Substitution & Elimination - Finite Mathematics

Given the system of equations: $\begin{cases} \frac{1}{2}x - \frac{1}{3}y = 1 \\ \frac{1}{4}x + \frac{2}{3}y = 6 \end{cases}$ What is the value of $x$ in the solution?
Answer: $6.4$ To solve the system, first clear the fractions from both equations. Multiply the first equation by 6 and the second equation by 12 (the least common multiple of the denominators in each equation). First equation: $6(\frac{1}{2}x - \frac{1}{3}y) = 6(1) \implies 3x - 2y = 6$ . Second equation: $12(\frac{1}{4}x + \frac{2}{3}y) = 12(6) \implies 3x + 8y = 72$ . Now we have a simpler system: (1) $3x - 2y = 6$ (2) $3x + 8y = 72$ Subtract equation (1) from equation (2): $(3x + 8y) - (3x - 2y) = 72 - 6 \implies 10y = 66 \implies y = 6.6$ . Substitute $y=6.6$ into equation (1): $3x - 2(6.6) = 6 \implies 3x - 13.2 = 6 \implies 3x = 19.2 \implies x = 6.4$ .
Consider the system of equations: $\begin{cases} 2x + 3y = -1 \\ 5x + 2y = 14 \end{cases}$ If $(x, y)$ is the solution to the system, what is the value of the expression $x - y$ ?
Answer: $7$ We can solve the system using the elimination method. Multiply the first equation by 2 and the second equation by $-3$ to eliminate $y$ . $2(2x + 3y = -1) \implies 4x + 6y = -2$ $-3(5x + 2y = 14) \implies -15x - 6y = -42$ Now, add the two new equations: $(4x + 6y) + (-15x - 6y) = -2 + (-42) \implies -11x = -44 \implies x = 4$ . Substitute $x = 4$ into the first original equation: $2(4) + 3y = -1 \implies 8 + 3y = -1 \implies 3y = -9 \implies y = -3$ . The solution is $(4, -3)$ . The question asks for the value of $x - y$ . So, $x - y = 4 - (-3) = 4 + 3 = 7$ .
The system $\begin{cases} ax + 2y = 6 \\ 3x + by = 9 \end{cases}$ has infinitely many solutions. If $a > 0$ , what is the value of $ab$ ?
Answer: $ab = 6$ When you encounter a system of linear equations with "infinitely many solutions," you're dealing with dependent equations—essentially the same line written in two different forms. This means one equation is a scalar multiple of the other. For the system to have infinitely many solutions, the second equation must be a multiple of the first. Let's find this relationship by comparing coefficients and constants. If we multiply the first equation $ax + 2y = 6$ by some constant $k$ , we get $kax + 2ky = 6k$ . For this to equal the second equation $3x + by = 9$ , we need: - $ka = 3$ - $2k = b$ - $6k = 9$ From the third equation: $k = \frac{9}{6} = \frac{3}{2}$ Substituting back: $a = \frac{3}{k} = \frac{3}{\frac{3}{2}} = 2$ and $b = 2k = 2 \cdot \frac{3}{2} = 3$ Therefore, $ab = 2 \times 3 = 6$ , which is answer choice D. Let's check why the other answers are wrong. Choice A ( $ab = 18$ ) would require much larger values that don't satisfy our constraint equations. Choice B ( $ab = 9$ ) might tempt you if you mistakenly thought $ab$ should equal the constant on the right side of the second equation. Choice C ( $ab = 12$ ) has no clear mathematical justification from our system. Study tip: For infinite solutions, remember the key relationship: corresponding coefficients and constants must have the same ratio. Set up the proportion $\frac{a}{3} = \frac{2}{b} = \frac{6}{9}$ and solve systematically.
Consider solving $\begin{cases} 5x + 2y = 23 \\ 3x - 4y = 1 \end{cases}$ by elimination. To eliminate $y$ , the first equation should be multiplied by $m$ and the second by $n$ . If $m$ and $n$ are chosen to be positive integers with no common factors, what is $m + n$ ?
Answer: 3 To eliminate $y$ , we need the coefficients to be opposites. The LCM of 2 and 4 is 4. Multiply the first equation by 2 to get $10x + 4y = 46$ , and the second equation by 1 to get $3x - 4y = 1$ . Since we need positive integers with no common factors and the $y$ coefficients to cancel, we use $m = 2, n = 1$ . Thus $m + n = 3$ . Choice B uses $m = 1, n = 4$ . Choice C uses $m = 2, n = 4$ (but these have a common factor). Choice D uses $m = 4, n = 3$ .
Sarah is solving the system $\begin{cases} 2x + 3y = 16 \\ 5x - 2y = 7 \end{cases}$ using elimination. She multiplies the first equation by 2 and the second equation by 3, then adds them. What is the coefficient of $x$ in her resulting equation?
Answer: 19 Multiplying the first equation by 2: $4x + 6y = 32$ . Multiplying the second equation by 3: $15x - 6y = 21$ . Adding these equations: $(4x + 6y) + (15x - 6y) = 32 + 21$ , which gives $19x = 53$ . The coefficient of $x$ is 19. Choice A only considers the first equation's coefficient after multiplication. Choice B only considers the second equation's coefficient after multiplication. Choice D incorrectly adds the constant terms instead of the coefficients.
A company produces two products, X and Y. The total production cost is $3x + 2y = 1200$ where $x$ and $y$ are the number of units produced. The profit constraint is $5x + 4y = 2200$ . If the company decides to reduce product X production by 50 units while maintaining the same total cost, how many additional units of product Y must be produced?
Answer: 75 units From the cost equation $3x + 2y = 1200$ , if $x$ decreases by 50, then $3(-50) + 2(\Delta y) = 0$ , so $-150 + 2(\Delta y) = 0$ , giving $\Delta y = 75$ . The profit constraint is irrelevant to this calculation since we're only maintaining the same total cost. Choice B assumes a 1:2 ratio error. Choice C uses the profit equation coefficients incorrectly. Choice D assumes a 1:3 ratio based on the cost coefficients.
The following system of equations involves the variables $x$ , $y$ , and a parameter $k$ . What is the value of $k$ that satisfies the system? $\begin{cases} 3x - 5y = 11 \\ x + y = 3k \\ y - x = k - 2 \end{cases}$
Answer: $-3/7$ This is a system of three equations with three unknowns ( $x, y, k$ ). We can use the second and third equations to express $x$ and $y$ in terms of $k$ . Let's rewrite the third equation as $-x + y = k-2$ . Now we have a 2x2 system for $x$ and $y$ : (2) $x + y = 3k$ (3) $-x + y = k - 2$ Add the two equations: $(x+y) + (-x+y) = 3k + (k-2) \implies 2y = 4k - 2 \implies y = 2k - 1$ . Subtract the third equation from the second: $(x+y) - (-x+y) = 3k - (k-2) \implies 2x = 2k + 2 \implies x = k + 1$ . Now substitute these expressions for $x$ and $y$ into the first equation: $3(k+1) - 5(2k-1) = 11$ . $3k + 3 - 10k + 5 = 11$ $-7k + 8 = 11$ $-7k = 3$ $k = -3/7$ .
Marcus attempts to solve $\begin{cases} 3x + 4y = 12 \\ 6x + 8y = 30 \end{cases}$ by substitution. From the first equation, he solves for $x = 4 - \frac{4y}{3}$ and substitutes into the second equation. What equation does he obtain?
Answer: $24 - \frac{24y}{3} + 8y = 30$ Substituting $x = 4 - \frac{4y}{3}$ into $6x + 8y = 30$ : $6(4 - \frac{4y}{3}) + 8y = 30$ , which expands to $24 - \frac{24y}{3} + 8y = 30$ . Note that this system actually has no solution since the equations are inconsistent. Choice A incorrectly cancels terms prematurely. Choice C shows the substitution but doesn't distribute the 6. Choice D has the wrong constant term on the right side.
Lisa solves $\begin{cases} 2x - y = 5 \\ x + 3y = 4 \end{cases}$ and gets $x = 2$ . She substitutes back to find $y$ but makes an error and uses the wrong equation's rearranged form. If she incorrectly uses $y = 4 - x$ instead of the correct form, what value of $y$ does she calculate?
Answer: $y = 2$ Lisa incorrectly uses $y = 4 - x$ (which comes from incorrectly rearranging $x + 3y = 4$ ). With $x = 2$ , she calculates $y = 4 - 2 = 2$ . The correct form should be $y = \frac{4-x}{3}$ , which would give $y = \frac{4-2}{3} = \frac{2}{3}$ , but that's not what she calculates due to her error. Choice A represents the correct $y$ value if properly solved. Choice B might result from other algebraic errors. Choice D comes from using $y = x + 1$ incorrectly.
Which of the following statements best describes the geometric interpretation of the system of equations and its solution set? $\begin{cases} 2x - y = 3 \\ -4x + 2y = -6 \end{cases}$
Answer: The equations represent the same line, so there are infinitely many solutions. To understand the relationship between the two equations, we can try to manipulate one to look like the other. If we multiply the first equation, $2x - y = 3$ , by $-2$ , we get $-2(2x - y) = -2(3)$ , which simplifies to $-4x + 2y = -6$ . This is identical to the second equation. Since both equations are equivalent, they represent the same line in the coordinate plane. When two equations in a system represent the same line, every point on that line is a solution, meaning there are infinitely many solutions.
A system of equations has the solution $(x,y) = (3,-2)$ . If both equations in the system are multiplied by different non-zero constants and then one equation is replaced by the sum of the two resulting equations, which statement about the new system is true?
Answer: The new system will have the same unique solution as the original system When you encounter questions about transforming systems of equations, remember that certain operations preserve the solution set while others don't. The key insight here is understanding which algebraic operations maintain equivalence between systems. Let's think through what happens step by step. You start with a system where $(x,y) = (3,-2)$ is the solution. When you multiply each equation by different non-zero constants, you're performing elementary row operations that don't change the solution set - they just scale the equations. Then, replacing one equation with the sum of these two scaled equations is another elementary row operation (adding one equation to another). These operations - multiplying by non-zero constants and adding equations together - are fundamental techniques used in methods like Gaussian elimination. They create an equivalent system, meaning the solution set remains identical to the original system. Choice B is correct because elementary row operations always preserve the solution set, so the new system has the same unique solution $(3,-2)$ . Choice A is wrong because the original system had a unique solution, and equivalent transformations can't change a unique solution into infinitely many solutions. Choice C reflects a common misconception - while the equations look different, algebraic equivalence means the solution set is unchanged. Choice D suggests the solution would change, but elementary row operations specifically preserve solutions. Study tip: Remember that multiplying equations by non-zero constants and adding/subtracting equations are solution-preserving operations. When you see these transformations described, the system remains equivalent to the original.
For which value of $c$ does the following system of equations have infinitely many solutions? $\begin{cases} 3x - 2y = c \\ -9x + 6y = -15 \end{cases}$
Answer: $5$ A system of linear equations has infinitely many solutions if the two equations represent the same line. This occurs when one equation is a constant multiple of the other. Let's multiply the first equation by $-3$ to match the coefficient of $x$ in the second equation: $-3(3x - 2y = c) \implies -9x + 6y = -3c$ . For this equation to be identical to $-9x + 6y = -15$ , their constant terms must be equal. Therefore, we must have $-3c = -15$ . Dividing by $-3$ gives $c = 5$ .
Consider the system of linear equations: $\begin{cases} x + y + z = 4 \\ 2x - y + z = 8 \\ x + 2y - z = -3 \end{cases}$ If $(x_0, y_0, z_0)$ is the unique solution to the system, what is the value of the product $x_0 y_0$ ?
Answer: $-2$ First, solve the system of three linear equations. Let the equations be (1), (2), and (3). Add equation (1) and (3) to eliminate $z$ : $(x+y+z) + (x+2y-z) = 4 + (-3) \implies 2x+3y=1$ . Let this be equation (4). Add equation (2) and (3) to eliminate $z$ : $(2x-y+z) + (x+2y-z) = 8 + (-3) \implies 3x+y=5$ . Let this be equation (5). From equation (5), solve for $y$ : $y = 5-3x$ . Substitute this expression for $y$ into equation (4): $2x + 3(5-3x) = 1 \implies 2x + 15 - 9x = 1 \implies -7x = -14 \implies x_0 = 2$ . Now find $y_0$ : $y_0 = 5 - 3(2) = 5 - 6 = -1$ . Finally, find $z_0$ using equation (1): $2 + (-1) + z_0 = 4 \implies 1 + z_0 = 4 \implies z_0 = 3$ . The solution is $(2, -1, 3)$ . The question asks for the product $x_0 y_0$ , which is $(2)(-1) = -2$ .
A system of two linear equations in two variables has a unique solution $(x_0, y_0)$ . If the constant terms of both equations are doubled while the coefficients of the variables remain unchanged, what is the solution to the new system?
Answer: $(2x_0, 2y_0)$ Let the original system be: $a_1x + b_1y = c_1$ $a_2x + b_2y = c_2$ We are given that $(x_0, y_0)$ is the unique solution, which means that $a_1x_0 + b_1y_0 = c_1$ and $a_2x_0 + b_2y_0 = c_2$ are true statements. The new system is formed by doubling the constant terms: $a_1x + b_1y = 2c_1$ $a_2x + b_2y = 2c_2$ Let the new solution be $(x_n, y_n)$ . We can substitute the expressions for $c_1$ and $c_2$ from the original solution into the new system: $a_1x_n + b_1y_n = 2(a_1x_0 + b_1y_0) = a_1(2x_0) + b_1(2y_0)$ $a_2x_n + b_2y_n = 2(a_2x_0 + b_2y_0) = a_2(2x_0) + b_2(2y_0)$ By comparing the left and right sides of these equations, we can see that if we set $x_n = 2x_0$ and $y_n = 2y_0$ , both equations are satisfied. Since the coefficients of the variables are unchanged, the slopes of the lines are the same, and the new system also has a unique solution. Thus, the new solution is $(2x_0, 2y_0)$ .
To operate at full capacity, meaning all available time for machining and finishing is used, what is the total number of products (Standard and Deluxe) the company must produce per week?
Answer: 70 Let $s$ be the number of Standard models and $d$ be the number of Deluxe models. We can set up a system of linear equations based on the available time. Machining time: $3s + 2d = 180$ Finishing time: $s + 2d = 100$ To solve this system, we can use elimination. Subtract the second equation from the first: $(3s + 2d) - (s + 2d) = 180 - 100 \implies 2s = 80 \implies s = 40$ . Now substitute the value of $s$ into the second equation: $40 + 2d = 100 \implies 2d = 60 \implies d = 30$ . The company must produce 40 Standard models and 30 Deluxe models. The question asks for the total number of products, which is $s + d = 40 + 30 = 70$ .
To obtain the desired mixture, how many more liters of Solution A than Solution B are needed?
Answer: 20 Let $A$ be the number of liters of Solution A and $B$ be the number of liters of Solution B. We can set up a system of two linear equations. The first equation is for the total volume: $A + B = 100$ . The second equation is for the amount of pure acid: $0.20A + 0.70B = 0.40(100) \implies 0.2A + 0.7B = 40$ . From the first equation, we can write $A = 100 - B$ . Substitute this into the second equation: $0.2(100 - B) + 0.7B = 40$ . $20 - 0.2B + 0.7B = 40$ . $20 + 0.5B = 40$ . $0.5B = 20$ . $B = 40$ liters. Now find the value of $A$ : $A = 100 - B = 100 - 40 = 60$ liters. The question asks for how many more liters of Solution A than Solution B are needed, which is the value of $A - B$ . $A - B = 60 - 40 = 20$ liters.
A student solves $\begin{cases} 4x + 3y = 25 \\ 2x - y = 3 \end{cases}$ using substitution. From the second equation, she correctly finds $y = 2x - 3$ . However, when substituting into the first equation, she writes $4x + 3(2x - 3) = 25$ but then incorrectly simplifies the left side. If her error is distributing the 3 to only the first term, what equation does she write?
Answer: $4x + 6x - 3 = 25$ This problem tests your understanding of the distributive property and how common mistakes occur when solving systems of equations by substitution. When you substitute $y = 2x - 3$ into $4x + 3y = 25$ , you get $4x + 3(2x - 3) = 25$ . The correct way to distribute the 3 would give you $4x + 6x - 9 = 25$ . However, the student makes a specific error: she only distributes the 3 to the first term inside the parentheses. If you distribute 3 to only the first term $2x$ , you get $3 \cdot 2x = 6x$ , but you leave the second term unchanged as $-3$ instead of multiplying it by 3. This gives you $4x + 6x - 3 = 25$ , which is answer choice D. Let's see why the other options don't match this specific error. Choice A ( $4x + 2x - 9$ ) would result if you forgot to distribute entirely to the first term but somehow still distributed to the second term. Choice B ( $4x + 6x - 9$ ) is actually the correct distribution with no errors. Choice C ( $4x + 5x - 3$ ) doesn't match any logical mistake pattern from the given setup. Remember that the distributive property requires you to multiply the outside number by every term inside the parentheses. When practicing substitution problems, always double-check your distribution step by ensuring you've multiplied the coefficient by each term, including any negative signs.
For what value of the constant $k$ does the following system of linear equations have no solution? $\begin{cases} kx + 8y = 4 \\ 2x + ky = -2 \end{cases}$
Answer: $k = 4$ A system of two linear equations has no solution if the lines are parallel but not coincident. For a system $a_1x+b_1y=c_1$ and $a_2x+b_2y=c_2$ , this condition is given by the relationship $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ . For the given system, we have $\frac{k}{2} = \frac{8}{k} \neq \frac{4}{-2}$ . The first part of the relationship, $\frac{k}{2} = \frac{8}{k}$ , implies $k^2 = 16$ , so $k = 4$ or $k = -4$ . We must check these values against the second part of the condition. If $k=4$ , we have $\frac{4}{2} = \frac{8}{4} \neq \frac{4}{-2}$ , which simplifies to $2 = 2 \neq -2$ . This is true, so the system has no solution when $k=4$ . If $k=-4$ , we have $\frac{-4}{2} = \frac{8}{-4} \neq \frac{4}{-2}$ , which simplifies to $-2 = -2 = -2$ . In this case, the ratios are all equal, meaning the lines are coincident and there are infinitely many solutions.

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