Algebraic Manipulation: Foundation - Go Teach Maths: Handcrafted ... - Free Printable

Here is the solution to all nine problems in the grid, with a clear explanation for each.

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Top Row

1. Simplify: `c + 2c × 2c - 2c`

* Step 1: Follow the order of operations (BIDMAS/BODMAS). Multiplication comes before addition and subtraction.
* `2c × 2c = 4c²`
* Step 2: Substitute back into the expression:
* `c + 4c² - 2c`
* Step 3: Combine like terms (`c` and `-2c`):
* `(c - 2c) + 4c² = -c + 4c²`
* Step 4: Write in standard form (highest power first):
* `4c² - c`

✔ Answer: `4c² - c`

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2. Simplify: `3(2a²b)³`

* Step 1: Apply the exponent to everything inside the parentheses. Remember `(xy)ⁿ = xⁿyⁿ` and `(xᵐ)ⁿ = xᵐⁿ`.
* `(2a²b)³ = 2³ × (a²)³ × b³ = 8 × a⁶ × b³ = 8a⁶b³`
* Step 2: Multiply by the 3 outside:
* `3 × 8a⁶b³ = 24a⁶b³`

✔ Answer: `24a⁶b³`

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3. Factorise: `6xy³ - 9x³y²`

* Step 1: Find the Highest Common Factor (HCF) of the coefficients and variables.
* Coefficients: HCF of 6 and 9 is 3.
* Variable `x`: Lowest power is `x¹` (from `6xy³`).
* Variable `y`: Lowest power is `y²` (from `9x³y²`).
* So, the HCF is `3xy²`.
* Step 2: Factor out `3xy²` from each term.
* `6xy³ ÷ 3xy² = 2y`
* `-9x³y² ÷ 3xy² = -3x²`
* Step 3: Write the factored form:
* `3xy²(2y - 3x²)`

✔ Answer: `3xy²(2y - 3x²)`

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Middle Row

4. Multiply out: `3a²b³(2a + 3b²)`

* Step 1: Distribute `3a²b³` to each term inside the parentheses.
* `3a²b³ × 2a = (3×2) × (a²×a) × b³ = 6a³b³`
* `3a²b³ × 3b² = (3×3) × a² × (b³×b²) = 9a²b⁵`
* Step 2: Combine the results:
* `6a³b³ + 9a²b⁵`

✔ Answer: `6a³b³ + 9a²b⁵`

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5. Factorise: `x² + 10x - 75`

* Step 1: We need two numbers that multiply to give `-75` and add to give `+10`.
* Factors of 75: 1 & 75, 3 & 25, 5 & 15.
* Since the product is negative and the sum is positive, the larger number must be positive.
* Try `15` and `-5`: `15 × (-5) = -75` and `15 + (-5) = 10`. Perfect!
* Step 2: Write the factored form using these numbers:
* `(x + 15)(x - 5)`

✔ Answer: `(x + 15)(x - 5)`

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6. Expand & Simplify: `2(x + 3)(x - 2)`

* Step 1: First, expand the two binomials `(x + 3)(x - 2)`.
* Use FOIL or distributive property:
* `x × x = x²`
* `x × (-2) = -2x`
* `3 × x = 3x`
* `3 × (-2) = -6`
* Combine: `x² - 2x + 3x - 6 = x² + x - 6`
* Step 2: Now multiply the result by 2:
* `2 × (x² + x - 6) = 2x² + 2x - 12`

✔ Answer: `2x² + 2x - 12`

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Bottom Row

7. Simplify: `3(x - 2) - 2(3x - 4)`

* Step 1: Distribute the numbers outside the parentheses.
* `3(x - 2) = 3x - 6`
* `-2(3x - 4) = -6x + 8` (Remember to distribute the negative sign!)
* Step 2: Combine the expressions:
* `3x - 6 - 6x + 8`
* Step 3: Combine like terms:
* `(3x - 6x) + (-6 + 8) = -3x + 2`

✔ Answer: `-3x + 2`

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8. Make `x` the subject: `y = √(x/3)`

* Step 1: To isolate `x`, we need to reverse the operations applied to it. The square root is the last operation applied to `x/3`.
* Square both sides to eliminate the square root:
* `y² = (√(x/3))²`
* `y² = x/3`
* Step 2: Multiply both sides by 3 to isolate `x`:
* `3y² = x`
* Step 3: Write the final answer with `x` as the subject:
* `x = 3y²`

✔ Answer: `x = 3y²`

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9. Factorise: `x² - 100`

* Step 1: Recognize this as a difference of squares. The formula is `a² - b² = (a + b)(a - b)`.
* Here, `x²` is `a²` and `100` is `b²` (since `10² = 100`).
* Step 2: Apply the formula:
* `x² - 100 = (x + 10)(x - 10)`

✔ Answer: `(x + 10)(x - 10)`

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All problems have been solved and explained step-by-step.