Manipulation of Formulae Revision Notes for Junior Cert Mathematics

Manipulation of Formulae

Manipulating formulae means rearranging an equation to make a different variable the subject (this means getting that variable by itself on one side of the equation).

This is an important skill in maths because it allows you to solve problems where you need to know one specific thing, like the value of a variable.

Importance in Junior Cycle Maths:

Understanding how to manipulate formulae is a key part of the Junior Cycle Maths curriculum. It's essential because:

It helps you solve equations in algebra.
You'll use it in other topics, like geometry (e.g., finding the radius of a circle if you know the area) and physics (e.g., rearranging formulas for speed, distance, and time).

Steps for Manipulating Formulae

Get rid of any brackets, fractions, or square roots. Simplify the equation as much as possible. This might mean multiplying out brackets, eliminating fractions by multiplying both sides by the denominator, or squaring both sides to remove a square root.
Move all terms with the variable you want to the Left-Hand Side (LHS). The Left-Hand Side (LHS) is just the side of the equation where you want your variable to end up. Move everything else to the Right-Hand Side (RHS). This involves adding or subtracting terms to shift them across the equation.
Factor out the variable (if necessary). If the variable you want is in more than one term on the LHS, factor it out. (This means removing the variable from each term and placing the variable in front of a bracket.)

Divide to solve for the variable.

Finally, divide both sides by whatever is left on the LHS to get your variable by itself.

Worked Examples

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

Step-by-Step Solution**:**

Start with the given equation:

The equation is $3x + 2y = 12$ .

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 $2y = 12 - 3x$ .

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 $y = 6 - \frac{3x}{2}$ .

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.

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

Step-by-Step Solution:

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: $y = 5x + 3x^2$ .

Rearrange the equation to set it to zero:

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

Rearrange to the standard form of a quadratic equation:

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

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.

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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.

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

Step-by-Step Solution:

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:
$\frac{p}{r - q} = p^2 .$

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:
$p = p^2(r - q) .$

Expand the RHS:

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

Move all terms involving $r$ to one side:

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

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 $p^2$ :
$r = \frac{p}{p^2} + \frac{p^2q}{p^2}$ .
$r = \frac{1}{p} + q .$

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.

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

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

Step-by-Step Solution:

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:
$A(4t - 5) = 2t + 3 .$

Expand the LHS:

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

Move all terms involving $t$ to one side:

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

Factor out $t$ from the LHS:

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

Solve for $t$ :

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

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 $t .$

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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.

Summary of Key Points:

Simplify first: Remove any brackets, fractions, or square roots to make the equation easier to work with.
Move terms with care: When moving terms across the equation, remember to reverse the operation (e.g., addition becomes subtraction).
Factor and divide: If your variable appears in more than one term, factor it out before dividing.

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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.

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

Step-by-step solutions for the above problems:

Problem 1:

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

Step-by-Step Solution:

Start with the given equation:

The equation is $4x + 7y = 28$ .

Move the $4x$ term to the RHS:

$4x + 7y - 4x = 28 - 4x .$
On the left-hand side, $4x - 4x$ cancels out, leaving:
$7y = 28 - 4x$ .

Solve for $y$ :

Now, we want $y$ by itself. To do that, we divide every term on both sides by $7$ (because $7$ is multiplied by $y$ ):
$\frac{7y}{7} = \frac{28}{7} - \frac{4x}{7}$ .
$\frac{7y}{7}$ simplifies to $y$ because $7 \div 7 = 1$ , so:
$y = \frac{28}{7} - \frac{4x}{7}$ .
$\frac{28}{7} = 4$ , and $\frac{4x}{7}$ remains as it is:
$y = 4 - \frac{4x}{7}$ .

Final Answer:

The rearranged equation is $y = 4 - \frac{4x}{7} .$ Explanation**:** In this problem, each step involves simplifying the equation by isolating $y$ . We moved the $x$ term to the right-hand side, then divided all terms by 7 to get $y$ alone.

Problem 2: Rearranging with Fractions

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

Step-by-Step Solution:

Start with the given equation:

The equation is $\frac{5}{x} + 3 = 7 .$

Isolate the fraction:

We want to isolate the fraction $\frac{5}{x}$ . To do this, we subtract $3$ from both sides:
$\frac{5}{x} + 3 - 3 = 7 - 3$ .
On the left-hand side, $3 - 3$ cancels out, leaving:
$\frac{5}{x} = 4$ .

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

Now, we want to remove the fraction. To do this, multiply both sides of the equation by $x$ :
$x \times \frac{5}{x} = 4 \times x$ .
On the left-hand side, $x$ cancels out with the $x$ in the denominator of $\frac{5}{x}$ (because any number multiplied by its reciprocal cancels out the fraction), leaving:
$5 = 4x$ .

Solve for $x$ :

Finally, divide both sides by 4 to get $x$ by itself:
$\frac{5}{4} = x$ .

Final Answer:

The rearranged equation is $x = \frac{5}{4}$ . Explanation**:** The fraction was eliminated by multiplying both sides by $x$ , and then we divided by $4$ to solve for $x$ . This problem highlights the importance of understanding how to cancel out terms when dealing with fractions.

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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.

Problem 3: Square Roots

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

Step-by-Step Solution:

Start with the given equation:

The equation is $A = \sqrt{2h}$ .

Eliminate the square root by squaring both sides:

To remove the square root, we square both sides of the equation:
$A^2 = (\sqrt{2h})^2 .$
Squaring the left side gives $A^2$ , and squaring the right side cancels out the square root (because the square root of a number squared is just the number itself):
$A^2 = 2h$ .

Solve for $h$ :

Now, divide both sides by 2 to isolate $h$ :
$\frac{A^2}{2} = h$ .

Final Answer:

The rearranged equation is $h = \frac{A^2}{2}$ . Explanation: The square root was eliminated by squaring both sides, and then we divided by $2$ to solve for $h$ . This problem emphasises the importance of correctly handling square roots and squaring.

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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.

Problem 4: Complex Rearrangement

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

Step-by-Step Solution:

Start with the given equation:

The equation is $S = \frac{2t + 1}{t - 3}$ .

Eliminate the fraction by multiplying both sides by $t - 3$ :

To remove the fraction, we multiply both sides by $t - 3$ :
$S \times (t - 3) = \frac{2t + 1}{t - 3} \times (t - 3)$ .
On the right-hand side, $t - 3$ is in both the numerator and the denominator, so they cancel each other out (because any number divided by itself equals $1$ ):
$S(t - 3) = 2t + 1 .$ Explanation of cancellation:
Consider that $\frac{2t + 1}{t - 3}$ is a fraction, and multiplying it by $t - 3$ is the same as multiplying by $\frac{t - 3}{1}$ .
This allows the $t - 3$ in the numerator and denominator to cancel out, leaving just $2t + 1$ .

Expand the left side:

Next, distribute $S$ across the bracket on the left-hand side:
$S \times t = St$
$S \times -3 = -3S$
So, $S(t - 3)$ becomes $St - 3S = 2t + 1$ .

Move all terms involving $t$ to one side:

To group all the $t$ terms together, subtract $2t$ from both sides:
$St - 2t = 3S + 1$ .

Factor out $t$ on the left side:

Now, factor out $t$ from the left side:
$t(S - 2) = 3S + 1$ .

Solve for $t$ :

Finally, divide both sides by $S - 2$ to isolate $t$ :
$t = \frac{3S + 1}{S - 2} .$

Final Answer:

The rearranged equation is $t = \frac{3S + 1}{S - 2}$ . Explanation: Every step was carefully managed to isolate $t$ , starting by removing the fraction, expanding, and then grouping and factoring terms to solve for $t$ .

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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.

Certainly! Here's a more concise version that maintains all the details and explanations while making it easier to read:

c) Solving Linear Equations with Brackets

When solving linear equations with brackets, follow these two steps:

Expand the brackets to simplify the equation.
Solve the equation by isolating the unknown variable $x$ on one side, then doing the opposite of what the equation says.

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

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

Top Tip: We could have divided both sides by $4$ at the start, since $4$ is a factor of $12$ .

:::

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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.

Manipulation of Formulae (Junior Cert Mathematics): Revision Notes