4n = 4.
Determining the exact linear multipliers needed to match the target expression.
4
Verify xx-coefficient consistency and compute the target value
xx-coefficient: 2(32)+4(1)=3+4=72\left(\frac{3}{2}\right) + 4(1) = 3 + 4 = 7. Value: 2(5)+4(2)=10+8=182(5) + 4(2) = 10 + 8 = 18.
Applying the scalars m=2m = 2 and n=4n = 4 to the right-hand side constants gives the exact numerical value of 7x23y7x - \frac{2}{3}y.

Anahtar Kavram

Linear combinations of dependent systems of equations

Alternatif Yöntem

Multiply the first equation by 2 to clear fractions: 3x2y+4z=103x - 2y + 4z = 10. Multiply the second equation by 3 to clear fractions: 3x+y3z=63x + y - 3z = 6. Eliminate zz by forming 3(3x2y+4z)+4(3x+y3z)=3(10)+4(6)    21x2y=543(3x - 2y + 4z) + 4(3x + y - 3z) = 3(10) + 4(6) \implies 21x - 2y = 54. Dividing both sides of 21x2y=5421x - 2y = 54 by 3 directly gives 7x23y=187x - \frac{2}{3}y = 18.
Tahmini Süre:2m 0s
Soru 13Soru

An electronics manufacturer produces three types of circuit boards: Alpha, Beta, and Gamma. Production requires processing across three specialized workstations: Solder, Component Placement, and Inspection.

- Each Alpha board requires 2 hours of Solder, 3 hours of Component Placement, and 1 hour of Inspection.
- Each Beta board requires 1 hour of Solder, 4 hours of Component Placement, and 2 hours of Inspection.
- Each Gamma board requires 3 hours of Solder, 2 hours of Component Placement, and 4 hours of Inspection.

During a given production cycle, the Solder station was operated for 55 hours, the Component Placement station for 85 hours, and the Inspection station for 65 hours. If all three workstations were operated at full capacity with no downtime, what was the total number of circuit boards produced?

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Cevap: 28

Cevap

The total number of circuit boards produced is 28.
Setting up equations for total machine hours yields 2x+y+3z=552x + y + 3z = 55, 3x+4y+2z=853x + 4y + 2z = 85, and x+2y+4z=65x + 2y + 4z = 65, where xx, yy, and zz represent the quantities of Alpha, Beta, and Gamma boards produced, respectively. Subtracting the third equation from the first equation gives (2x+y+3z)(x+2y+4z)=5565(2x + y + 3z) - (x + 2y + 4z) = 55 - 65, which simplifies to xyz=10x - y - z = -10, or x=y+z10x = y + z - 10. Substituting x=y+z10x = y + z - 10 into the second and third equations produces the 2x2 system 7y+5z=1157y + 5z = 115 and 3y+5z=753y + 5z = 75. Subtracting these two equations eliminates zz, giving 4y=404y = 40, so y=10y = 10. Substituting y=10y = 10 into 3y+5z=753y + 5z = 75 gives 30+5z=7530 + 5z = 75, so z=9z = 9. Substituting y=10y = 10 and z=9z = 9 into x=y+z10x = y + z - 10 gives x=9x = 9. The total number of circuit boards produced is x+y+z=9+10+9=28x + y + z = 9 + 10 + 9 = 28.

Adım Adım Çözüm

1
Set up the 3x3 system of linear equations based on workstation hours.
2x+y+3z=552x + y + 3z = 55, 3x+4y+2z=853x + 4y + 2z = 85, and x+2y+4z=65x + 2y + 4z = 65
Each equation models the total operational hours used across the three product types.
2
Subtract the third equation from the first equation to isolate xx in terms of yy and zz.
x=y+z10x = y + z - 10
Eliminating terms directly reduces coefficient complexity.
3
Substitute x=y+z10x = y + z - 10 into the second and third equations to construct a 2x2 system.
3y+5z=753y + 5z = 75 and 7y+5z=1157y + 5z = 115
Reducing to a two-variable system allows direct elimination.
4
Subtract the two reduced equations to solve for yy and zz.
y=10y = 10 and z=9z = 9
The 5z5z terms cancel out upon subtraction.
5
Determine xx and compute the total sum x+y+zx + y + z.
x=9x = 9, total =9+10+9=28= 9 + 10 + 9 = 28
The question asks for the total quantity of circuit boards produced.

Anahtar Kavram

Systems of Linear Equations
Soru 14Soru
Consider the following system of three linear equations in variables xx, yy, and zz, where kk is a real constant:
2x+3yz=11x2y+4z=34xy+7z=k\begin{aligned} 2x + 3y - z &= 11 \\ x - 2y + 4z &= -3 \\ 4x - y + 7z &= k \end{aligned}
For what value of kk does the system have at least one solution (x,y,z)(x, y, z)?
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Cevap: 55

Cevap

55
The left-hand side of the third equation is a linear combination of the first two equations: 1(2x+3yz)+2(x2y+4z)=4xy+7z1 \cdot (2x + 3y - z) + 2 \cdot (x - 2y + 4z) = 4x - y + 7z. For the linear system to be consistent and possess at least one solution, the same linear combination must hold for the constant terms on the right-hand side: 1(11)+2(3)=116=51(11) + 2(-3) = 11 - 6 = 5. Therefore, the value of kk must be 55.

Adım Adım Çözüm

1
Analyze the variable coefficients across the three equations for linear dependence.
Notice that the coefficients of the third equation can be expressed as a linear combination of the first two equations.
If the left-hand side of the third equation is a linear combination of the first two equations, the system will only be consistent if the right-hand side constants satisfy the exact same linear combination.
2
Determine the multiplier needed to produce the third equation's left-hand side.
Multiply the second equation by 22 and add it to the first equation: (2x+3yz)+2(x2y+4z)=4xy+7z(2x + 3y - z) + 2(x - 2y + 4z) = 4x - y + 7z.
This yields the exact expression 4xy+7z4x - y + 7z present on the left-hand side of the third equation.
3
Apply the identical combination to the right-hand side constant terms.
The combined constant value is 11+2(3)=116=511 + 2(-3) = 11 - 6 = 5.
For the system to have at least one solution (i.e., to avoid contradiction and be consistent), kk must equal this computed value of 55.

Anahtar Kavram

Linear Dependence and Consistency in 3x3 Linear Systems
Tahmini Süre:2m 0s
Soru 15Soru
If xx and yy are real numbers such that x>y>0x > y > 0 and they satisfy the following system of equations:
3x+y+4xy=114\frac{3}{x+y} + \frac{4}{x-y} = \frac{11}{4}
5x+y2xy=14\frac{5}{x+y} - \frac{2}{x-y} = \frac{1}{4}
what is the value of xx?
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Cevap: 3

Cevap

The value of xx is 3.
Substituting u=1x+yu = \frac{1}{x+y} and v=1xyv = \frac{1}{x-y} transforms the non-linear looking equations into the linear system 3u+4v=1143u + 4v = \frac{11}{4} and 5u2v=145u - 2v = \frac{1}{4}. Solving this system yields u=14u = \frac{1}{4} and v=12v = \frac{1}{2}. Consequently, x+y=4x + y = 4 and xy=2x - y = 2. Adding these two equations gives 2x=62x = 6, so x=3x = 3.

Adım Adım Çözüm

1
Introduce auxiliary variables to linearize the system.
Let u=1x+yu = \frac{1}{x+y} and v=1xyv = \frac{1}{x-y}. The system becomes 3u+4v=1143u + 4v = \frac{11}{4} and 5u2v=145u - 2v = \frac{1}{4}.
Replacing non-linear reciprocal terms with simple variables allows elimination or substitution methods for linear systems.
2
Solve the system of linear equations for uu and vv using elimination.
Multiply 5u2v=145u - 2v = \frac{1}{4} by 2 to get 10u4v=1210u - 4v = \frac{1}{2}. Add this to 3u+4v=1143u + 4v = \frac{11}{4}: 13u=114+24=134    u=1413u = \frac{11}{4} + \frac{2}{4} = \frac{13}{4} \implies u = \frac{1}{4}. Then 4v=1143(14)=2    v=124v = \frac{11}{4} - 3\left(\frac{1}{4}\right) = 2 \implies v = \frac{1}{2}.
Eliminating vv yields a single equation in uu, which provides the values of both auxiliary variables.
3
Convert auxiliary values back to equations in xx and yy.
Since u=1x+y=14u = \frac{1}{x+y} = \frac{1}{4}, we get x+y=4x + y = 4. Since v=1xy=12v = \frac{1}{x-y} = \frac{1}{2}, we get xy=2x - y = 2.
Inverting the fractions restores the original variables in a standard 2x2 linear system.
4
Solve for xx by adding the two linear equations.
(x+y)+(xy)=4+2    2x=6    x=3(x + y) + (x - y) = 4 + 2 \implies 2x = 6 \implies x = 3.
Adding the equations eliminates yy directly, isolating xx.

Anahtar Kavram

Solving systems of linear equations using substitution variables for algebraic simplification
Soru 16Soru
Consider the following system of linear equations in variables xx, yy, and zz, where aa is a real constant:
x+y+z=6x+2y+3z=10x+2y+(a21)z=a+8\begin{aligned} x + y + z &= 6 \\ x + 2y + 3z &= 10 \\ x + 2y + (a^2 - 1)z &= a + 8 \end{aligned}

Which of the following statements must be true? Select all that apply.

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Cevap: If a=2a = 2, the system has infinitely many solutions.; If a=2a = -2, the system has no solution.; If a=2a = 2, every solution to the system satisfies 2x+y=82x + y = 8.

Cevap

The correct statements are that a=2a = 2 yields infinitely many solutions, a=2a = -2 results in no solution, and for a=2a = 2 every solution satisfies 2x+y=82x + y = 8.
The reduced equation (a24)z=a2(a^2 - 4)z = a - 2 determines the behavior of the system. Setting a=2a = 2 gives 0=00 = 0, leading to infinitely many solutions where x=z+2x = z + 2 and y=42zy = 4 - 2z, which identically satisfies 2x+y=82x + y = 8. Setting a=2a = -2 gives 0=40 = -4, an inconsistency yielding no solutions.

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1
Eliminate xx and yy using elimination between the second and third equations.
(x+2y+(a21)z)(x+2y+3z)=(a+8)10    (a24)z=a2(x + 2y + (a^2 - 1)z) - (x + 2y + 3z) = (a + 8) - 10 \implies (a^2 - 4)z = a - 2
Isolating the parameter dependence onto a single variable zz reveals existence and uniqueness conditions.
2
Analyze the equation (a2)(a+2)z=a2(a - 2)(a + 2)z = a - 2 for key parameter values.
If a=2a = 2, 0z=00 \cdot z = 0 (infinitely many solutions). If a=2a = -2, 0z=40 \cdot z = -4 (no solution). If a±2a \neq \pm 2, z=1a+2z = \frac{1}{a + 2} (unique solution).
Determining system consistency depends on whether the leading coefficient and right-hand side evaluate to zero.
3
Express xx and yy in terms of zz for the consistent case a=2a = 2.
Subtracting the first equation from the second gives y+2z=4    y=42zy + 2z = 4 \implies y = 4 - 2z. Substituting into the first gives x=z+2x = z + 2.
Parameterizing the solution set allows verification of linear combinations.
4
Evaluate the linear combination 2x+y2x + y when a=2a = 2.
2x+y=2(z+2)+(42z)=2z+4+42z=82x + y = 2(z + 2) + (4 - 2z) = 2z + 4 + 4 - 2z = 8.
Verifies that 2x+y=82x + y = 8 is an invariant across all parametric solutions.

Anahtar Kavram

Parametric Analysis of 3x3 Systems of Linear Equations
Soru 17Soru
A financial analyst models a company's weekly metrics using three variables—revenue RR, operating cost CC, and advertising expenditure AA, measured in thousands of dollars. The metrics satisfy the following system of linear equations, where kk is a real constant:
4R3C+2A=18R+5C6A=145R+2C4A=k\begin{aligned} 4R - 3C + 2A &= 18 \\ R + 5C - 6A &= 14 \\ 5R + 2C - 4A &= k \end{aligned}
If this system of linear equations is consistent (has at least one solution), what is the value of kk?
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Cevap: 32

Cevap

32
Adding the left-hand sides of the first two equations yields (4R3C+2A)+(R+5C6A)=5R+2C4A(4R - 3C + 2A) + (R + 5C - 6A) = 5R + 2C - 4A, which is identical to the left-hand side of the third equation. For the linear system to have at least one solution (to be consistent), the right-hand side constant must satisfy the exact same linear combination: k=18+14=32k = 18 + 14 = 32.

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1
Examine the linear combination of the left-hand sides of the first two equations
(4R3C+2A)+(R+5C6A)=5R+2C4A(4R - 3C + 2A) + (R + 5C - 6A) = 5R + 2C - 4A
Observing that the sum of the coefficients of the first two equations matches the left-hand side of the third equation.
2
Apply the condition for system consistency
Right-hand side of Equation 3 must equal Right-hand side of Equation 1 + Right-hand side of Equation 2
For a system with linearly dependent left-hand sides to be consistent, the same linear combination must hold for the right-hand constants.
3
Calculate the value of kk
k=18+14=32k = 18 + 14 = 32
Adding the constants from the right-hand side of the first two equations gives the consistent value for kk.

Anahtar Kavram

Consistency and Linear Dependence in Systems of Linear Equations
Soru 18Soru
For how many real values of the constant aa does the following system of linear equations in xx, yy, and zz have no solution?
x+yz=3x+(a1)y+3z=5x+4y+(a+1)z=a+2\begin{aligned} x + y - z &= 3 \\ x + (a-1)y + 3z &= 5 \\ x + 4y + (a+1)z &= a + 2 \end{aligned}
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Cevap: Exactly one

Cevap

Exactly one
To find when the system has no solution, we first eliminate xx by subtracting the first equation from the second and third equations. This produces a two-variable system in yy and zz: (a2)y+4z=2(a-2)y + 4z = 2 and 3y+(a+2)z=a13y + (a+2)z = a - 1. The determinant of this system's coefficients is (a2)(a+2)12=a216(a-2)(a+2) - 12 = a^2 - 16. Setting the determinant to zero yields two critical values: a=4a = 4 and a=4a = -4. Testing a=4a = 4 simplifies both reduced equations to y+2z=1y + 2z = 1, which means the system is consistent with infinitely many solutions. Testing a=4a = -4 yields 3y+2z=1-3y + 2z = 1 and 3y+2z=5-3y + 2z = 5, which is impossible (1=51 = 5), making the system inconsistent. Thus, there is exactly one real value of aa (a=4a = -4) for which the system has no solution.

Adım Adım Çözüm

1
Eliminate the variable xx from the second and third equations using the first equation.
Subtracting the first equation x+yz=3x + y - z = 3 from the second equation yields:
(a2)y+4z=2(a-2)y + 4z = 2
Subtracting the first equation from the third equation yields:
3y+(a+2)z=a13y + (a+2)z = a - 1
Reducing the 3×33 \times 3 system to a 2×22 \times 2 system in yy and zz simplifies the analysis of linear dependence and consistency.
2
Determine the values of aa for which the reduced 2×22 \times 2 system lacks a unique solution by setting its coefficient determinant to zero.
The determinant of the coefficient matrix is:
D=(a2)(a+2)(3)(4)=a2412=a216D = (a-2)(a+2) - (3)(4) = a^2 - 4 - 12 = a^2 - 16
Setting D=0D = 0 yields a2=16a^2 = 16, which gives a=4a = 4 or a=4a = -4.
A system of linear equations has either a unique solution (when the determinant is non-zero) or non-unique behavior—either no solution or infinitely many solutions—when the determinant is zero.
3
Test a=4a = 4 in the reduced system.
Substituting a=4a = 4 into the reduced equations gives:
2y+4z=2    y+2z=12y + 4z = 2 \implies y + 2z = 1
3y+6z=3    y+2z=13y + 6z = 3 \implies y + 2z = 1
Since both equations are identical, the system is consistent and has infinitely many solutions.
When equation ratios match completely including constant terms, the equations represent identical hyperplanes, yielding infinitely many solutions.
4
Test a=4a = -4 in the reduced system.
Substituting a=4a = -4 into the reduced equations gives:
6y+4z=2    3y+2z=1-6y + 4z = 2 \implies -3y + 2z = 1
3y2z=5    3y+2z=53y - 2z = -5 \implies -3y + 2z = 5
Comparing these gives 1=51 = 5, which is a contradiction. Thus, for a=4a = -4, the system has no solution.
When parallel equations have equal coefficient ratios but unequal constant ratios, the system is inconsistent.

Anahtar Kavram

Parametric Systems of Linear Equations and Consistency Conditions
Soru 19Soru

Two water pumps, Pump A and Pump B, working simultaneously at their respective constant rates, can fill an empty storage tank in 44 hours. If Pump A operates alone at its constant rate for 22 hours and then Pump B operates alone at its constant rate for 77 hours, the tank is also filled completely. How many hours would it take Pump A, working alone at its constant rate, to fill the entire storage tank?

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Cevap: 203\frac{20}{3} hours

Cevap

203\frac{20}{3} hours (or 6236\frac{2}{3} hours)
The correct answer is derived by setting up two linear equations representing the total work accomplished: 4rA+4rB=14r_A + 4r_B = 1 and 2rA+7rB=12r_A + 7r_B = 1. Solving this system gives Pump A's rate rA=320r_A = \frac{3}{20} tanks per hour. The time required for Pump A working alone is the reciprocal of its rate, which equals 203\frac{20}{3} hours.

Adım Adım Çözüm

1
Define variables and formulate the system of linear equations
Let rAr_A be the rate of Pump A (tanks/hour) and rBr_B be the rate of Pump B (tanks/hour).
Combined work equation: 4(rA+rB)=1    4rA+4rB=14(r_A + r_B) = 1 \implies 4r_A + 4r_B = 1
Sequential work equation: 2rA+7rB=12r_A + 7r_B = 1
Work done equals rate multiplied by time, and completing one full tank corresponds to total work =1= 1.
2
Express rBr_B in terms of rAr_A using the first equation
rA+rB=14    rB=14rAr_A + r_B = \frac{1}{4} \implies r_B = \frac{1}{4} - r_A
Simplifying the combined rate equation allows substitution into the second linear equation.
3
Substitute rBr_B into the second equation and solve for rAr_A
2rA+7(14rA)=1    2rA+747rA=1    5rA=174=34    rA=3202r_A + 7\left(\frac{1}{4} - r_A\right) = 1 \implies 2r_A + \frac{7}{4} - 7r_A = 1 \implies -5r_A = 1 - \frac{7}{4} = -\frac{3}{4} \implies r_A = \frac{3}{20}
Eliminating rBr_B isolates rAr_A as a single-variable linear equation.
4
Calculate the time required for Pump A to fill the tank alone
TimeA=1rA=1320=203 hours\text{Time}_A = \frac{1}{r_A} = \frac{1}{\frac{3}{20}} = \frac{20}{3}\text{ hours}
The total time to complete 1 unit of work is the reciprocal of the unit work rate.

Anahtar Kavram

Formulating and solving systems of two linear equations in two variables derived from work-rate relationships.
Soru 20Soru
Consider the following system of linear equations in variables xx, yy, and zz, where aa, bb, and cc are real constants:
2xy+3z=ax+2yz=b7x+4y+3z=c\begin{aligned} 2x - y + 3z &= a \\ x + 2y - z &= b \\ 7x + 4y + 3z &= c \end{aligned}
Which of the following statements must be true? Select all such statements.

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Cevap: If c=2a+3bc = 2a + 3b, the system has infinitely many solutions.; There exist no real values of aa, bb, and cc for which the system has a unique solution.

Cevap

The correct statements are: 'If c=2a+3bc = 2a + 3b, the system has infinitely many solutions' and 'There exist no real values of aa, bb, and cc for which the system has a unique solution.'
The correct options accurately reflect the structural properties of the system. First, scaling the first equation by 2 and the second by 3 yields 7x+4y+3z=2a+3b7x + 4y + 3z = 2a + 3b. Comparing this with the third equation, 7x+4y+3z=c7x + 4y + 3z = c, shows that when c=2a+3bc = 2a + 3b, the third equation provides no new constraints, leaving 2 independent equations in 3 variables and thus producing infinitely many solutions. Second, because the coefficient matrix has linearly dependent rows, its rank is 2 (less than the 3 variables), making a unique solution impossible regardless of the constants aa, bb, and cc.

Adım Adım Çözüm

1
Analyze the linear dependence of the left-hand sides of the equations.
Observe that 2(2xy+3z)+3(x+2yz)=(4x+3x)+(2y+6y)+(6z3z)=7x+4y+3z2(2x - y + 3z) + 3(x + 2y - z) = (4x + 3x) + (-2y + 6y) + (6z - 3z) = 7x + 4y + 3z.
Finding a linear combination of the first two equations that produces the left-hand side of the third equation allows us to analyze system consistency.
2
Determine the condition for consistency.
The system is consistent if and only if 2a+3b=c2a + 3b = c.
If c=2a+3bc = 2a + 3b, the third equation is a linear combination of the first two, resulting in a system of 2 independent equations in 3 variables, which yields infinitely many solutions.
3
Evaluate the possibility of a unique solution.
The rank of the coefficient matrix is 2, which is strictly less than the number of variables (3).
A system of linear equations has a unique solution if and only if the rank of the coefficient matrix equals the number of variables. Thus, no choice of a,b,ca, b, c can produce a unique solution.
4
Verify specific numerical options.
For a=0,b=0,c=0a=0, b=0, c=0, c=2(0)+3(0)=0c = 2(0)+3(0)=0, giving infinitely many solutions. For a=1,b=2,c=7a=1, b=2, c=7, 2(1)+3(2)=872(1)+3(2)=8 \neq 7, giving zero solutions.
Testing specific constant values confirms consistency or inconsistency based on whether c=2a+3bc = 2a + 3b is satisfied.

Anahtar Kavram

Consistency and Number of Solutions in 3x3 Systems of Linear Equations
Sayfa 1 / 2Sonraki