Algebra |
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1. Simplifying Expressions
Show that \(2x + 2y + 4 + 3x -3y +2\) simplifies to \(5x -y +6\)
Question 1 — Simplifying Expressions (Worked Solutions)
Idea: Combine like terms (all are multiples of \(x\)).
\[ 3x + 4x - 2x \;=\; (3+4-2)x \;=\; 5x. \]
Explanation: Coefficients add and subtract in the usual way because each term is the same variable \(x\) to the same power \(1\).
Idea: Group like terms in \(a\) and in \(b\) separately.
\[ 5a + 2b - 2a - 4b \;=\; (5a - 2a) + (2b - 4b) \;=\; 3a - 2b. \]
Explanation: Only like terms combine: \(5a-2a=3a\) and \(2b-4b=-2b\).
Idea: Combine like terms by degree: the \(x^{2}\) terms, the \(x\) terms, and the constants.
\[ 2x^{2} - 3x + 4 - x^{2} + 5x - 3 \;=\; (2x^{2} - x^{2}) + (-3x + 5x) + (4 - 3) \;=\; x^{2} + 2x + 1. \]
Check (optional): \(x^{2} + 2x + 1 = (x+1)^{2}\), which is neatly factorised; this is a useful sanity check but not required.
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2. Expanding
Question 2 — Removing Brackets & Multiplying (Worked Solutions)
Method: Distribute over each bracket, then collect like terms.
\[ 3(2x - 1) + 5(x + 2) \;=\; (6x - 3) + (5x + 10) \;=\; (6x + 5x) + (-3 + 10) \;=\; 11x + 7. \]
Explanation: Use the distributive property \(a(b+c)=ab+ac\). Combine coefficients of \(x\) and combine constants separately.
Answer: \(\boxed{11x + 7}\)
Area/array plan: Use a \(2\times 2\) grid with rows \(2x\), \(-3\) and columns \(3x\), \(4\).
| \(3x\) | \(4\) | |
|---|---|---|
| \(2x\) | \(6x^{2}\) | \(8x\) |
| \(-3\) | \(-9x\) | \(-12\) |
Combine the four cells: \(6x^{2} + 8x - 9x - 12 = 6x^{2} - x - 12.\)
Answer: \(\boxed{6x^{2} - x - 12}\)
Area/array plan: Use a \(2\times 3\) grid with rows \(x\), \(2\) and columns \(x^{2}\), \(-3x\), \(5\).
| \(x^{2}\) | \(-3x\) | \(5\) | |
|---|---|---|---|
| \(x\) | \(x^{3}\) | \(-3x^{2}\) | \(5x\) |
| \(2\) | \(2x^{2}\) | \(-6x\) | \(10\) |
Group by degree: \(x^{3} + (-3x^{2}+2x^{2}) + (5x-6x) + 10 = x^{3} - x^{2} - x + 10.\)
Answer: \(\boxed{x^{3} - x^{2} - x + 10}\)
Worked steps (array/area model)
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3. Substituting
If \( x = 2 \), explain why \( x^2 + 3x - 4 = 6 \).
Question 3 — Substituting into Expressions (Worked Solutions)
Given \(x = 3\) and \(y = -2\) evaluate:
Method: Substitute \(x=3\), \(y=-2\). Evaluate powers first, then multiply, then add/subtract.
\[ 2x^{2} - 4xy + y^{2} = 2(3)^{2} - 4(3)(-2) + (-2)^{2} = 2\cdot 9 - 4\cdot(-6) + 4. \] \[ = 18 + 24 + 4 = 46. \]
Note: Since \(xy = 3\cdot(-2) = -6\), the term \(-4xy = -4\cdot(-6) = +24\).
Answer: \(\boxed{46}\)
Method: Substitute, then simplify numerator and denominator separately.
\[ \frac{3x + 2y}{x - y} = \frac{3(3) + 2(-2)}{3 - (-2)} = \frac{9 - 4}{3 + 2} = \frac{5}{5} = 1. \]
Note: Subtracting a negative gives a plus: \(x - y = 3 - (-2) = 5\).
Answer: \(\boxed{1}\)
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4. Solving Equations
Show that if \( 2x + 2 = x + 9 \) then \(x = 7\).
Question 4 — Solving Equations (Worked Solutions)
Do the same to both sides; write the simplified result on the next line:
\[ \begin{aligned} 3x + 7 &= 25 \\ \text{Subtract }7\text{ from both sides: } 3x + 7 - 7 &= 25 - 7 \\ 3x &= 18 \\ \text{Divide both sides by }3\,(>0):\ \dfrac{3x}{3} &= \dfrac{18}{3} \\ x &= 6 \end{aligned} \]
Answer: \(\boxed{x=6}\)
Do the same to both sides; write the simplified result on the next line:
\[ \begin{aligned} 4(2x - 3) &= 3x + 7 \\ \text{Expand the bracket: } 8x - 12 &= 3x + 7 \\ \text{Subtract }3x\text{ from both sides: } 8x - 12 - 3x &= 3x + 7 - 3x \\ 5x - 12 &= 7 \\ \text{Add }12\text{ to both sides: } 5x - 12 + 12 &= 7 + 12 \\ 5x &= 19 \\ \text{Divide both sides by }5\,(>0):\ \dfrac{5x}{5} &= \dfrac{19}{5} \\ x &= \dfrac{19}{5} \end{aligned} \]
Answer: \(\boxed{x=\tfrac{19}{5}}\) \((=3.8)\)
Do the same to both sides; write the simplified result on the next line:
\[ \begin{aligned} 2x + 3 &= 4x - 5 \\ \text{Subtract }2x\text{ from both sides: } 2x + 3 - 2x &= 4x - 5 - 2x \\ 3 &= 2x - 5 \\ \text{Add }5\text{ to both sides: } 3 + 5 &= 2x - 5 + 5 \\ 8 &= 2x \\ \text{Divide both sides by }2\,(>0):\ \dfrac{8}{2} &= \dfrac{2x}{2} \\ 4 &= x \end{aligned} \]
Answer: \(\boxed{x=4}\)
5. Dividing
Show using the area model that \( \dfrac{x^2 + 5x + 6}{x + 2} = x + 3 \)
Question 5 — Algebraic Division (Area / Array Model)
Goal: Find the quotient so that (divisor) × (quotient) reconstructs the dividend.
Step 1 — match leading term: To make \(x^{2}\) from \(x+2\), start with a quotient term \(x\) because \(x \cdot x = x^{2}\).
| \(x\) | \(+\;3\) | |
|---|---|---|
| \(x\) | \(x^{2}\) | \(3x\) |
| \(2\) | \(2x\) | \(6\) |
Step 2 — complete middle term: The grid so far gives \(x^{2} + (2x + 3x) + 6 = x^{2} + 5x + 6\). The extra column header \((+\,3)\) is chosen so that \(x\cdot3=3x\) combines with \(2x\) to make \(5x\).
Conclusion: Quotient \(=\; x + 3\). Check: \((x+2)(x+3)=x^{2}+5x+6.\)
Answer: \(\boxed{x+3}\)
Plan: Choose quotient terms so column sums by degree match the dividend.
Step 1 — leading term: Use \(2x^{2}\) so that \(x \cdot 2x^{2} = 2x^{3}\).
Step 2 — match the \(x^{2}\) term: From the first column we also get \(3 \cdot 2x^{2} = 6x^{2}\). We need overall \(+1x^{2}\), so include \(-5x\) as the next quotient term: \(x\cdot (-5x) = -5x^{2}\). Net \(x^{2}\): \(6x^{2} + (-5x^{2}) = +x^{2}\) ✔.
Step 3 — finish linear and constant terms: Choose \(+2\) as the final quotient term.
| \(2x^{2}\) | \(-5x\) | \(2\) | |
|---|---|---|---|
| \(x\) | \(2x^{3}\) | \(-5x^{2}\) | \(2x\) |
| \(3\) | \(6x^{2}\) | \(-15x\) | \(6\) |
Combine by degree: \(2x^{3}\) (only one), \(x^{2}\!: -5x^{2}+6x^{2}=+x^{2}\); \(x\!:\; 2x-15x=-13x\); constant \(=6\). This reproduces \(2x^{3} + x^{2} - 13x + 6\).
Answer: \(\boxed{2x^{2} - 5x + 2}\)
6. Solving inequalities
Show that if \( 3x - 5 < x + 7 \), then \( x < 5\).
Question 6 — Solving Inequalities (Worked Solutions)
Do the same to both sides, simplifying on the next line:
\[ \begin{aligned} 3x - 2 &< 10 \\ \text{Add }2\text{ to both sides: } 3x - 2 + 2 &< 10 + 2 \\ 3x &< 12 \\ \text{Divide both sides by }3\,(>0):\ \dfrac{3x}{3} &< \dfrac{12}{3} \\ x &< 4 \end{aligned} \] Since \(x \in \mathbb{N} = \{1,2,3,\dots\}\), the solutions are \(\{1,2,3\}\).
Answer: \(\boxed{x \in \{1,2,3\}}\).
Do the same to both sides, simplifying on the next line:
\[ \begin{aligned} 4 - 2x &> 6 \\ \text{Subtract }4\text{ from both sides: } 4 - 2x - 4 &> 6 - 4 \\ -2x &> 2 \\ \text{Divide both sides by }{-2}\,(\!<0)\text{ and reverse the inequality: } \dfrac{-2x}{-2} &< \dfrac{2}{-2} \\ x &< -1 \end{aligned} \] With \(x \in \mathbb{Z}\), this means \(x \le -2\) (…, \(-5,-4,-3,-2\)).
Answer: \(\boxed{x \in \mathbb{Z},\; x \le -2}\).
Do the same to both sides, simplifying on the next line:
\[ \begin{aligned} 5x + 3 &\le 18 \\ \text{Subtract }3\text{ from both sides: } 5x + 3 - 3 &\le 18 - 3 \\ 5x &\le 15 \\ \text{Divide both sides by }5\,(>0):\ \dfrac{5x}{5} &\le \dfrac{15}{5} \\ x &\le 3 \end{aligned} \] Hence the solution set is \((-\infty,\,3]\).
Answer: \(\boxed{x \le 3}\).
