CTJan27 Online Year 7 - Solving Multistep Equations

Questions for: Solving Two-Step Equations Review

The Mystery Box

Imagine someone gives you a locked box. Inside is a prize, let's call it '$x$'. To get the prize, you have to undo two locks. Maybe first you have to use a key, and then you have to enter a code. You have to do these steps in the right order, right? Solving an equation is exactly the same!

An equation like $2x + 5 = 15$ is a mystery box. The '$x$' is the prize. The operations (multiplying by 2, adding 5) are the locks. To find '$x$', we have to undo those locks in the reverse order.

From Real Life to Algebra

Let's make this real. You buy a movie ticket for 10 dollars and two identical snacks. Your total cost is 18 dollars. How much did each snack cost?

• Concrete: You have the total cost (18 dollars). You know the ticket's cost (10 dollars). The rest of the money was for the two snacks.

• Pictorial: Think of a balance scale.

  [Image showing a balance scale. On the left side are two bags labeled 's' and a weight labeled '10'. On the right side is a weight labeled '18'.]

  To find the weight of one bag ('$s$'), you first remove the 10-unit weight from both sides to keep it balanced. Now you have two 's' bags on the left and an 8-unit weight on the right. If two bags weigh 8, then one bag must weigh 4.

• Abstract: We can write this as an equation. Let '$s$' be the cost of one snack.

  $2s + 10 = 18$

The Golden Rule: SADMEP

You've probably learned the Order of Operations: PEMDAS (Parentheses, Exponents, Multiplication/Division, Addition/Subtraction). It tells you the order to put the locks on.

To solve an equation, we need to take the locks off in the reverse order. So, we reverse PEMDAS!

Meet SADMEP:

1. Subtraction / Addition
2. Division / Multiplication
3. Exponents
4. Parentheses

For the two-step equations we're reviewing, we only need to worry about the first two. Always handle Addition or Subtraction first, then handle Multiplication or Division.

Worked Example + Your Turn

Let's solve that snack problem using SADMEP: $2s + 10 = 18$

Step 1: Undo Addition/Subtraction. The equation has + 10. The opposite of adding 10 is subtracting 10. To keep the scale balanced, we must do it to both sides.

$2s + 10 - 10 = 18 - 10$ $2s = 8$

Step 2: Undo Multiplication/Division. The variable '$s$' is being multiplied by 2. The opposite of multiplying by 2 is dividing by 2. Let's do that to both sides.

$\frac{2s}{2} = \frac{8}{2}$ $s = 4$

So, each snack cost 4 dollars. It works!

✨ Your Turn!

A fitness club charges a 20 dollar sign-up fee and then 15 dollars per month. If your total cost for a certain number of months was 110 dollars, how many months did you sign up for? Try writing the equation first, then solve it. The equation is $15m + 20 = 110$.

Another Example: Division in Disguise

What about an equation like $\frac{x}{3} - 4 = 2$?

Step 1: Undo Addition/Subtraction. The equation has - 4. The opposite of subtracting 4 is adding 4 to both sides.

$\frac{x}{3} - 4 + 4 = 2 + 4$ $\frac{x}{3} = 6$

Step 2: Undo Multiplication/Division. The variable '$x$' is being divided by 3. The opposite of dividing by 3 is multiplying by 3.

$\frac{x}{3} \times 3 = 6 \times 3$ $x = 18$

✨ Your Turn!

Solve this one: $\frac{k}{5} + 7 = 11$.

Remember the two simple steps, and you can unlock any two-step equation!

Questions for: Combining Like Terms to Solve Equations

The Messy Backpack Analogy

Imagine your backpack after a long week. You've got pencils, pens, maybe a few highlighters, and some erasers all jumbled together. If someone asked, "How many writing tools do you have?" you wouldn't just count every single item. You'd probably group them first, right? You'd say, "I have 5 pencils, 3 pens, and 2 highlighters."

Algebra can be a lot like that! An equation like $3x + 5 + 2x = 15$ is a bit like that messy backpack. The terms $3x$ and $2x$ are like your pencils. They're the same kind of thing. The number $5$ is like an eraser – a different kind of thing. To make sense of it all, our first step is always to tidy up by grouping similar items together. In algebra, we call this Combining Like Terms.

What Exactly Are 'Like Terms'?

Before we can combine them, what should we be looking for? Let's see if you can spot the pattern.

[Image showing a two-column table. Left column titled 'Like Terms' shows pairs: 7x and 2x; 5y and 8y; 4 and 10. Right column titled 'Unlike Terms' shows pairs: 3x and 3y; 5x and 5x²; 6a and 9.]

Did you figure it out? Here's the secret:

Memory Aid: Terms are 'alike' if they have the exact same variable part, raised to the exact same power.

Think of the variable as the type of fruit. An '$x$' is an apple. A '$y$' is a banana. An '$x^2$' is a basket of apples! You can combine a single apple ($x$) with other single apples, but you can't just add a single apple to a whole basket of apples ($x^2$) and call them the same thing.

The Tidying-Up Process: A Step-by-Step Guide

Let's tackle this equation together: $4x + 9 + 3x = 23$

1. Step 1: Identify the 'Like Terms'. Think of it as color-coding. Let's find all the terms with an '$x$' and all the plain number terms (we call these constants). In $4x + 9 + 3x = 23$, the terms $4x$ and $3x$ are like terms.

2. Step 2: Combine the 'Like Terms'. This is just simple addition or subtraction. If you have 4 apples and get 3 more apples, how many do you have? You have 7 apples! So, $4x + 3x$ becomes $7x$. Our equation now looks much tidier: $7x + 9 = 23$.

3. Step 3: Solve the Simplified Equation. This is just a two-step equation, which you already know how to solve! Our goal is to get '$x$' all by itself.

   • First, undo the addition. Subtract 9 from both sides. $7x + 9 - 9 = 23 - 9$ $7x = 14$
   • Now, undo the multiplication. Divide both sides by 7. $\frac{7x}{7} = \frac{14}{7}$ $x = 2$

And we've found our answer!

Worked Example + Your Turn

Worked Example: Solve $8y - 3y - 2 = 18$.

• Identify: The like terms on the left side are $8y$ and $-3y$. Notice we keep the sign that's in front of the term!
• Combine: $8y - 3y = 5y$. The equation becomes $5y - 2 = 18$.
• Solve:
  • Add 2 to both sides: $5y - 2 + 2 = 18 + 2$, which gives $5y = 20$.
  • Divide both sides by 5: $\frac{5y}{5} = \frac{20}{5}$, which gives $y = 4$.

Your Turn: Now you try one! Solve $6p + 2p + 5 = 21$. (Pause and work it out before moving on. The answer is $p=2$.)

Real-World Connection: Snack Time!

Imagine you're buying snacks. You buy 3 bags of chips and a 2-dollar drink. Your friend joins you and buys 2 more bags of the same chips. The total bill is 17 dollars. How much did one bag of chips cost?

• Let '$c$' be the cost of one bag of chips.
• Your items: $3c + 2$.
• Your friend's items: $2c$.
• The total cost equation is: $3c + 2 + 2c = 17$.

Can you solve it from here?

• Combine like terms: $3c + 2c$ is $5c$.
• The new equation is: $5c + 2 = 17$.
• Solve: Subtract 2 from both sides ($5c = 15$), then divide by 5. You'll find that $c = 3$.

So, each bag of chips cost 3 dollars! See? You use this logic all the time without even realizing you're doing algebra.

Questions for: Solving Equations with Variables on Both Sides

The Seesaw Problem

Imagine you and a friend are on a seesaw. To make it perfectly balanced, the weight on both sides must be equal, right? What if you had a mix of known weights (like bricks) and mystery bags (whose weight you don't know)?

[Image showing a perfectly balanced seesaw. On the left side, there are 2 mystery bags labeled 'x' and 5 small bricks. On the right side, there is 1 mystery bag labeled 'x' and 9 small bricks.]

This is exactly like an algebraic equation with variables on both sides! The equal sign = is the pivot point of the seesaw. Our goal is to figure out the weight of a single mystery bag (the value of x) while keeping the seesaw balanced at all times.

Let's turn our seesaw into an equation:

The left side: $2x + 5$

The right side: $x + 9$

Since it's balanced, we have: $2x + 5 = x + 9$

How Do We Solve This?

Our mission is to isolate one mystery bag (x) on one side of the seesaw to find its weight. The golden rule is: Whatever you do to one side, you MUST do to the other to keep it balanced.

Step 1: Gather the Mystery Bags (Variables)

To get all the bags on one side, let's remove the single bag (x) from the right side. To keep things balanced, we must also remove one bag from the left side.

In Algebra: We subtract x from both sides. $2x - x + 5 = x - x + 9$ This simplifies to: $x + 5 = 9$

Now our seesaw has one mystery bag and 5 bricks on the left, and 9 bricks on the right. It's still perfectly balanced!

Step 2: Gather the Bricks (Constants)

Next, we want to get the mystery bag all by itself. Let's remove the 5 bricks from the left side. To stay balanced, we have to remove 5 bricks from the right side too.

In Algebra: We subtract 5 from both sides. $x + 5 - 5 = 9 - 5$ This simplifies to: $x = 4$

Success! We've found that each mystery bag has a weight of 4. We can check our answer by plugging 4 back into the original equation:

Left side: $2(4) + 5 = 8 + 5 = 13$ Right side: $(4) + 9 = 13$

Since $13 = 13$, our answer is correct! The seesaw is balanced.

Worked Example + Your Turn

Worked Example: Solve for a in the equation $7a - 4 = 3a + 20$.

1. Gather Variables: The variable a is on both sides. Let's move the smaller term, 3a, to keep our variable term positive. Subtract 3a from both sides. $7a - 3a - 4 = 3a - 3a + 20$ $4a - 4 = 20$

2. Gather Constants: Now, let's move the -4. The opposite of subtracting 4 is adding 4. Add 4 to both sides. $4a - 4 + 4 = 20 + 4$ $4a = 24$

3. Isolate the Variable: The a is being multiplied by 4. To undo this, we divide both sides by 4. $\frac{4a}{4} = \frac{24}{4}$ $a = 6$

Your Turn: Solve for p in the equation $8p + 2 = 5p + 11$. (Answer: $p=3$)

What if It Gets More Complicated?

Sometimes, the sides of the equation are messy. Before you start balancing, you need to tidy up!

Memory Aid: C.M.S. (Clean up, Move, Solve)

1. Clean Up: Simplify each side of the equation first. This means using the distributive property and combining any like terms.
2. Move: Use addition or subtraction to move all variable terms to one side and all constant terms to the other.
3. Solve: Use multiplication or division to isolate the variable.

Example: Solve $5(x + 2) = 3x + 18$

1. Clean Up: Distribute the 5 on the left side. $5x + 10 = 3x + 18$

2. Move: Subtract 3x from both sides. $2x + 10 = 18$ Then, subtract 10 from both sides. $2x = 8$

3. Solve: Divide by 2. $x = 4$

Special Cases: Weird Results!

Sometimes, you'll try to solve an equation and get a strange result.

• Infinitely Many Solutions: If you solve an equation and the variables disappear, leaving you with a true statement like $7 = 7$ or $0 = 0$, it means any number will work as a solution! Example: $2(x+3) = 2x+6$. This simplifies to $2x+6 = 2x+6$. If you subtract $2x$ from both sides, you get $6 = 6$. This is always true, so there are infinitely many solutions.

• No Solution: If you solve an equation and the variables disappear, leaving you with a false statement like $5 = 9$ or $0 = 2$, it means no number will work as a solution. The equation can never be true. Example: $3y + 5 = 3y - 1$. If you subtract $3y$ from both sides, you get $5 = -1$. This is false, so there is no solution.

Real-World Relevance: Choosing a Plan

When do you need to solve equations with variables on both sides in real life? All the time! Think about choosing between two options.

Scenario: You want to join a gym.

• Gym A charges 50 dollars to sign up, plus 20 dollars per month. Cost = $20m + 50$.
• Gym B charges no sign-up fee, but costs 30 dollars per month. Cost = $30m$.

You want to know after how many months (m) the total cost will be the same. You set the costs equal to each other:

$20m + 50 = 30m$

Solving this equation will tell you the exact point where the costs are identical!

(Can you solve it? Subtract $20m$ from both sides: $50 = 10m$. Then divide by 10: $m = 5$. The cost is the same after 5 months!)

Questions for: Using the Distributive Property to Solve Equations

Your New Algebraic Superpower!

Have you ever been to a party and received a goodie bag? Imagine you're the host. You have several groups of friends arriving, and you want to give each person in every group a slice of pizza. You wouldn't just give pizza to the first person in each group, right? You'd make sure everyone in the group gets a slice.

The Distributive Property in algebra works just like that! The number outside the parentheses is the 'pizza'. The terms inside the parentheses are your 'friends in the group'. You must 'distribute' the pizza to every single friend.

[Image showing a party host handing out slices of pizza to a group of three friends standing together. An arrow points from the host to each friend.]

From Pictures to Problems

Let's think about an area. Imagine you have a rectangular garden split into two sections: one for carrots and one for lettuce.

[Image of a rectangle divided into two smaller rectangles. The total width is labeled '4'. The length of the first section is 'x' and the length of the second section is '2'. The total length is 'x+2'.]

How could we find the total area of the garden?

Method 1: Add First, Then Multiply Total Area = Width × Total Length Total Area = $4 \times (x + 2)$

Method 2: Multiply Each Part, Then Add Total Area = (Area of Carrot Patch) + (Area of Lettuce Patch) Total Area = $(4 \times x) + (4 \times 2)$

Since both methods calculate the same total area, they must be equal!

$4(x+2) = 4x + 8$

This is the Distributive Property in action! The '4' has been distributed to both the '$x$' and the '$2$'.

The Official Rule: $a(b+c) = ab + ac$

Using Your Superpower to Solve Equations

Now, let's use this to crack some codes... I mean, solve equations!

What if we know the total area of our garden is 36 square meters? We can set up an equation:

$4(x+2) = 36$

Step-by-Step Guide:

1. Distribute! Share the number outside the parentheses with every term inside.

   • $4 \times x = 4x$
   • $4 \times 2 = 8$
   • Our equation becomes: $4x + 8 = 36$

2. Isolate the Variable Term. Now it's a simple two-step equation. We want to get the '$4x$' term by itself. Undo the addition by subtracting 8 from both sides.

   • $4x + 8 - 8 = 36 - 8$
   • $4x = 28$

3. Solve for the Variable. Undo the multiplication by dividing both sides by 4.

   • $\frac{4x}{4} = \frac{28}{4}$
   • $x = 7$

4. Check Your Work (The Hero's Final Move). Always plug your answer back into the original equation to see if it makes sense.

   • $4(7+2) = 36$
   • $4(9) = 36$
   • $36 = 36$. It works! The unknown length of our carrot patch was 7 meters.

Worked Example:

Solve the equation $-3(p-5) = 21$. Watch out for the negative signs!

• Step 1: Distribute.

  • $-3 \times p = -3p$
  • $-3 \times -5 = +15$ (A negative times a negative is a positive!)
  • Equation becomes: $-3p + 15 = 21$

• Step 2: Isolate the variable term.

  • $-3p + 15 - 15 = 21 - 15$
  • $-3p = 6$

• Step 3: Solve.

  • $\frac{-3p}{-3} = \frac{6}{-3}$
  • $p = -2$

• Step 4: Check.

  • $-3(-2 - 5) = 21$
  • $-3(-7) = 21$
  • $21 = 21$. Correct!

Your Turn:

Now you try! Solve the equation $5(a+2) = 30$. Follow the steps above. Don't forget to check your answer!

(Answer: $a=4$)

Pause and Ponder

Why is it called the distributive property? What exactly is being 'distributed'? Think about our party and garden examples. How does the name help you remember what to do?

This superpower is essential for tackling more complex equations you'll see in science, engineering, and even when planning a budget. Let's practice putting it to use!

Questions for: Solving Equations Containing Fractions

The Messy Room Analogy

Imagine trying to solve a puzzle where the pieces are all different sizes and shapes—some are big, some are small, some are halves, some are thirds. It's confusing! That's what solving an equation with lots of fractions can feel like.

[Image showing a jumble of different-sized puzzle pieces on the left, and the same pieces neatly sorted into piles of the same size on the right.]

Now, what if you had a magic way to make all the pieces the same standard size? The puzzle would become much easier to solve, right? In algebra, we have a similar magic trick for dealing with fractions. It's called clearing the denominators, and it turns a messy fraction equation into a simple, clean one you already know how to handle.

The Magic Wand: The Least Common Denominator (LCD)

So, what's our magic wand? It's the Least Common Denominator (LCD).

Think of the denominators (the bottom numbers of the fractions) as different languages. If you have denominators of 2, 3, and 4, it's like having people speaking half-language, third-language, and quarter-language. It's hard to get them to agree! The LCD is the one common language they can all translate into perfectly.

• Definition: The LCD is the smallest number that all the denominators in your equation can divide into evenly. It's just the Least Common Multiple (LCM) of all the Denominators.

For example, in the equation $\frac{x}{2} + \frac{x}{3} = 10$, the denominators are 2 and 3. What's the smallest number that both 2 and 3 divide into? It's 6! So, our LCD is 6.

The 'Fraction-Busting' Method: A Step-by-Step Guide

Ready to bust some fractions? It's a simple three-step process.

Step 1: Find the Boss! Look at all the denominators in the equation and find their LCD. This number is the 'boss'—it's going to tell everyone else what to do.

Step 2: Multiply Everyone! This is the most important step! Multiply every single term in the equation (on both sides of the equals sign) by the LCD. Don't leave anyone out, not even the whole numbers!

Step 3: Simplify and Solve! As you multiply, the denominators will cancel out, and the fractions will disappear. Poof! You'll be left with a straightforward equation. Solve it like you normally would.

Worked Example 1: The Basics

Let's solve: $\frac{x}{4} + \frac{1}{2} = \frac{5}{4}$

1. Find the Boss: The denominators are 4 and 2. The smallest number that both 4 and 2 divide into is 4. So, LCD = 4.

2. Multiply Everyone: We multiply every piece of the equation by 4. $4 \cdot (\frac{x}{4}) + 4 \cdot (\frac{1}{2}) = 4 \cdot (\frac{5}{4})$

3. Simplify and Solve: Now watch the magic happen.

   • For the first term, the 4s cancel: $x$
   • For the second term, $4 \div 2 = 2$, so we have: $2 \cdot 1 = 2$
   • For the third term, the 4s cancel: $5$

   Our new, clean equation is: $x + 2 = 5$ Subtract 2 from both sides, and you get: $x = 3$

Your Turn 1: Solve the equation: $\frac{y}{6} + \frac{2}{3} = \frac{7}{6}$ (Hint: The LCD is 6. The answer should be y=3)

Worked Example 2: Mixing in Whole Numbers

Imagine you're planning a trip. You spend one-third of your budget on a flight and 100 dollars on a hotel. You have one-half of your budget left. What was your original budget, $b$?

Equation: $\frac{b}{3} + 100 = \frac{b}{2}$

1. Find the Boss: Denominators are 3 and 2. The LCD is 6.

2. Multiply Everyone: Remember to multiply the 100 too! $6 \cdot (\frac{b}{3}) + 6 \cdot (100) = 6 \cdot (\frac{b}{2})$

3. Simplify and Solve:

   • $6 \div 3 = 2$, so we have: $2b$
   • $6 \cdot 100 = 600$
   • $6 \div 2 = 3$, so we have: $3b$

   Our new equation is: $2b + 600 = 3b$ Subtract $2b$ from both sides: $600 = b$ Your original budget was 600 dollars!

Your Turn 2: Solve the equation: $\frac{a}{5} - 2 = \frac{a}{7}$ (Hint: The LCD is 35. The answer should be a=35)

Watch Out! The Parentheses Package Trap

What happens when the numerator (the top part) has more than one term, like in $\frac{x+5}{3} = 4$?

Think of the numerator as a package deal. It's like it's already wrapped in parentheses: $\frac{(x+5)}{3} = 4$. You can't separate the parts until you've dealt with the fraction.

Let's solve $\frac{x-2}{3} + \frac{x}{4} = 10$

1. Find the Boss: Denominators are 3 and 4. The LCD is 12.

2. Multiply Everyone: $12 \cdot \frac{(x-2)}{3} + 12 \cdot (\frac{x}{4}) = 12 \cdot (10)$

3. Simplify and Solve:

   • $12 \div 3 = 4$, so we have: $4(x-2)$
   • $12 \div 4 = 3$, so we have: $3x$
   • $12 \cdot 10 = 120$

   Our new equation: $4(x-2) + 3x = 120$ Now, distribute the 4: $4x - 8 + 3x = 120$ Combine like terms: $7x - 8 = 120$ Add 8 to both sides: $7x = 128$ Divide by 7: $x = \frac{128}{7}$

Your Turn 3: Solve: $\frac{y+6}{2} - \frac{y}{3} = 5$ (Hint: The LCD is 6. Be careful with the distribution! The answer should be y=12)

Final Reflection

The 'Fraction-Busting' method is your secret weapon for making complex equations simple.

Think about it: Why is Step 2, 'Multiply Everyone', the most critical part of this process? What would happen if you forgot to multiply a whole number term by the LCD? Discuss it with a friend or write down your thoughts.

The Detective's Toolkit: From Clues to Cases

Have you ever felt like a detective when you read a math word problem? The words are like clues scattered around a scene. Your job is to gather these clues, piece them together, and build a case—an algebraic equation! The 'suspect' you're looking for is the unknown value, which we call a variable.

Think of it this way:

• The Story: A word problem describing a situation.
• The Clues: Key numbers and phrases like "more than," "half of," or "is the same as."
• The Unknown: The one piece of information you don't have. This is your variable (let's call it $x$).
• The Case File: The equation you build by translating the clues into math symbols.
• Solving the Case: Using your algebra skills to find the value of $x$.

Let's put on our detective hats and learn how to translate these word-clues into rock-solid equations.

From Pictures to Symbols: A Step-by-Step Mission

Let's start with something you can touch and see.

Concrete Scenario: Imagine you buy two identical sandwiches and a drink that costs 3 dollars. Your total bill is 15 dollars. What's the cost of one sandwich?

Pictorial Translation: We can draw this out. Let's use blocks to represent the items.

[Image showing a balance scale. On the left side are two identical blocks labeled 'S' and one smaller block labeled '3 dollars'. On the right side is a single large block labeled '15 dollars'.]

This picture shows that two sandwiches (S) plus 3 dollars is the same as 15 dollars.

Abstract Equation: Now, let's translate our picture into the language of algebra. Let $s$ be the cost of one sandwich.

• Two sandwiches: $2s$
• Plus a 3 dollar drink: $2s + 3$
• Is the same as 15 dollars: $2s + 3 = 15$

Look at that! We just built an equation. Solving it tells us that each sandwich costs 6 dollars.

The R.I.V.E.R. Method: A Flow for Solving Word Problems

To make sure you don't miss any clues, you can follow a simple, four-step method. Think of it like following a river from the source (the words) to the sea (the solution). Let's call it R.I.V.E.R.

1. Read & Understand: Read the entire problem. What's the story about? What are you being asked to find?
2. Identify the Unknown: Pinpoint the exact piece of information you need to find. This will be your variable. Give it a letter, like $x$ or $c$.
3. Verbalize & Translate: Translate the key phrases into mathematical operations. Create a 'verbal model' first, then substitute symbols.
4. Equation & Solve: Write the final equation and use your algebra skills to solve for the variable. Don't forget to Review and check if your answer makes sense in the context of the story!

Here are some common clues to look out for:

Worked Example: The Gaming Marathon

Problem: You are saving for a new video game that costs 89 dollars. You have already saved 25 dollars from your birthday. You can earn 8 dollars an hour by helping a neighbor with yard work. How many hours do you need to work to have exactly enough to buy the game?

Let's use the R.I.V.E.R. method.

1. Read & Understand: The goal is to find the number of hours needed to work to afford an 89 dollar game, given some starting money and an hourly wage.
2. Identify the Unknown: The question asks, "How many hours?" So, our unknown is the number of hours to work.
   • Let $h$ = the number of hours worked.
3. Verbalize & Translate:
   • "Money you earn" is "8 dollars an hour" times "the number of hours." -> $8h$
   • "Total money saved" is "money you already have" plus "money you earn." -> $25 + 8h$
   • The "total money saved" must equal the "cost of the game." -> $25 + 8h = 89$
4. Equation & Solve:
   • Our equation is: $25 + 8h = 89$
   • Subtract 25 from both sides: $8h = 64$
   • Divide by 8: $h = 8$
   • Review: Does it make sense? 8 hours at 8 dollars/hour is 64 dollars. Add the 25 dollars you started with, and you get $64 + 25 = 89$ dollars. Perfect!

Answer: You need to work for 8 hours.

Your Turn: The Concert Ticket

Problem: A concert ticket costs 120 dollars. You have 30 dollars saved. You get a weekly allowance of 15 dollars. How many weeks will it take you to save enough for the ticket? Let $w$ be the number of weeks.

Pause here and try to write the equation yourself before reading the solution below.

Solution Equation: $30 + 15w = 120$

Challenge Zone: Variables on Both Sides

Sometimes, the mystery is more complex. What if you're comparing two different situations?

Problem: A local gym offers two membership plans. Plan A has a 50 dollar sign-up fee and costs 20 dollars per month. Plan B has no sign-up fee but costs 30 dollars per month. After how many months will the total cost of both plans be the same?

1. Read & Understand: We need to find the number of months when the total cost of Plan A equals the total cost of Plan B.
2. Identify the Unknown: The question is "After how many months?"
   • Let $m$ = the number of months.
3. Verbalize & Translate: We need to write an expression for the cost of each plan.
   • Cost of Plan A = "Sign-up fee" + "monthly cost" $\times$ "number of months" -> $50 + 20m$
   • Cost of Plan B = "monthly cost" $\times$ "number of months" -> $30m$
   • We want to know when they are the same: Cost of Plan A = Cost of Plan B.
4. Equation & Solve:
   • Our equation is: $50 + 20m = 30m$
   • Subtract $20m$ from both sides: $50 = 10m$
   • Divide by 10: $5 = m$
   • Review: Let's check. After 5 months: Plan A costs $50 + 20(5) = 50 + 100 = 150$ dollars. Plan B costs $30(5) = 150$ dollars. They match!

Answer: The total cost will be the same after 5 months.

Now you have the tools to decode almost any word problem that comes your way. Remember the R.I.V.E.R. method and always check that your final answer makes sense in the real world!

Unlocking the Secrets of Formulas

Have you ever tried to rearrange the furniture in your room? You might move your bed to a different wall or shift your desk to get better light. You're not getting new furniture; you're just changing its arrangement to suit a new purpose.

Solving a literal equation is a lot like that! A literal equation is just an equation with two or more variables. Think of them as the formulas you see all the time in math and science, like for area, distance, or temperature. Rearranging a formula means you're solving for a different variable – a different piece of furniture – to make the formula useful in a new way.

Our goal is always the same: isolate the variable we want to solve for. Just like when you solve a regular equation, we'll use inverse operations to get that variable all by itself on one side of the equals sign.

The Toolbox: One and Two-Step Rearranging

Let's start with a formula most people know: distance equals rate (speed) times time.

Visualizing the Formula:

This gives us the abstract formula: $d = rt$.

But what if you drove 300 miles at a speed of 60 miles per hour and wanted to know how long it took? We need to rearrange the formula to solve for time ($t$).

Worked Example 1: Solving for Time

Our goal is to get $t$ by itself.

1. Start with the formula: $d = rt$
2. Identify the operation: The variable we want, $t$, is being multiplied by $r$.
3. Use the inverse operation: The opposite of multiplication is division. We'll divide both sides by $r$. $\frac{d}{r} = \frac{rt}{r}$
4. Simplify: The $r$ on the right side cancels out. $\frac{d}{r} = t$
5. The New Formula: $t = \frac{d}{r}$ (Time = Distance / Rate)

Your Turn 1: Force and Acceleration

The formula for force is $F = ma$, where $F$ is force, $m$ is mass, and $a$ is acceleration. Try rearranging this formula to solve for acceleration ($a$).

Now, what about a formula with more than one step? Think of it like unwrapping a present. You have to untie the ribbon and take off the paper (the addition/subtraction parts) before you can open the box (the multiplication/division part).

Memory Aid: To solve equations, we often use the reverse order of operations, sometimes called SADMEP (Subtraction/Addition, then Division/Multiplication, then Exponents, then Parentheses).

Worked Example 2: Perimeter of a Rectangle

The formula is $P = 2l + 2w$. Our goal is to solve for the width, $w$.

1. Start with the formula: $P = 2l + 2w$
2. Isolate the term with 'w': The term $2l$ is being added to our 'w' term. We'll use the inverse operation, subtraction. Subtract $2l$ from both sides. $P - 2l = 2l - 2l + 2w$ $P - 2l = 2w$
3. Isolate 'w': Now, $w$ is being multiplied by 2. The inverse is division. Divide both sides by 2. $\frac{P - 2l}{2} = \frac{2w}{2}$
4. The New Formula: $w = \frac{P - 2l}{2}$

Your Turn 2: Equation of a Line

A common formula in algebra is the slope-intercept form of a line: $y = mx + b$. Can you rearrange this formula to solve for the slope, $m$?

Level Up: Dealing with Parentheses

Sometimes the variable you want is 'trapped' inside parentheses. You have two options: divide the outside term first or use the distributive property to remove the parentheses.

Worked Example 3: Converting Temperature

The formula to convert degrees Fahrenheit ($F$) to degrees Celsius ($C$) is $C = \frac{5}{9}(F - 32)$. Let's rearrange it to solve for $F$. Our goal is to free the $F$ from the parentheses.

1. Start with the formula: $C = \frac{5}{9}(F - 32)$
2. Clear the fraction: To get rid of the $\frac{5}{9}$, we can multiply by its reciprocal, $\frac{9}{5}$. $\frac{9}{5} \cdot C = \frac{9}{5} \cdot \frac{5}{9}(F - 32)$ $\frac{9}{5}C = F - 32$
3. Isolate 'F': Now it's a simple one-step equation. The number 32 is being subtracted from $F$. The inverse is addition. Add 32 to both sides. $\frac{9}{5}C + 32 = F - 32 + 32$
4. The New Formula: $F = \frac{9}{5}C + 32$

Your Turn 3: Area of a Trapezoid

The formula for the area of a trapezoid is $A = \frac{1}{2}h(b_1 + b_2)$. Try to solve this equation for the base, $b_1$.

Key Takeaway: Rearranging literal equations uses the exact same skills as solving regular equations. The only difference is that your final answer will be an expression made of other variables instead of a single number. Always focus on the goal: isolate your target variable, step by step!