Algebra 1, Lesson 83: Word Problems with Two Statements of Equality

Lesson 83 focuses on solving word problems by identifying and establishing two distinct statements of equality. This builds upon foundational skills introduced in Lesson 82 and other previous lessons regarding systems of linear equations.
Systems of linear equations have appeared in every practice set since Lesson 61, emphasizing their importance as a core Algebra 1 skill with numerous real-world applications.
While specific word problems (such as counting coins or dividing rope) may seem artificial or less efficient than just counting, their primary purpose is to develop the structural ability to compare two or more mathematical patterns.
Learning to solve these systems provides a mathematical "toolbox" for observing, interpreting, and modeling nature and creation. Mastery of these patterns prevents difficulty when applying these skills to complex real-world situations in the future.

The most critical step in Lesson 83 is searching for two statements of equality within the text of the word problem.
Keywords such as "is" typically signal an equality, while terms like "sum," "difference," "ratio," "more than," and "less than" dictate the operations required to build the equations.
Students are encouraged to use descriptive variables rather than generic ones to keep information organized. For example, using $l$ for larger and $s$ for smaller, or $n$ for numerator and $d$ for denominator.
Descriptive variables help track what each number represent throughout the substitution or elimination process.

The problem states: "The sum of two numbers is 50 and their difference is 18. Find the numbers."
Equation for the sum: $l + s = 50$ .
Equation for the difference: $l - s = 18$ .
Solving by elimination:
- Adding the two equations together: $2l = 68$ .
- Dividing by 2: $l = 34$ .
Solving for the smaller number:
- Substitute $l$ back into the sum equation: $34 + s = 50$ .
- Subtracting 34 from both sides: $s = 16$ .
The two numbers are 34 and 16.

The problem states: "The ratio of two numbers is 2 to 3 and their sum is 110. What are the two numbers?"
First equation (ratio): $\frac{n}{d} = \frac{2}{3}$ .
Second equation (sum): $n + d = 110$ .
Solving using the substitution method:
- Rearrange the ratio equation to solve for $n$ : $n = \frac{2}{3}d$ .
- Substitute this into the sum equation: $\frac{2}{3}d + d = 110$ .
- Convert $d$ to a common denominator: $\frac{2}{3}d + \frac{3}{3}d = 110$ .
- Simplify: $\frac{5}{3}d = 110$ .
- Multiply by 3: $5d = 330$ .
- Divide by 5: $d = 66$ .
Solve for the numerator:
- $n = \frac{2}{3} \times 66$ .
- Simplifying $66/3$ to 22: $n = 2 \times 22 = 44$ .
Verification: $\frac{44}{66}$ simplifies to $\frac{2}{3}$ , and $44 + 66 = 110$ .

The problem states: "The larger of two numbers is 15 greater than the smaller number. When added together, their sum is 7 less than three times the small number. Find the numbers."
Equation 1: $l = s + 15$ .
Equation 2: $l + s = 3s - 7$ .
Substitution process:
- Replace $l$ in the second equation with $(s + 15)$ : $(s + 15) + s = 3s - 7$ .
- Combine like terms: $15 + 2s = 3s - 7$ .
- Rearrange variables: Subtract $2s$ from both sides: $15 = s - 7$ .
- Add 7 to both sides: $s = 22$ .
Solving for the larger number:
- $l = 22 + 15 = 37$ .
The numbers are 22 and 37.

The problem states: "A handyman cut a 125 foot rope into two pieces. The short piece was 18 feet shorter than the long piece. How long was each piece?"
Total quantity equation: $l + s = 125\,\text{feet}$ .
Relationship equation: $s = l - 18$ .
Substitution:
- $l + (l - 18) = 125$ .
- Combine terms: $2l - 18 = 125$ .
- Add 18 to both sides: $2l = 143$ .
- Divide by 2: $l = 71.5\,\text{feet}$ .
Solving for short piece:
- $s = 71.5 - 18 = 53.5\,\text{feet}$ .
Verification: $71.5 + 53.5 = 125$ .

These problems are common on standardized tests like the SAT and ACT. Test creators use them to verify that a student can read multiple sentences, distill them into equations, and solve for multiple unknowns.
Using subscripted variables (e.g., $n_b$ and $n_f$ ) is presented as a professional algebraic technique to maintain clarity during complex problems.

The problem states: "Together, Bubba and Forrest processed 300 pounds of shrimp. If Bubba processed 36 more pounds than Forrest, how many pounds did they each process?"
Variables: $n_b$ (pounds Bubba processed), $n_f$ (pounds Forrest processed).
Equation 1: $n_b + n_f = 300$ .
Equation 2: $n_b = n_f + 36$ .
Substitution:
- $(n_f + 36) + n_f = 300$ .
- Combine terms: $2n_f + 36 = 300$ .
- Subtract 36: $2n_f = 264$ .
- Divide by 2: $n_f = 132\,\text{lb}$ .
Solving for Bubba:
- $n_b = 132 + 36 = 168\,\text{lb}$ .

This problem is derived from Leonhard Euler's book Elements of Algebra (1765), which serves as the foundation for modern Algebra 1 and 2 courses.
Problem: "Find two numbers in the proportion of 2 to 1 so that if 4 be added to each, the two sums shall be in the proportion of 3 to 2."
Initial Proportion: $\frac{n}{d} = \frac{2}{1} \rightarrow n = 2d$ .
Subsequent Proportion: $\frac{n+4}{d+4} = \frac{3}{2}$ .
Solving the system:
- Cross-multiply the second equation: $2(n + 4) = 3(d + 4)$ .
- Expand: $2n + 8 = 3d + 12$ .
- Substitute $n = 2d$ into the expanded equation: $2(2d) + 8 = 3d + 12$ .
- Simplify: $4d + 8 = 3d + 12$ .
- Subtract $3d$ and subtract 8: $d = 4$ .
Solve for numerator:
- $n = 2 \times 4 = 8$ .
Check: Initial ratio is $\frac{8}{4} = \frac{2}{1}$ . Adding 4 gives $\frac{12}{8}$ , which reduces to $\frac{3}{2}$ .