Systems of Linear Equations - Pre Algebra

1. Sarah is solving the system $x + 2y = 8$ and $3x - y = 1$ using substitution. She solves the first equation for $x$ and gets $x = 8 - 2y$. When she substitutes this into the second equation, what equation does she get?
   Answer: $3(8 - 2y) - y = 1$ When using substitution, we replace the variable we solved for (x) with its expression (8 - 2y) in the other equation. The second equation is 3x - y = 1. Substituting x = 8 - 2y gives us 3(8 - 2y) - y = 1. This can be simplified to 24 - 6y - y = 1, then 24 - 7y = 1, giving y = 23/7. Choice B incorrectly substitutes the expression for y instead of x. Choice C substitutes into the wrong equation (the first one we already used). Choice D omits the coefficient 3 that multiplies x in the second equation.
2. The solution to the system $ax + by = 12$ and $2x + 3y = 8$ is $(6, -\frac{4}{3})$. What is the value of $a + b$?
   Answer: $a + b = 10$ Since $(6, -\frac{4}{3})$ satisfies both equations, we can substitute into the first equation: $a(6) + b(-\frac{4}{3}) = 12$, which gives $6a - \frac{4b}{3} = 12$. Multiplying by 3: $18a - 4b = 36$, or $9a - 2b = 18$. We can verify the point satisfies the second equation: $2(6) + 3(-\frac{4}{3}) = 12 - 4 = 8$ ✓. From $9a - 2b = 18$, we need another relationship. Since this is a unique system with a specific intersection point, we can work with the constraint. Rearranging: $b = \frac{9a - 18}{2}$. For the answer choices, if $a + b = 10$, then $a + \frac{9a - 18}{2} = 10$, giving $\frac{2a + 9a - 18}{2} = 10$, so $11a - 18 = 20$, thus $a = \frac{38}{11}$ and $b = \frac{72}{11}$. Check: $\frac{38}{11} + \frac{72}{11} = \frac{110}{11} = 10$ ✓.
3. A rectangle has a perimeter of 36 units and an area of 77 square units. If the length is $x$ and the width is $y$, what is the value of $x + y$?
   Answer: 18 units From the given information: 2x + 2y = 36 (perimeter) and xy = 77 (area). From the first equation: x + y = 18. We can verify this works: if x + y = 18 and xy = 77, then we can solve for x and y individually. Let x = 18 - y, so (18 - y)y = 77, giving 18y - y² = 77, or y² - 18y + 77 = 0. Using the quadratic formula: y = (18 ± √(324 - 308))/2 = (18 ± √16)/2 = (18 ± 4)/2. So y = 11 or y = 7, meaning the dimensions are 11 and 7. Check: perimeter = 2(11 + 7) = 36 ✓, area = 11 × 7 = 77 ✓. Therefore x + y = 18. Choice B would give xy = 56, choice C would give xy = 27, and choice D would give xy = 14, none of which equal 77.
4. The solution to the system $2x + 3y = 12$ and $4x - 6y = k$ depends on the value of $k$. For which value of $k$ will the system have infinitely many solutions?
   Answer: $k = -24$ For a system to have infinitely many solutions, the equations must be equivalent (one is a multiple of the other). The first equation is 2x + 3y = 12. Multiplying by 2: 4x + 6y = 24. For the second equation 4x - 6y = k to be equivalent, we need 4x - 6y = k to represent the same line. This happens when 4x - 6y = k is equivalent to -(4x + 6y) = -24, which gives 4x - 6y = -24. Therefore k = -24. Choice B (k = 0) would give 4x - 6y = 0, making the lines intersect at one point. Choice C (k = 12) would make the system inconsistent. Choice D (k = 24) would make 4x - 6y = 24, which is also inconsistent with 2x + 3y = 12.
5. The lines $y = 2x - 1$ and $y = mx + 3$ intersect at the point $(2, 3)$. What is the value of $m$?
   Answer: $m = 0$ Since both lines pass through (2, 3), we can substitute this point into both equations. For the first line: 3 = 2(2) - 1 = 4 - 1 = 3 ✓. For the second line: 3 = m(2) + 3, so 3 = 2m + 3, which gives 0 = 2m, therefore m = 0. We can verify: the second line becomes y = 0x + 3 = 3, which is a horizontal line through (2, 3). The intersection point (2, 3) satisfies both equations when m = 0. Choice A gives y = -x + 3, which passes through (0, 3) and (3, 0), not (2, 3). Choice C gives y = x + 3, passing through (2, 5), not (2, 3). Choice D gives y = 2x + 3, passing through (2, 7), not (2, 3).
6. A theater sells adult tickets for $12 and student tickets for $8. On a particular day, they sold 150 tickets and collected $1,560 in revenue. If they had sold 10 more adult tickets and 10 fewer student tickets instead, what would their revenue have been?
   Answer: $1,600 Let a = adult tickets and s = student tickets. We have: a + s = 150 and 12a + 8s = 1560. From the first equation: s = 150 - a. Substituting: 12a + 8(150 - a) = 1560, so 12a + 1200 - 8a = 1560, giving 4a = 360, so a = 90. Therefore s = 60. Originally: 90 adult, 60 student tickets. With the change: 100 adult, 50 student tickets. New revenue = 12(100) + 8(50) = 1200 + 400 = $1,600. Choice A assumes they sold fewer of both types. Choice B assumes no change in revenue, which ignores that adult tickets cost more than student tickets. Choice D overcalculates the increase.
7. The system $3x - 2y = 7$ and $6x - 4y = 14$ has how many solutions?
   Answer: Infinitely many solutions To determine the number of solutions, we need to analyze the relationship between the equations. Notice that the second equation 6x - 4y = 14 is exactly twice the first equation 3x - 2y = 7. When we multiply the first equation by 2: 2(3x - 2y) = 2(7), we get 6x - 4y = 14, which is identical to the second equation. This means the two equations represent the same line, so there are infinitely many solutions. Choice A would be correct if the equations were parallel but distinct (like 3x - 2y = 7 and 6x - 4y = 15). Choice B would be correct if the lines had different slopes. Choice C is impossible for a system of two linear equations.

Start this quiz