Equations With Two Variables - PSAT Math

A store sells apples for $a$ dollars per pound and grapes for $g$ dollars per pound. A customer buys 2 pounds of apples and 3 pounds of grapes for $18$ , so $2a+3g=18$ . What is $a$ when $g=4$ ?
Answer: $a=3$ This problem gives $2a + 3g = 18$ and asks for a when $g = 4$ . Substituting $g = 4$ : $2a + 3(4) = 18$ , which gives $2a + 12 = 18$ . Solving for a: $2a = 18 - 12 = 6$ , so $a = 6 ÷ 2 = 3$ . Let me verify: $2(3) + 3(4) = 6 + 12 = 18$ ✓. This represents a customer buying apples at $3 per pound and grapes at $4 per pound. A common error is making arithmetic mistakes in substitution or division. When solving linear equations with two variables, substitute carefully and solve systematically.
A runner's distance is modeled by $d=1.2t$ , where $d$ is miles and $t$ is minutes. Which statement is true about the relationship?
Answer: Runner goes 1.2 mi/min This problem asks about the meaning of the relationship d = 1.2t, where d is miles and t is minutes. The coefficient 1.2 represents the rate of change: 1.2 miles per minute. This means the runner covers 1.2 miles for each minute of running, which is the runner's speed. The relationship shows that distance increases as time increases, with a constant rate. A common error is confusing the units or the direction of the relationship. In rate equations of the form distance = rate × time, the coefficient represents the speed or rate of travel.
The relationship between temperature in Celsius $C$ and Fahrenheit $F$ is $F=\frac{9}{5}C+32$ . What is $F$ when $C=25$ ?
Answer: $77$ This problem uses the temperature conversion formula F = (9/5)C + 32 and asks for F when C = 25. Substituting C = 25: F = (9/5)(25) + 32 = 9 × 5 + 32 = 45 + 32 = 77. This converts 25 degrees Celsius to Fahrenheit. The formula shows that each degree Celsius equals 9/5 degrees Fahrenheit, plus a 32-degree offset. A common error is making arithmetic mistakes with fractions or forgetting to add the constant term. When using conversion formulas, substitute carefully and follow the order of operations.
If $k(x-2)+3y=0$ and the graph of the equation passes through $(6,4)$ , what is the value of $k$ ?
Answer: $-3$ When you encounter a linear equation with an unknown parameter and are told the graph passes through a specific point, you need to use the fact that the coordinates of that point must satisfy the equation. Since the graph passes through $(6,4)$ , you can substitute $x = 6$ and $y = 4$ into the equation $k(x-2) + 3y = 0$ : $k(6-2) + 3(4) = 0$ $k(4) + 12 = 0$ $4k + 12 = 0$ $4k = -12$ $k = -3$ Let's examine why each answer choice leads to different conclusions. Choice A gives us $k = -3$ , which we just verified works perfectly. Choice B suggests $k = -4$ , which would give us $-4(4) + 12 = -16 + 12 = -4 \neq 0$ . Choice C proposes $k = 3$ , leading to $3(4) + 12 = 12 + 12 = 24 \neq 0$ . Choice D suggests $k = 4$ , which yields $4(4) + 12 = 16 + 12 = 28 \neq 0$ . The incorrect choices likely result from sign errors or arithmetic mistakes. For instance, you might get choice C if you forgot the negative sign when solving $4k = -12$ , or choice D if you incorrectly moved the 12 to the other side as positive. Strategy tip: When finding unknown parameters in equations, always substitute the given point's coordinates directly into the original equation and solve step-by-step. Double-check your arithmetic, especially with negative signs, as these are common sources of error on the PSAT.
A charity buys pens and notebooks. Each pen costs $0.50, and each notebook costs $1.20. The total cost is modeled by $0.50p+1.20n=T$ , where $p$ is the number of pens and $n$ is the number of notebooks. If the charity spends $72 and buys 60 pens, how many notebooks did it buy?
Answer: 35 When you encounter a linear equation with multiple variables and specific values given, you're dealing with substitution to solve for an unknown. This equation models a real-world scenario where you need to find one variable when others are known. Start by substituting the given values into the equation $0.50p + 1.20n = T$ . You know the charity spent $72 (so $T = 72$ ) and bought 60 pens (so $p = 60$ ). Substituting these values: $0.50(60) + 1.20n = 72$ . Calculate the cost of pens: $0.50 \times 60 = 30$ . This gives you $30 + 1.20n = 72$ . Subtract 30 from both sides: $1.20n = 42$ . Finally, divide by 1.20: $n = 42 \div 1.20 = 35$ notebooks. Looking at the wrong answers: (A) 25 would result from incorrectly calculating $30 \div 1.20$ instead of $42 \div 1.20$ . (C) 42 is the trap answer you get if you forget to divide by the notebook price—this is just the remaining money after buying pens, not the number of notebooks. (D) 60 might tempt you if you assume the charity bought equal quantities of each item, but this ignores the different prices entirely. Strategy tip: In substitution problems, always double-check your final answer by plugging all values back into the original equation. Here: $0.50(60) + 1.20(35) = 30 + 42 = 72$ ✓. This verification step catches arithmetic errors and builds confidence.
In a science lab, the amount of solution (in liters) is $S$ , made from water $w$ and concentrate $c$ . The relationship is $S=w+c$ . Rewrite the equation to solve for $c$ in terms of $S$ and $w$ .
Answer: $c=S-w$ The question requires rewriting the equation S = w + c to solve for c in terms of S and w. The equation describes the total solution S as the sum of water w and concentrate c. To isolate c, subtract w from both sides: c = S - w. This manipulation maintains the relationship between the variables. A common error is to add or multiply instead of subtracting, leading to incorrect expressions. Always check by plugging in numbers; for example, if S=5, w=3, c=2, and c =5-3=2 works. For test-taking, remember that solving for one variable in a two-variable equation involves isolating it using inverse operations.
A music streaming service charges $p$ dollars per month and also charges a one-time setup fee of $10$ dollars. The total paid after $m$ months is $T=pm+10$ . Which expression gives the monthly price $p$ in terms of $T$ and $m$ ?
Answer: $p=\dfrac{T-10}{m}$ The question requires expressing the monthly price p in terms of total T and months m, given T = p m + 10. This equation models total payment T as monthly charge p times m plus a fixed 10 dollars. To solve for p, subtract 10 from both sides: T - 10 = p m. Then, divide by m: p = (T - 10)/m. A common error is to add the 10 instead of subtracting or to invert the division. Check with numbers: if p=5, m=2, T=5*2 +10=20, then (20-10)/2=5, yes. Isolate the variable by reversing the operations in the correct order for such problems.
A fundraiser has the constraint $x+y=75$ , where $x$ is the number of online donations and $y$ is the number of in-person donations. If $y$ decreases by 12, how must $x$ change to keep the total number of donations the same?
Answer: Increase by 12 This problem involves the constraint x + y = 75, where if y decreases by 12, we need to find how x must change to maintain the total. If y decreases by 12, then to keep x + y = 75, x must increase by 12. The constraint equation requires that any decrease in one variable be offset by an equal increase in the other variable. A common error is thinking both variables change in the same direction or getting the magnitude wrong. In constraint equations where variables sum to a constant, changes in the variables must be equal and opposite.
If $2(x+y)=x-3y$ , what is $y$ in terms of $x$ ?
Answer: $y=\dfrac{-x}{5}$ This question tests your ability to solve linear equations for one variable in terms of another—a fundamental algebra skill that appears frequently on standardized tests. To isolate $y$ in terms of $x$ , start by distributing the 2 on the left side: $2x + 2y = x - 3y$ . Now collect all terms containing $y$ on one side and all terms with $x$ on the other. Subtract $2x$ from both sides: $2y = x - 3y - 2x$ , which simplifies to $2y = -x - 3y$ . Next, add $3y$ to both sides: $2y + 3y = -x$ , giving you $5y = -x$ . Finally, divide both sides by 5: $y = \frac{-x}{5}$ . Looking at the wrong answers: Choice B gives $y = \frac{x}{5}$ , which would result from forgetting the negative sign when moving terms—a common algebraic error. Choice C, $y = \frac{-5}{x}$ , incorrectly places $x$ in the denominator, likely from confusing which variable should be isolated. Choice D, $y = -5x$ , represents a sign error combined with incorrectly multiplying by 5 instead of dividing. You can verify the correct answer A by substituting back into the original equation: if $y = \frac{-x}{5}$ , then $2(x + \frac{-x}{5}) = 2(\frac{4x}{5}) = \frac{8x}{5}$ and $x - 3(\frac{-x}{5}) = x + \frac{3x}{5} = \frac{8x}{5}$ . ✓ Study tip: When solving for one variable in terms of another, always substitute your answer back into the original equation to check your work—it catches algebraic mistakes quickly.
The equation $9x-6y=30$ relates $x$ and $y$ . Which equation is equivalent and solved for $x$ ?
Answer: $x=\frac{6y+30}{9}$ This problem asks to solve $9x - 6y = 30$ for x. Adding 6y to both sides: $9x = 30 + 6y$ . Dividing by 9: $x = \frac{30 + 6y}{9}$ . This can be written as $x = \frac{6y + 30}{9}$ , which matches choice A. The equation expresses x in terms of y by isolating x on one side. A common error is making sign errors when rearranging terms or mistakes in division. When solving for one variable in terms of another, perform operations systematically to isolate the desired variable.
The relationship between the number of adult tickets $a$ and student tickets $s$ sold for a show is given by $12a+8s=480$ , where 12 and 8 are the ticket prices in dollars and 480 is the total revenue. Rewrite the equation to solve for $s$ in terms of $a$ .
Answer: $s=\frac{480-12a}{8}$ This problem asks to solve $12a + 8s = 480$ for s in terms of a. Starting with $12a + 8s = 480$ , subtract 12a from both sides: $8s = 480 - 12a$ . Then divide both sides by 8: $s = \frac{480 - 12a}{8}$ . This can also be written as $s = 60 - 1.5a$ , but the fraction form matches choice B exactly. A common error is incorrectly rearranging the terms or making sign errors when moving terms across the equal sign. When isolating a variable, perform the same operation to both sides of the equation systematically.
A scientist models a relationship by $3p+q=20$ , where $p$ and $q$ are variables. Which expression gives $q$ in terms of $p$ ?
Answer: $q=20-3p$ This problem asks to solve $3p + q = 20$ for q in terms of p. Starting with $3p + q = 20$ , subtract $3p$ from both sides: $q = 20 - 3p$ . This isolates q on one side and expresses it in terms of p. The relationship shows that as p increases, q decreases at a rate of 3 units per unit increase in p. A common error is making sign errors when moving terms across the equal sign. When solving for one variable in terms of another, isolate the desired variable by performing inverse operations systematically.
A delivery company's total earnings $E$ (in dollars) are related to the number of deliveries $d$ by $E=120+8d$ . Which statement best describes the meaning of the coefficient 8 in this context?
Answer: $8 per delivery The question asks which statement best describes the meaning of the coefficient 8 in E = 120 + 8d, where E is earnings and d is deliveries. This is a linear equation with intercept 120 and slope 8. The coefficient of d, which is 8, represents the earnings per delivery, so $8 per delivery. The constant 120 might be a fixed amount, not related to deliveries. A common error is to misinterpret the constant as the per-unit rate or vice versa. In context, identify the slope as the rate of change with respect to the independent variable. For test strategy, relate coefficients to real-world meanings in word problems.
A school orders $x$ pizzas and $y$ salads. Each pizza costs $12$ and each salad costs $5$ , with total cost $C$ . Which equation represents the total cost?
Answer: $C=12x+5y$ This problem asks for an equation representing the total cost when ordering x pizzas ($12 each) and y salads ($5 each). The total cost C equals the sum of the cost of pizzas and the cost of salads: C = 12x + 5y. Each pizza contributes $12 to the total cost, and each salad contributes $5. A common error is mixing up the prices or incorrectly writing the cost structure. When writing cost equations, multiply the quantity of each item by its unit price, then sum all the costs.
A concert venue has $r$ regular seats and $v$ VIP seats with the constraint $r+v=350$ . If the number of VIP seats is $v=120$ , what is $r$ ?
Answer: $230$ This problem gives the constraint r + v = 350 with v = 120, asking for r. Substituting v = 120: r + 120 = 350. Solving for r: r = 350 - 120 = 230. This represents a concert venue where the total number of regular and VIP seats is fixed. With 120 VIP seats, there are 230 regular seats. A common error is adding instead of subtracting or making arithmetic mistakes. In constraint equations where variables sum to a constant, one variable equals the constant minus the other.
A gym charges a one-time sign-up fee $s$ dollars and $m$ dollars per month for a membership. After $t$ months, the total cost is $C$ dollars, modeled by $C=s+mt$ . If the monthly fee $m$ increases by 3 while $s$ stays the same, by how much does $C$ change after 8 months?
Answer: $24$ This problem asks how the total cost C changes when the monthly fee increases by 3 dollars over 8 months. The original cost equation is $C = s + mt$ , and the new equation becomes $C = s + (m + 3)t$ . The change in cost is the difference: $[s + (m + 3)t] - [s + mt] = 3t$ . After 8 months, the change is $3 \times 8 = 24$ dollars. A common error would be calculating just the increase in monthly fee (3) without multiplying by the number of months. When dealing with linear relationships involving time, always multiply rate changes by the time period to find total change.
A landscaper's total charge $T$ (in dollars) is related to the number of labor hours $h$ by $T = 120 + 35h$ . What is the value of $T$ when $h=6$ ?
Answer: $330$ We need to find T when h = 6 in the equation T = 120 + 35h. Substituting h = 6 gives T = 120 + 35(6) = 120 + 210 = 330 dollars. The equation represents a linear relationship where 120 is the base charge and 35 is the hourly rate. A common error is arithmetic mistakes when multiplying 35 × 6. When evaluating linear functions, carefully perform the multiplication before adding the constant term.
The relationship between two variables is $y=7-2x$ . Which of the following is the value of $y$ when $x=-3$ ?
Answer: $13$ This problem gives y = 7 - 2x and asks for y when x = -3. Substituting x = -3: y = 7 - 2(-3) = 7 - (-6) = 7 + 6 = 13. The key is handling the double negative correctly: subtracting a negative number is the same as adding. A common error is making sign mistakes when dealing with negative numbers, especially with double negatives. When substituting negative values into expressions, be careful with signs and follow the order of operations.
The equation $y=\frac{1}{3}x-5$ models the relationship between $x$ and $y$ . If $x$ increases by 9, by how much does $y$ change?
Answer: $3$ This problem involves the equation y = (1/3)x - 5 and asks how y changes when x increases by 9. In a linear equation, the change in y equals the coefficient of x times the change in x: Δy = (1/3) × 9 = 3. The coefficient 1/3 represents the slope or rate of change. When x increases by 9, y increases by 3. A common error is forgetting to multiply by the coefficient or confusing the direction of change. In linear relationships of the form y = mx + b, when x changes by Δx, y changes by m × Δx.
A line passes through the points $(0,3)$ and $(2,-1)$ . Which equation represents the line?
Answer: $y=-2x+3$ This problem asks for the equation of a line passing through (0,3) and (2,-1). The y-intercept is 3 (since the line passes through (0,3)). The slope is m = (-1-3)/(2-0) = -4/2 = -2. Therefore, the equation is y = -2x + 3. Let me verify with both points: at (0,3): y = -2(0) + 3 = 3 ✓; at (2,-1): y = -2(2) + 3 = -4 + 3 = -1 ✓. A common error is making mistakes in slope calculation or misidentifying the y-intercept. When one point has x = 0, that gives the y-intercept directly.

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