The question asks for the value of s when f=4 in the equation 2f + s = 11. This is a linear equation with two variables, constraining the amounts of flour f and sugar s. Substitute f=4: 24 + s = 11, which is 8 + s = 11. Then, subtract 8 from both sides: s = 11 - 8 = 3. A common error is to multiply instead of adding or solving for f instead. Verify by plugging back: 24 + 3 = 8+3=11, correct. When given a value for one variable, substitute directly to find the other in two-variable equations.

This problem asks to solve 5x + y = 3 for y. Subtracting 5x from both sides: y = 3 - 5x. This can also be written as y = -5x + 3, but the form y = 3 - 5x matches choice B exactly. The equation shows that y decreases by 5 units for each unit increase in x. A common error is making sign errors when moving terms across the equal sign. When isolating variables, perform the same operation to both sides systematically.

The question asks for the equation that solves for n in terms of T and s, given T = 4n + s. This equation models the total cost T as 4 times the price per notebook n plus shipping s. To solve for n, subtract s from both sides: T - s = 4n. Then, divide both sides by 4: n = (T - s)/4. A common error is to divide by something else or forget to subtract s first. Check by substituting back; if T=20, s=4, n=4, then 20=4*4 +4, yes, and (20-4)/4=4. Focus on inverse operations to isolate the desired variable in equations with multiple variables.

This problem involves y = (5/2)x - 1, where if x increases by 4, we need to find how y changes. The change in y equals the coefficient of x times the change in x: Δy = (5/2) × 4 = 5 × 2 = 10. The coefficient 5/2 represents the slope or rate of change. When x increases by 4, y increases by 10. A common error is making mistakes with fraction arithmetic or forgetting to multiply by the coefficient. In linear relationships, changes in the dependent variable equal the slope times changes in the independent variable.

This problem asks how much C changes when m increases from 40 to 52 in the equation C = 75 + 1.25m. The change in m is 52 - 40 = 12 miles. The change in C equals the coefficient of m times the change in m: ΔC = 1.25 × 12 = 15. The cost increases by $15 for the additional 12 miles driven. A common error is calculating the wrong change in distance or making arithmetic mistakes with decimals. When finding changes in linear expressions, multiply the coefficient by the change in the variable.

This problem asks for the equation of a line passing through (2,1) and (6,9). First, find the slope: m = (9-1)/(6-2) = 8/4 = 2. Using point-slope form with point (2,1): y - 1 = 2(x - 2), which gives y - 1 = 2x - 4, so y = 2x - 3. Let me verify with both points: at (2,1): y = 2(2) - 3 = 1 ✓; at (6,9): y = 2(6) - 3 = 9 ✓. A common error is making arithmetic mistakes in the slope calculation or when converting to slope-intercept form. When finding line equations, calculate slope first, then use point-slope form to find the equation.

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.

This problem involves the relationship d = 55t, where if t decreases by 0.5 hour, we need to find how d changes. The change in d equals the rate (55) times the change in t: Δd = 55 × (-0.5) = -27.5. Since the change is negative, d decreases by 27.5 miles. The relationship shows that distance is directly proportional to time with a rate of 55 miles per hour. A common error is forgetting the negative sign when t decreases, or confusing the direction of change. In linear relationships, the change in the dependent variable equals the coefficient times the change in the independent variable.

This is a linear equation application where you're given specific values and need to solve for an unknown variable. When you see a real-world scenario with a formula already provided, substitute the known values and solve algebraically.

You're told the charity spends $72 total and buys 60 pens. Substitute these values into the equation $$0.50p + 1.20n = T$$:

$$0.50(60) + 1.20n = 72$$

First, calculate the cost of the pens: $$0.50 \times 60 = 30$$

So the equation becomes: $$30 + 1.20n = 72$$

Subtract 30 from both sides: $$1.20n = 42$$

Divide by 1.20: $$n = \frac{42}{1.20} = 35$$

The charity bought 35 notebooks.

Looking at the wrong answers: Choice (A) 25 would result from incorrectly calculating $$\frac{30}{1.20}$$ instead of $$\frac{42}{1.20}$$, mixing up which dollar amount to divide. Choice (C) 42 comes from forgetting to divide by the notebook price—this is just the remaining dollars after buying pens, not the number of notebooks. Choice (D) 60 assumes they bought the same number of notebooks as pens, ignoring the price difference entirely.

Strategy tip: In substitution problems, work methodically through each step and always check your answer makes sense. Here, 35 notebooks at $1.20 each plus 60 pens at $0.50 each should equal $72: $$35(1.20) + 60(0.50) = 42 + 30 = 72$$ ✓

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.