PSAT/NMSQT Math · Lesson 1 of 4
Linear models and systems
Turn a constant-rate situation into a linear model, and solve systems that represent two conditions at once.
Slope as a rate
If a rental costs $29 for 3 hours and $47 for 6 hours, the change in cost over 3 hours is $18, so the hourly rate is $6. The total is not simply hours times a single charge when a fixed fee is also present. Substituting one point into C = 6h + b gives the fee.
The slope is change in output over change in input. Units belong in the interpretation: dollars per hour, not just “6.” A negative slope means the output decreases as the input increases.
Two constraints
A system is the right tool when one equation cannot carry all the information. “Three notebooks and two pens cost $11” and “one notebook and two pens cost $7” differ by two notebooks and $4, so each notebook costs $2. Then each pen costs $2.50.
Graphically, the solution is the intersection. If the lines are parallel and distinct, there is no solution. If they are the same line, every point on that line works.
Worked example
C = 6h + b, and C = 29 when h = 3. Find the fixed fee b.
- Substitute the known point: 29 = 6 × 3 + b.
- 29 = 18 + b, so b = 11.
- Check the other point if it is given. For h = 6, C = 36 + 11 = 47.
Why this works. Use a rate from two points, then one point to recover the starting fee.
Check your understanding
- Compute a slope from two points before guessing a formula.
- Include units when you describe the slope.
- Verify a solution of a system in both equations.