Linear Equations, Inequalities, and Simultaneous Equations

Worked Example: Application of Linear Equations

Worked Example: Cracking the Code

1. 1
   We want to isolate the term with $h$. The opposite of adding 10 is subtracting 10. So, let's subtract 10 from both sides to keep things balanced.

   $5h + 10 - 10 = 40 - 10$
   $5h = 30$
2. 2
   Now, $h$ is being multiplied by 5. The inverse operation is division. Let's divide both sides by 5 to get $h$ on its own.

   $\frac{5h}{5} = \frac{30}{5}$
   $h = 6$
3. 3
   Boom! You've solved it. Your hourly rate is £6. Not bad!

   $h = 6$

Worked Example: Expanding Brackets to Solve Equations

Worked Example: Unpacking the Brackets

1. 1
   First, let's expand the brackets. We multiply the 4 by the $x$ and by the $-5$. Don't forget that minus sign!

   $(4 \times x) + (4 \times -5) = 12$
   $4x - 20 = 12$
2. 2
   Now it's a standard two-step equation. Let's get rid of the $-20$ by adding 20 to both sides.

   $4x - 20 + 20 = 12 + 20$
   $4x = 32$
3. 3
   Finally, divide both sides by 4 to get $x$ by itself.

   $\frac{4x}{4} = \frac{32}{4}$
   $x = 8$

Worked Example: Equations with Variables on Both Sides

Worked Example: The Group Selfie

1. 1
   We have $7x$ on the left and $3x$ on the right. The smaller one is $3x$, so let's subtract $3x$ from both sides to get all the $x$'s on the left.

   $7x - 3x - 4 = 3x - 3x + 16$
   $4x - 4 = 16$
2. 2
   Look at that! It's a simple two-step equation now. We know what to do. Add 4 to both sides to isolate the $x$ term.

   $4x - 4 + 4 = 16 + 4$
   $4x = 20$
3. 3
   Last step! Divide by 4 to find out what $x$ is.

   $\frac{4x}{4} = \frac{20}{4}$
   $x = 5$

Worked Example: Solving and Representing Inequalities

Worked Example: Solving an Inequality

1. 1
   Treat it just like an equation. The inverse of subtracting 5 is adding 5. Let's do that to both sides.

   $3x - 5 + 5 > 10 + 5$
   $3x > 15$
2. 2
   Now, divide both sides by 3. Since 3 is positive, we don't need to flip the sign. Phew!

   $\frac{3x}{3} > \frac{15}{3}$
   $x > 5$
3. 3
   The solution is any number greater than 5. On a number line, you'd draw an open circle at 5 (because it's not 'equal to') and an arrow pointing to the right.

   Solution: $x > 5$

Worked Example: Finding Equal Terms in Sequences

Worked Example: Sequence Showdown

1. 1
   To find when the terms are the same, we literally set the two expressions equal to each other. This is the key step!

   $5n - 2 = 3n + 8$
2. 2
   It's an equation with variables on both sides! Let's subtract the smaller $n$ term, which is $3n$, from both sides.

   $5n - 3n - 2 = 3n - 3n + 8$
   $2n - 2 = 8$
3. 3
   Now it's a simple two-step equation. Add 2 to both sides.

   $2n = 10$
4. 4
   Finally, divide by 2. This tells us that the 5th term in both sequences will be identical.

   $n = 5$

Solving Simultaneous Equations by Substitution

Worked Example: The Mystery Point

1. 1
   We know $y$ is the same as $2x + 1$. So, let's substitute $(2x + 1)$ into the second equation wherever we see $y$.

   $x + (2x + 1) = 7$
2. 2
   Now we have an equation with only one variable. Let's simplify and solve for $x$.

   $3x + 1 = 7$
   $3x = 6$
   $x = 2$
3. 3
   We're halfway there! We know $x=2$. Now substitute this value back into one of the original equations to find $y$. The first one looks easiest.

   $y = 2(2) + 1$
   $y = 4 + 1$
   $y = 5$
4. 4
   So the solution is $x=2$ and $y=5$. This is the point $(2, 5)$ where the two lines would meet on a graph.

   Solution: $(2, 5)$

Worked Example: Solution by Elimination

Worked Example: Elimination Takedown

1. 1
   Look at the $y$ terms. We have a $+y$ and a $-y$. The coefficients are basically 1 and -1. Since the signs are different, we can ADD the two equations together to eliminate $y$.

   $(3x + y) + (2x - y) = 11 + 4$
2. 2
   Let's simplify that. The $y - y$ cancels out to zero. That's the elimination!

   $5x = 15$
3. 3
   Now we have a super simple equation. Just divide by 5 to solve for $x$.

   $x = \frac{15}{5} = 3$
4. 4
   We found $x=3$. Now substitute this back into either of the original equations to find $y$. Let's use Equation 1.

   $3(3) + y = 11$
   $9 + y = 11$
   $y = 11 - 9$
   $y = 2$
5. 5
   And we're done! The solution is $x=3$ and $y=2$. You just owned that problem. ✅

   Solution: $(3, 2)$

Worked Example: Expanding and Simplifying a Combined Expression

Let's Solve This Thing! 🚀

1. 1
   First, we need to expand both sets of brackets separately. Let's start with the double brackets, $(3x+2)(x-5)$, using the FOIL method (First, Outer, Inner, Last).

   $(3x \times x) + (3x \times -5) + (2 \times x) + (2 \times -5)$
2. 2
   Now, let's calculate each of those products to get the expanded form of the first part.

   $3x^2 - 15x + 2x - 10$
3. 3
   Next, we expand the second bracket, $4(x-1)$. Remember, the 4 multiplies with both the $x$ and the $-1$.

   $(4 \times x) + (4 \times -1) = 4x - 4$
4. 4
   Let's combine everything into one long expression. We have the result from the double brackets and the result from the single bracket.

   $3x^2 - 15x + 2x - 10 + 4x - 4$
5. 5
   Finally, it's time to simplify by collecting all the like terms. We'll group the $x^2$ term, all the $x$ terms, and all the constant numbers.

   $3x^2 + (-15x + 2x + 4x) + (-10 - 4)$
6. 6
   Add and subtract the coefficients of the like terms to get our final, fully simplified answer. Looking much neater!

   $3x^2 - 9x - 14$

Worked Example: Solving Simultaneous Linear Equations

The Case of the Mysterious Milkshake & Fries 🥤🍟

1. 1
   First, we need to translate the word problem into two algebraic equations. Your order is Equation 1, and your friend's order is Equation 2.

   $(1): \quad 2m + f = 7$ $(2): \quad m + f = 5$
2. 2
   We're going to use the elimination method. Notice that both equations have a '$+f$'. If we subtract the second equation from the first one, the '$f$' terms will cancel each other out. Game on.

   $(2m + f) - (m + f) = 7 - 5$
3. 3
   Now, let's simplify that subtraction. $2m - m$ is $m$, and $f - f$ is 0. This leaves us with a super simple equation to find the cost of a milkshake.

   $m = 2$
4. 4
   Boom! A milkshake ($m$) costs £2. Now we can substitute this value back into either of the original equations to find the cost of the fries ($f$). Let's use the simpler one, Equation 2.

   $m + f = 5 \implies (2) + f = 5$
5. 5
   To get '$f$' by itself, we just need to subtract 2 from both sides of the equation. This will give us the final cost for the fries.

   $f = 5 - 2 \implies f = 3$
6. 6
   So there we have it! The mystery is solved. We can write our final answer clearly. You can even double-check it with the first equation: $2(£2) + £3 = £4 + £3 = £7$. Perfect!

   A milkshake costs £2 and an order of fries costs £3.

Worked Example: Rearranging a Formula and Representing an Inequality

Saving Up for that New Skin Pack 🎮

1. 1
   First, let's tackle part (a) and change the subject to $h$. Our goal is to get $h$ all by itself on one side of the equals sign. We start with the original formula.

   $T = 500 + 15h$
2. 2
   The 500 is added to the term with $h$, so we do the inverse: subtract 500 from both sides of the equation. This is like peeling back the first layer.

   $T - 500 = 15h$
3. 3
   Now, $h$ is being multiplied by 15. The inverse of multiplying is dividing. So, we divide both sides by 15 to finally get $h$ alone. And boom, we've remixed the formula!

   $\frac{T - 500}{15} = h$
4. 4
   For part (b), we need our total savings $T$ to be at least \$950. The phrase 'at least' means greater than or equal to ($\ge$). So we set up our inequality using the original formula.

   $500 + 15h \ge 950$
5. 5
   Now, we solve this just like we rearranged the formula. First, subtract 500 from both sides.

   $15h \ge 450$
6. 6
   Finally, divide by 15 to find the minimum hours you need to work. Then, we draw this on a number line. Since it's $\ge$, we use a closed circle on 30 and draw the arrow to the right, showing all the numbers that are 30 or more.

   $h \ge 30$

Worked Example: Finding the nth Term of a Linear Sequence

Let's Find That Formula & Get This Bread 🥖

1. 1
   First, let's find the term-to-term rule, which is the common difference. This tells us what the sequence is going up by each time.

   $28 - 20 = 8$ $36 - 28 = 8$
2. 2
   The common difference is 8. This means the formula for the nth term will start with $8n$, just like the 8 times table.

   $n^{th} \text{ term} = 8n + \text{...something}$
3. 3
   Now, let's compare our sequence to the 8 times table ($8n$). We need to figure out the 'adjustment' we need to make to get from $8n$ to our actual term.

   For n=1: $8 \times 1 = 8$. But our first term is 20. We need to add 12. ($8+12=20$)
   For n=2: $8 \times 2 = 16$. But our second term is 28. We need to add 12. ($16+12=28$)
4. 4
   The adjustment is always '+12'. So we can complete our formula for the nth term. This is the answer for part (a).

   $n^{th} \text{ term} = 8n + 12$
5. 5
   For part (b), we need to find the savings after 12 weeks. This means we need to find the 12th term. We just substitute $n=12$ into our shiny new formula.

   $\text{Savings} = 8(12) + 12 = 96 + 12 = 108$
6. 6
   Finally, write down the answer clearly. After 12 weeks, you will have saved £108. You're getting closer to that console! 🎉

   $\text{Answer: } £108$

Worked Example: Interpreting a Distance-Time Graph

The Journey to the Gig 🎶

1. 1
   For part (a), we need to find Leo's cycling speed. Speed is the gradient of the first section of the graph. We'll use the formula $Speed = \frac{\text{change in distance}}{\text{change in time}}$. The first section goes from (0, 0) to (0.5 hours, 10 km).

   $Speed = \frac{10 - 0 \text{ km}}{0.5 - 0 \text{ h}} = \frac{10}{0.5} = 20 \text{ km/h}$
2. 2
   For part (b), we look for the flat, horizontal part of the graph. This is where the distance from home doesn't change, meaning Leo is stationary. The flat section starts at 0.5 hours and ends at 0.75 hours.

   $\text{Time stopped} = 0.75 \text{ h} - 0.5 \text{ h} = 0.25 \text{ hours} \text{ (or 15 minutes)}$
3. 3
   For part (c), we need the speed of the car. This is the gradient of the final, steepest section. The section starts at the point (0.75 h, 10 km) and ends at the point (1.25 h, 60 km).

   $Speed = \frac{60 - 10 \text{ km}}{1.25 - 0.75 \text{ h}} = \frac{50 \text{ km}}{0.5 \text{ h}} = 100 \text{ km/h}$
4. 4
   Let's quickly compare the speeds. The cycling speed was 20 km/h, and the car's speed was 100 km/h. This makes perfect sense – the car section is way steeper on the graph because the speed (the rate of change) is much higher. You got this! 💪

Worked Example: Constructing and Substituting into a Formula

1. 1
   For (a), fixed part is $4$, variable part is $1.20$ per km so $1.20k$. Add them.

   $C = 4 + 1.20k$
2. 2
   For (b), sub $k = 7$. BODMAS: multiply first.

   $C = 4 + 1.20 \times 7 = 4 + 8.40 = \$12.40$
3. 3
   For (c), bracket the negative before squaring — this is the whole point of the question.

   $2 \times (-3)^2 - 5 = 2 \times 9 - 5$
4. 4
   BODMAS: multiply then subtract.

   $= 18 - 5 = 13$

Worked Example: nth Term of a Cubic Sequence

1. 1
   Stare at the numbers. $1 = 1^3$, $8 = 2^3$, $27 = 3^3$, $64 = 4^3$, $125 = 5^3$. Every term is a cube. 🎲

   $1 = 1^3,\ 8 = 2^3,\ 27 = 3^3,\ 64 = 4^3,\ 125 = 5^3$
2. 2
   The position number $n$ is literally the thing being cubed. Too easy.

   $\text{n}^{\text{th}} \text{ term} = n^3$
3. 3
   Sanity check: next term should be $6^3 = 216$. ✅

   $n = 6 \Rightarrow 6^3 = 216$