Example 1: Basic Rearrangement Problem: Rearrange the equation $3x + 2y = 12$ to make $y$ the subject.

Step-by-Step Solution**:**

1. Start with the given equation:

• The equation is $3x + 2y = 12$.

2. Move the $3x$ term to the RHS:

• To isolate $y$, subtract $3x$ from both sides of the equation: $3x + 2y - 3x = 12 - 3x$

• This simplifies to $2y = 12 - 3x$.

3. Solve for $y$:

• Now, divide every term by $2$ to isolate $y$:
• $\frac{2y}{2} = \frac{12}{2} - \frac{3x}{2} .$
• Simplifying this, we get $y = 6 - \frac{3x}{2}$.

4. Final Answer:

• The rearranged equation is $y = 6 - \frac{3x}{2}$.

Explanation: Every step involved either adding, subtracting, or dividing terms to gradually isolate $y$ on one side of the equation.

Example 2: Dealing with Fractions Problem: Make $x$ the subject in $\frac{y}{x} = 5 + 3x .$

Step-by-Step Solution:

5. Eliminate the fraction by multiplying both sides by $x$:

• Start by multiplying both sides by $x$ to remove the fraction:
• $x \times \frac{y}{x} = x \times (5 + 3x)$.
• On the LHS, $x$ and $\frac{y}{x}$ cancel out, leaving $( y )$.
• On the RHS, distribute $x$across the bracket:
• $x \times 5 = 5x$
• $x \times 3x = 3x^2$
• This gives us the equation: $y = 5x + 3x^2$.

6. Rearrange the equation to set it to zero:

• Subtract $y$ from both sides to move all terms to one side:
• $0 = 3x^2 + 5x - y$.

7. Rearrange to the standard form of a quadratic equation:

• We can rewrite this as $3x^2 + 5x - y = 0$.

8. Solve for $x$ (using the quadratic formula or factorising):

• To solve for $x$, you would typically use the quadratic formula:
• $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Substitute $a = 3 , b = 5$, and $c = -y$ into the quadratic formula to find the values of $x$.

Explanation**:** We carefully eliminated the fraction by multiplying, expanded the equation, and then rearranged it into a standard quadratic form to prepare for solving.

Exam Tip: When you multiply both sides by $x$ to eliminate a fraction, make sure you multiply every term on the RHS by $x$ as well. It's common to accidentally leave out one of the terms.

Example 3: Working with Square Roots Problem: Rearrange $\sqrt{\frac{p}{r - q}} = p$ to make $r$ the subject.

Step-by-Step Solution:

9. Remove the square root by squaring both sides:

• To eliminate the square root, square both sides of the equation:
• $\left(\sqrt{\frac{p}{r - q}}\right)^2 = (p)^2$.
• Squaring the LHS cancels out the square root, leaving:
• $\frac{p}{r - q} = p^2 .$

10. Clear the fraction by multiplying both sides by $r - q$:

• Multiply both sides by $r - q$ to get rid of the fraction:
• $\frac{p}{r - q} \times (r - q) = p^2 \times (r - q)$.
• On the LHS, $r - q$ cancels out, leaving:
• $p = p^2(r - q) .$

11. Expand the RHS:

• Distribute $p^2$ across $r - q$:
• $p = p^2r - p^2q .$

12. Move all terms involving $r$ to one side:

• To isolate $r$, add $p^2q$ to both sides:
• $p + p^2q = p^2r .$

13. Solve for $r$:

• Divide both sides by $p^2$ to isolate $r$:
• $\frac{p + p^2q}{p^2} = r .$
• Simplify by dividing each term in the numerator by $p^2$:
• $r = \frac{p}{p^2} + \frac{p^2q}{p^2}$.
• $r = \frac{1}{p} + q .$

14. Final Answer:

• The rearranged equation is $r = \frac{1}{p} + q$.

Explanation**:** Each step carefully worked to remove the square root, eliminate the fraction, and isolate the variable $r .$ The key was to square both sides and then rearrange terms accordingly.

Exam Tip: When dealing with square roots, always remember to square every term on both sides of the equation.

Example 4: More Complex Rearrangement Problem: Rearrange $A = \frac{2t + 3}{4t - 5}$ to make $t$ the subject.

Step-by-Step Solution:

15. Eliminate the fraction by multiplying both sides by $4t - 5$:

• Multiply both sides by $4t - 5$ to remove the denominator:
• $A \times (4t - 5) = \frac{2t + 3}{4t - 5} \times (4t - 5)$.
• On the RHS, $4t - 5$ cancels out, leaving:
• $A(4t - 5) = 2t + 3 .$

16. Expand the LHS:

• Distribute $A$ across $4t - 5$:
• $A \times 4t = 4At$ and $A \times -5 = -5A .$
• This gives:
• $4At - 5A = 2t + 3 .$

17. Move all terms involving $t$ to one side:

• Subtract $2t$ from both sides to group the $t$ terms:
• $4At - 2t = 5A + 3 .$

18. Factor out $t$ from the LHS:

• Factor $t$ from the LHS:
• $t(4A - 2) = 5A + 3 .$

19. Solve for $t$:

• Divide both sides by $4A - 2$ to isolate $t$:
• $t = \frac{5A + 3}{4A - 2} .$

20. Final Answer:

• The rearranged equation is $t = \frac{5A + 3}{4A - 2}$.

Explanation**:** The solution required careful elimination of the fraction, expansion, grouping like terms, and factoring to isolate $t .$

Exam Tip: When working with more complex equations, always take your time to expand and simplify carefully. Mistakes often happen when terms are moved across the equation incorrectly.

Hey! We know that Maths can sometimes feel like a tough climb, but remember this: every step you take gets you closer to the top. The challenges you're facing now are making you stronger and smarter. Don't be too hard on yourself—you're doing better than you think! With each problem you tackle, you're building the skills that will make everything easier over time.

Try it out! Question 1**:** Rearrange the equation $4x + 7y = 28$ to make $y$ the subject.

Question 2**:** Make $x$ the subject of the equation $\frac{5}{x} + 3 = 7$.

Question 3**:** Rearrange the formula$A = \sqrt{2h}$ to make $h$ the subject.

Question 4**:** Make $t$ the subject of the equation $S = \frac{2t + 1}{t - 3} .$

Solutions:

Solution 1: $y = 4 - \frac{4x}{7}$

Solution 2: $x = \frac{5}{4}$

Solution 3: $h = \frac{A^2}{2}$

Solution 4: $t = \frac{3S + 1}{S - 2}$

Question**:** Rearrange the equation $4x + 7y = 28$ to make $y$ the subject.

Question**:** Make $x$ the subject of the equation $\frac{5}{x} + 3 = 7$.

Exam Tip: Remember that multiplying by the variable $x$ cancels it out in the fraction because $x \times \frac{1}{x} = 1$. This step can sometimes be tricky, so make sure you practice it.

Question**:** Rearrange the formula $A = \sqrt{2h}$ to make $h$ the subject.

Exam Tip: When you square both sides, make sure that you apply the square to every part of the equation. Forgetting to square both sides is a common mistake.

Question**:** Make $t$ the subject of the equation $S = \frac{2t + 1}{t - 3}$.

Exam Tip: When eliminating fractions by multiplying, always remember to expand and simplify the equation step by step. This will help you avoid mistakes, especially with complex expressions.

Example: Solving $4(x - 2) = 12$ Here's how to solve this equation step by step:

21. Expand the brackets. Multiply everything inside the brackets by $4$: $4(x - 2) = 12$ This gives: $4x - 8 = 12$

22. Solve the equation by isolating $x$.

• First, add 8 to both sides to eliminate $-8$: $4x - 8 (+8) = 12 (+8)$

Which simplifies to: $4x = 20$

• Then, divide both sides by $4$ to solve for $x$: $\frac{4x}{4} = \frac{20}{4}$ Result: $x = 5$ Check Your Work:

Substitute $x = 5$ back into the original equation to confirm it works:

• Original equation: $4(x - 2) = 12$
• Substitute $x = 5$: $4(5 - 2) = 4(3) = 12$ Since both sides are equal, our solution x = 5 is correct!

Summary of the Example:

• Original equation: $4(x - 2) = 12$
• Step 1: Expand the brackets: $4x - 8 = 12$
• Step 2: Add 8 to both sides: $4x - 8 (+8) = 12 (+8)$ Result: $4x = 20$
• Step 3: Divide both sides by 4: $\frac{4x}{4} = \frac{20}{4}$ Result: x = 5
• Check: Substitute $x = 5$ back into the original equation to confirm.