--- title: "Substituting into formulas" subtitle: "Algebra Foundations · Expressions · Lesson 2" --- # Substituting into formulas **Name:** {{field-24}} **Date:** {{field-15}} **Class:** {{field-15}} > **Key question:** A formula is a recipe for a number. Put in the values you know, and it gives you the one you want: a perimeter, a cost, a speed. :::teacher-note **In one sentence:** Students substitute values into formulas, writing in the hidden × signs and using the order of operations, including decimals, division and brackets (AC9M7A01, node 2). **Students should be able to say:** I can replace each letter in a formula with its value, writing in the × signs. I can work out the result in the right order, including brackets and division. I can show my substitution as a line of working, with units. **Width:** the unmarked parts are the floor every student does (predict, run, investigate and make). Typical adds the modify practice; wider adds the real-world contexts. They add contexts, not difficulty. Students work in the online tutorial alongside this sheet. ::: ## Predict $$P = 2l + 2w$$ What does $2l$ mean? Write the formula again with the × signs in. :::keep-together **$P = 2l + 2w$. What is $P$ when $l = 5$ and $w = 3$?** ☐ 13 ☐ 16 ☐ 30 ☐ 48 My prediction, and why: {{lines-3}} ::: :::teacher-note **What happens:** 16. 2l means 2 × l = 2 × 5 = 10, and 2w = 2 × 3 = 6. So P = 10 + 6 = 16. (48 comes from reading 2l as 25 and 2w as 23.) ::: ## Run Try it on the board in the tutorial. Was your prediction right? What was different? {{lines-2}} ## Three steps :::keep-together ### Area Here is how to substitute, step by step. 1. $P = 2l + 2w$ — Write the formula first. 2. $l = 5, \; w = 3$ — Write down the values you know. 3. $P = 2 \times 5 + 2 \times 3$ — Replace each letter with its value. Write in the × signs that were hidden in $2l$ and $2w$. 4. $P = 10 + 6$ — Multiply before adding. 5. $P = 16 \text{ cm}$ — Add, and give the units. **The area of a rectangle is $A = lw$. Find $A$ when $l = 8$ and $w = 3$.** Answer: {{field-15}} ::: :::teacher-note **Answer:** 24 A = 8 × 3 = 24. ::: :::keep-together ### A plumber **A plumber charges $C = 5h + 20$ dollars. Find $C$ when $h = 3$.** Answer: {{field-15}} ::: :::teacher-note **Answer:** 35 C = 5 × 3 + 20 = 15 + 20 = 35 dollars. ::: :::keep-together ### Find the mistake **Leo works out a perimeter. Click the line where he goes wrong.** ``` 1 P = 2l + 2w, with l = 5 and w = 3. 2 P = 25 + 23 3 P = 48 ``` Line: {{field-15}} Why is it wrong? {{lines-2}} ::: :::teacher-note **Answer:** line 2 Line 2: 2l means 2 × l, so 2 × 5 = 10, not 25. With the × signs in: P = 2 × 5 + 2 × 3 = 10 + 6 = 16. ::: :::typical ## Other formulas **A decimal.** Each task loads a new formula. The board shows the substitution, but not the answer: work it out yourself. **Three letters.** **Divide.** **Brackets.** Work in the tutorial. Plan first: what will you change, and how will you test it? {{box-4}} ::: ## Substitute :::keep-together ### Tickets **Find $T = 12n + 6$ when $n = 4$.** Answer: {{field-15}} ::: :::teacher-note **Answer:** 54 T = 12 × 4 + 6 = 48 + 6 = 54. ::: :::keep-together ### Brackets again **Find $y = 3(x - 2)$ when $x = 7$.** Answer: {{field-15}} ::: :::teacher-note **Answer:** 15 y = 3 × (7 − 2) = 3 × 5 = 15. ::: :::keep-together ### A triangle **Find $A = \frac{bh}{2}$ (the area of a triangle) when $b = 6$ and $h = 5$.** Answer: {{field-15}} ::: :::teacher-note **Answer:** 15 A = 6 × 5 ÷ 2 = 30 ÷ 2 = 15. ::: :::keep-together ### Review **Review: in $V = 3t + 10$, which are the constants?** ☐ Only 10 ☐ 3 and 10 ☐ $V$ and $t$ ::: :::teacher-note **Answer:** 3 and 10 3 and 10 stay the same; V and t are variables that change. ::: ## Make: rectangles with the same area Find a rectangle with area 24 cm². Then find another one. How many can you find with whole-number sides? Stretch: which of your rectangles has the smallest perimeter? Use $P = 2l + 2w$ to check. Work in the tutorial. Plan first: what will you change, and how will you test it? {{box-4}} ## Exit ticket :::keep-together **Find C = 8h + 15 when h = 3, and V = lwh when l = 2, w = 5, h = 3.** {{lines-3}} ::: :::teacher-note **Answer:** C = 8 × 3 + 15 = 39. V = 2 × 5 × 3 = 30. :::