- \(\displaystyle \frac{T}{2\pi}=\sqrt{\frac{l}{g}}\)
- Square both sides: \(\displaystyle \left(\frac{T}{2\pi}\right)^2=\frac{l}{g}\)
- Hence \(\displaystyle l=\frac{gT^2}{4\pi^2}\).
Assume \(g>0\) and \(T>0\).
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- Definition of polynomials, degree, coefficients, and terms.
- Addition and subtraction of polynomials (combining like terms).
- Multiplication of polynomials (using the distributive law and expansions).
- Perfect squares and recognition of patterns.
- Division of polynomials by a monomial or binomial when possible.
Expand and Simplify
Expand and simplify: \(7(x^3 + 2x^2 - 5x) - 2(2 + 3x + 4x^2 - 2x^3)\)
Quadratic Perfect Square
Given that \(25x^2 + px + 16\) is a perfect square and \(p > 0\), find the value of \(p\).
Cubic Division
Divide:
(i) \(2x^3 + x^2 - 13x + 6\) by \(x + 3\)
(ii) \(4x^3 - 13x - 6\) by \(x - 2\).
Polynomial Expressions: Practice Questions
Expand both parts:
\(6x^3 - 6x^2 + 24x\) and \(-15 + 6x - 3x^2 + 3x^3\).
Combine like terms:
\( (6x^3+3x^3) + (-6x^2-3x^2) + (24x+6x) + (-15) \)
\(= 9x^3 - 9x^2 + 30x - 15\).
\(6x^3 - 6x^2 + 24x\) and \(-15 + 6x - 3x^2 + 3x^3\).
Combine like terms:
\( (6x^3+3x^3) + (-6x^2-3x^2) + (24x+6x) + (-15) \)
\(= 9x^3 - 9x^2 + 30x - 15\).
Using a 2×3 area model:
Collecting terms: \(3x^3 + 7x^2 - 22x - 8\).
| \(3x^2\) | \(-5x\) | \(-2\) | |
|---|---|---|---|
| \(x\) | \(3x^3\) | \(-5x^2\) | \(-2x\) |
| \(+4\) | \(12x^2\) | \(-20x\) | \(-8\) |
Compare with \((4x\pm3)^2\).
Case 1: \((4x+3)^2\)
Middle term \(= +24x \Rightarrow p=24\).
Case 2: \((4x-3)^2\)
Middle term \(= -24x \Rightarrow p=-24\).
Case 1: \((4x+3)^2\)
| \(4x\) | \(+3\) | |
|---|---|---|
| \(4x\) | \(16x^2\) | \(12x\) |
| \(+3\) | \(12x\) | \(9\) |
Case 2: \((4x-3)^2\)
| \(4x\) | \(-3\) | |
|---|---|---|
| \(4x\) | \(16x^2\) | \(-12x\) |
| \(-3\) | \(-12x\) | \(9\) |
Use a 2×3 area model (divisor \((x-2)\) × quotient \((2x^2 + x - 5)\)):
Adding terms gives the dividend \(2x^3 - 3x^2 - 7x + 10\). Hence the quotient is \(2x^2 + x - 5\).
| \(2x^2\) | \(+x\) | \(-5\) | |
|---|---|---|---|
| \(x\) | \(2x^3\) | \(x^2\) | \(-5x\) |
| \(-2\) | \(-4x^2\) | \(-2x\) | \(10\) |
- Understand independent and dependent variables in function notation.
- Model real-life contexts (e.g. area, volume, perimeter, and cost) using polynomial functions.
- Perform operations on functions — addition, subtraction, and multiplication.
- Evaluate a polynomial function by substitution.
Area and Perimeter Expressed as Functions
The length of a rectangle is \((2x+3)\,\text{cm}\). If the area is \(A(x)=2x^2+7x+6\), find:
(a) an expression for the width, (b) an expression for the perimeter \(P(x)\), (c) the minimum value of \(x\).
Quadratic Function in Context
A paint manufacturer knows that the daily cost (€ C) of producing \(x\) litres of paint is given by
\(C(x) = 0.001x^2 + 0.1x + 5.\)
(a) State the degree of \(C(x)\).
(b) Find the daily cost of producing (i) 100 ℓ of paint (ii) 400 ℓ of paint.
Adding and Subtracting Polynomial Functions
Given \(f(x) = 3x^3 - 4x^2 - 3x + 4\) and \(g(x) = 5x^3 + 14x^2 + 7x - 2\), find:
(a) \(2f(x) - g(x)\) and state its degree.
(b) \(f(x) + 2g(x)\) and state its degree.
Evaluating Linear Functions
Given the function \(f(x) = 2x - 4\) for all \(x \in \mathbb{R}\), find:
(a) \(f(3)\), \(f(-2)\), \(f(t)\)
(b) For what values of \(t\) is \(f(t) = t\)?
Polynomial Functions: Practice Questions
\(A(x) = x \cdot (x-5)\).
Area model (1×2):
So \(A(x) = x^2 - 5x\).
Area model (1×2):
| \(x\) | \(-5\) | |
|---|---|---|
| \(x\) | \(x^2\) | \(-5x\) |
Substitute \(x=100\):
\(C(100) = 0.001(100^2) + 0.1(100) + 5\).
\(= 0.001(10\,000) + 10 + 5\).
\(= 10 + 10 + 5 = 25\).
So the cost is **€25**.
\(C(100) = 0.001(100^2) + 0.1(100) + 5\).
\(= 0.001(10\,000) + 10 + 5\).
\(= 10 + 10 + 5 = 25\).
So the cost is **€25**.
- Recognise factorisation as the reverse of multiplication.
- Apply the main methods of factorisation:
- Highest common factor (HCF).
- Grouping in pairs.
- Difference of two squares.
- Quadratic trinomials.
- Sum and difference of two cubes.
- Use factorisation to simplify algebraic fractions.
Factorising Quadratic Expressions
Factorise: (i) \(3x^2+10x+8\) (ii) \(x^2-2\sqrt{2}\,x-6\)
Factorise: Difference of Two Squares
Factorise fully:
(i) \(x^4 - y^4\)
(ii) \(12x^2 - 75y^2\)
Factorise \(a^{4}-b^{4}\).
Factorise: Sum of Two Cubes
Using the sum of two cubes, factorise \[\; 8x^3 + y^3\]
Factorising: Practice Questions
Group terms: \((6x^2y + 3xy^2) - (12x + 6y)\).
Factorise each group: \(3xy(2x+y) - 6(2x+y)\).
Common factor \((2x+y)\) ⇒ \((2x+y)(3xy - 6)\).
Verification (2×2 area model for \((2x+y)(3xy-6)\)):
Sum \(\to 6x^2y + 3xy^2 - 12x - 6y\).
Factorise each group: \(3xy(2x+y) - 6(2x+y)\).
Common factor \((2x+y)\) ⇒ \((2x+y)(3xy - 6)\).
Verification (2×2 area model for \((2x+y)(3xy-6)\)):
| \(3xy\) | \(-6\) | |
|---|---|---|
| \(2x\) | \(6x^2y\) | \(-12x\) |
| \(y\) | \(3xy^2\) | \(-6y\) |
Difference of two squares: \((x-3y)(x+3y)\).
Area model (2×2) for \((x-3y)(x+3y)\):
Middle terms cancel \(\to x^2 - 9y^2\).
Area model (2×2) for \((x-3y)(x+3y)\):
| \(x\) | \(+3y\) | |
|---|---|---|
| \(x\) | \(x^2\) | \(3xy\) |
| \(-3y\) | \(-3xy\) | \(-9y^2\) |
Find numbers with product \(-18\) and sum \(3\): \(+6\) and \(-3\).
So \(x^2 + 3x - 18 = (x+6)(x-3)\).
Area model (2×2) for \((x+6)(x-3)\):
Collecting gives \(x^2 + 3x - 18\).
So \(x^2 + 3x - 18 = (x+6)(x-3)\).
Area model (2×2) for \((x+6)(x-3)\):
| \(x\) | \(-3\) | |
|---|---|---|
| \(x\) | \(x^2\) | \(-3x\) |
| \(+6\) | \(6x\) | \(-18\) |
Solve \(3x^2 - 17x + 20 = 0\) using \(a=3,\; b=-17,\; c=20\).
Quadratic formula: \[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} =\frac{-(-17)\pm\sqrt{(-17)^2-4(3)(20)}}{2\cdot 3} =\frac{17\pm\sqrt{289-240}}{6} =\frac{17\pm\sqrt{49}}{6} =\frac{17\pm 7}{6}. \] Hence \[ x_1=\frac{17+7}{6}=\frac{24}{6}=4,\qquad x_2=\frac{17-7}{6}=\frac{10}{6}=\frac{5}{3}. \] Roots \(x=4\) and \(x=\tfrac{5}{3}\) give factors \((x-4)\) and \(\bigl(x-\tfrac{5}{3}\bigr)\).
To keep integer coefficients: \[ 3x^2-17x+20=3\bigl(x-4\bigr)\Bigl(x-\tfrac{5}{3}\Bigr)=(x-4)(3x-5). \] Therefore, \(\boxed{\,3x^2-17x+20=(x-4)(3x-5)\,}\).
Quadratic formula: \[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} =\frac{-(-17)\pm\sqrt{(-17)^2-4(3)(20)}}{2\cdot 3} =\frac{17\pm\sqrt{289-240}}{6} =\frac{17\pm\sqrt{49}}{6} =\frac{17\pm 7}{6}. \] Hence \[ x_1=\frac{17+7}{6}=\frac{24}{6}=4,\qquad x_2=\frac{17-7}{6}=\frac{10}{6}=\frac{5}{3}. \] Roots \(x=4\) and \(x=\tfrac{5}{3}\) give factors \((x-4)\) and \(\bigl(x-\tfrac{5}{3}\bigr)\).
To keep integer coefficients: \[ 3x^2-17x+20=3\bigl(x-4\bigr)\Bigl(x-\tfrac{5}{3}\Bigr)=(x-4)(3x-5). \] Therefore, \(\boxed{\,3x^2-17x+20=(x-4)(3x-5)\,}\).
Recognise a sum of cubes: \(a^3+b^3=(a+b)(a^2-ab+b^2)\).
Here \(a=3x\) and \(b=y\). Thus \[ 27x^3 + y^3 = (3x)^3 + y^3 = (3x + y)\bigl((3x)^2 - (3x)y + y^2\bigr) = (3x + y)\bigl(9x^2 - 3xy + y^2\bigr). \] Therefore, \(\boxed{\,27x^3 + y^3 = (3x+y)(9x^2 - 3xy + y^2)\,}\).
Here \(a=3x\) and \(b=y\). Thus \[ 27x^3 + y^3 = (3x)^3 + y^3 = (3x + y)\bigl((3x)^2 - (3x)y + y^2\bigr) = (3x + y)\bigl(9x^2 - 3xy + y^2\bigr). \] Therefore, \(\boxed{\,27x^3 + y^3 = (3x+y)(9x^2 - 3xy + y^2)\,}\).
- Simplify by **factorising** numerator and denominator, then cancelling common factors (state restrictions).
- Find **common denominators** for addition and subtraction; simplify the result.
- For **multiplication/division**: multiply numerators and denominators (or invert and multiply), cancelling where possible.
Simplifying Algebraic Fractions
Simplify (i) \(\dfrac{5ax}{15a + 10a^2}\) (ii) \(\dfrac{t^2 + 3t - 4}{t^2 - 16}\) (iii) \(\dfrac{\frac{5}{8} + y}{\frac{1}{8}}\)
Adding and Subtracting Algebraic Fractions
Simplify each of the following
(i) \(\dfrac{6y}{x(x+4y)} - \dfrac{3}{2x}\) (ii) \(\dfrac{x-4}{x^2 - x - 2} - \dfrac{x-3}{x^2 - 4}\)
Dividing by Fractions
Simplify \[ \frac{y - \frac{x^2 + y^2}{y}}{\frac{1}{x} - \frac{1}{y}}. \]
Algebraic Fractions: Practice Questions
Factor the denominator:
\[
15a + 10a^2 = 5a(3 + 2a).
\]
Then
\[
\frac{5ax}{15a + 10a^2} = \frac{5ax}{5a(3+2a)} = \frac{x}{3+2a},
\]
valid for \(a \neq 0\).
Factor numerator and denominator:
\[
x^2 - 9y^2 = (x-3y)(x+3y),\qquad 3x+9y=3(x+3y).
\]
Cancel the common factor \(x+3y\) (when \(x\neq -3y\)):
\[
\frac{x^2 - 9y^2}{3x + 9y}=\frac{(x-3y)(x+3y)}{3(x+3y)}=\frac{x-3y}{3}.
\]
Write each part as a single fraction:
\[
\text{Numerator: } \frac{1}{x} + \frac{2}{y} = \frac{y + 2x}{xy},\qquad
\text{Denominator: } \frac{3}{x} - \frac{1}{y} = \frac{3y - x}{xy}.
\]
Divide by multiplying by the reciprocal:
\[
\frac{\frac{y+2x}{xy}}{\frac{3y-x}{xy}}
= \frac{y+2x}{xy}\cdot\frac{xy}{3y-x}
= \frac{y+2x}{3y-x},
\]
valid when \(x\neq 0\), \(y\neq 0\) and \(3y\neq x\).
- Use the notation \( \binom{n}{r} = \dfrac{n!}{r!(n-r)!} \).
- Apply the Binomial Theorem: \[ (a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r}b^r. \]
- Find specific terms and expansions for small powers of \( n \).
- Use Pascal’s triangle to obtain coefficients for small values of \( n \).
Combinations
Evaluate (i) \(\binom{7}{4}\) (ii) \(\binom{6}{2}\) (iii) \(\binom{6}{4}\) (iv) \(\binom{15}{4}\)
Using the Binomial Theorem
(i) Find the first 3 terms of the expansion of \((1 - 5y)^8\).
(ii) Find the fourth term of the expansion of \((3a + b)^7\).
Binomial Expansion: Practice Questions
First square \((x+2)\) using a 2×2 area model:
Hence \((x+2)^2 = x^2 + 4x + 4\). Now multiply by \((x+2)\):
\[
(x^2+4x+4)(x+2)
= x^3 + 6x^2 + 12x + 8.
\]
Therefore, \(\boxed{(x+2)^3 = x^3 + 6x^2 + 12x + 8}\).
| \(x\) | \(+2\) | |
|---|---|---|
| \(x\) | \(x^2\) | \(2x\) |
| \(+2\) | \(2x\) | \(4\) |
Use coefficients \(1,4,6,4,1\) (row 4 of Pascal’s triangle):
\[
(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4,
\quad a=2x,\ b=-3.
\]
\[
\begin{aligned}
(2x-3)^4 &=(2x)^4 + 4(2x)^3(-3) + 6(2x)^2(-3)^2 \\
&\quad + 4(2x)(-3)^3 + (-3)^4 \\
&=16x^4 - 96x^3 + 216x^2 - 216x + 81.
\end{aligned}
\]
Hence, \(\boxed{(2x-3)^4 = 16x^4 - 96x^3 + 216x^2 - 216x + 81}\).
General term: \(\displaystyle T_{k+1}=\binom{7}{k}(1)^{7-k}(-2x)^k\).
For \(x^3\), take \(k=3\):
\[
\binom{7}{3}(-2)^3 x^3 = 35(-8)x^3 = -280x^3.
\]
Coefficient of \(x^3\) is \(\boxed{-280}\).
General term: \(\displaystyle \binom{5}{k}(1)^{5-k}(3x)^k\).
For \(x^2\), take \(k=2\):
\[
\binom{5}{2}3^2 x^2 = 10\cdot 9\,x^2 = 90x^2.
\]
Coefficient of \(x^2\) is \(\boxed{90}\).
- Compare coefficients of like powers in algebraic expressions.
- Use known identities to determine unknown coefficients.
- Find unknown coefficients using given factors or roots.
Quadratic Identity
Find the values of \(a\) and \(b\) given that \((2x + a)^2 = 4x^2 + 12x + b,\) for all values of \(x.\)
Linear Identity
If \(3t^2x - 3px + c - 2t^3 = 0\) for all values of \(x,\) find \(c\) in terms of \(p.\)
Quadratic Identity with Fractions
Given \[ \frac{1}{(x + 1)(x - 2)} = \frac{A}{(x + 1)} + \frac{B}{(x - 2)}, \] for all values of \(x,\) find the values of \(A\) and \(B.\)
Undetermined Coefficients involving Division
Given that \((x - t)^2\) is a factor of \(x^3 + 3px + c,\) show that \(p = -t^2\) and \(c = 2t^3.\)
Division by Factor
\(2x - \sqrt{3}\) is a factor of \(4x^2 - 2(1+\sqrt{3})x + \sqrt{3}\;\); find the second factor.
Algebraic Identities: Practice Questions
Expand LHS: \((x+1)(ax+b)=ax^2+(a+b)x+b\).
Compare coefficients with \(3x^2+7x+2\): \[ a=3,\quad a+b=7,\quad b=2. \] From \(a=3\) and \(a+b=7\) we get \(b=4\), which contradicts \(b=2\).
Conclusion: As written, there is no solution for \(a,b\); the identity is inconsistent.
Compare coefficients with \(3x^2+7x+2\): \[ a=3,\quad a+b=7,\quad b=2. \] From \(a=3\) and \(a+b=7\) we get \(b=4\), which contradicts \(b=2\).
Conclusion: As written, there is no solution for \(a,b\); the identity is inconsistent.
Divide \(x^3+px^2+qx+r\) by \((x-2)\). The remainder is the value at \(x=2\):
\[ R = 2^3 + p\cdot 2^2 + q\cdot 2 + r = 8 + 4p + 2q + r. \] Since \((x-2)\) is a factor, \(R=0\). Therefore, \[ \boxed{\,4p + 2q + r + 8 = 0\,}. \]
\[ R = 2^3 + p\cdot 2^2 + q\cdot 2 + r = 8 + 4p + 2q + r. \] Since \((x-2)\) is a factor, \(R=0\). Therefore, \[ \boxed{\,4p + 2q + r + 8 = 0\,}. \]
- Rearrange algebraic and physical formulae to change the subject.
- Apply inverse operations carefully when isolating variables.
- Recognise and apply these manipulations in geometry and science contexts.
Manipulating Formula
(i) If \(v^2=u^2+2as\), express \(a\) in terms of \(v\), \(u\) and \(s\).
(ii) If \(\sqrt{\dfrac{x+y}{x-y}}=\dfrac12\), express \(y\) in terms of \(x\). Hence find the value of \(y\) when \(x=5\).
Conic Volume
A container in the shape of an inverted cone is used to hold liquid.
Given \(\tan\theta=\dfrac{r}{h}\),
express the volume, \(V\), in terms of the depth, \(h\), of the liquid and the angle \(\theta\).
Manipulating Formula — Requires Factorisation
Given \(x=\dfrac{t+4}{3t+1}\), find \(t\) in terms of \(x\).
Manipulating Formulae: Practice Questions
- Divide both sides by \(\pi r^2\) (with \(r\ne 0\)): \(\displaystyle \frac{V}{\pi r^2}=h\).
- Therefore \(\displaystyle h=\frac{V}{\pi r^2}\).
- Factorise the right-hand side: \(A=xy+xz=x(y+z)\).
- Provided \(y+z\ne 0\): \(\displaystyle x=\frac{A}{\,y+z\,}\).
- Recognise and describe linear patterns (first difference constant).
- Recognise and describe quadratic patterns (second difference constant).
- Represent patterns numerically, algebraically, and graphically.
- Find the general term for linear and quadratic sequences using pattern recognition or formulae.
Linear and Quadratic Sequences
Examine each of the following patterns of numbers and determine if there is a
linear or quadratic relationship between the terms. Write an algebraic expression for each set of numbers:
(a) \(-2, 1, 4, 7, \dots\)
(b) \(3, 5, 11, 21, \dots\)
Quadratic Matchstick Pattern
Single matchsticks were used to form a sequence of patterns as shown.
Find an algebraic quadratic expression for the number of matchsticks needed for each pattern.
How many matchsticks are needed for the 10th pattern?
Algebraic Patterns: Practice Questions
- First differences: \(7-3=4,\; 11-7=4,\; 15-11=4\).
- Next two terms: \(19,\; 23\).
- Rule (recursive): \(u_{n+1}=u_n+4\) with \(u_1=3\).
- Rule (explicit): \(u_n=4n-1\).
- First differences: \(3,5,7,9\) (increase by \(2\)).
- Continue differences: \(11,13\) ⇒ next terms \(37, 50\).
- Rule (explicit): \(u_n=n^2+1\).
- Alternative: “add successive odd numbers starting from \(3\)”.
- Understand that solving an equation means finding all values that satisfy it.
- Solve linear equations, including those containing fractions.
- Verify proposed solutions by substitution into the original equation.
- Recognise equations with no solution or an infinite number of solutions.
Solve the Linear Equation
Solve the linear equation \(\dfrac{2t-3}{5}+\dfrac{1}{20}=\dfrac{t-1}{4}.\)
Solving Equations: Practice Question
Multiply by 30: \(10x+12=15x-3\).
Rearrange: \(12+3=15x-10x\Rightarrow 15=5x\Rightarrow \boxed{x=3}\).
Rearrange: \(12+3=15x-10x\Rightarrow 15=5x\Rightarrow \boxed{x=3}\).
- Solve a system of two linear equations in two variables algebraically (by substitution or elimination).
- Interpret the solution graphically as the intersection point of two lines.
- Apply simultaneous equations to contextual problems (mixtures, ticket sales, age, motion, etc.).
- Solve systems involving three unknowns using elimination or substitution.
Simultaneous Linear Equations
Solve the equations \(3x-y=1\) and \(x-2y=-8.\)
Simultaneous Linear Equations
Solve the equations \(2x - 5y = 9\) and \(3x + 2y = 4.\)
Simultaneous Equations with Three Variables
Solve the simultaneous equations:
A: \(x + y + z = 6\)
B: \(2x + y - z = 1\)
C: \(4x - 3y + 2z = 4\).
Solving a System of Three Unknowns - Elimination Method
Solve the following system:
\[
\begin{cases}
2x + y - z = 9 \quad &(A)\\[2pt]
x + 2y + z = 6 \quad &(B)\\[2pt]
3x - y + 2z = 17 \quad &(C)
\end{cases}
\]
Step 1 — Eliminate \(z\)
First, eliminate \(z\) using (A) and (B), then using \((C) + 2(A)\).
\[ \text{(A) + (B):} \begin{array}{rcrcrcrl} 2x &+& y &-& z &=& 9 \\[3pt] +~x &+& 2y &+& z &=& 6 \\[3pt] \hline 3x &+& 3y& & &=& 15 \end{array} \Rightarrow x+y=5\quad\text{(D)} \] \[ \text{(C) + 2(A):} \begin{array}{rcrcrcrl} 3x &-& y &+& 2z &=& 17 \\[3pt] +~4x &+& 2y &-& 2z &=& 18 \\[3pt] \hline 7x &+& y & & &=& 35 \end{array} \Rightarrow 7x+y=35\quad\text{(E)} \]Step 2 — Eliminate \(y\)
Subtract (D) from (E).
\[ \begin{array}{rcrcrl} 7x &+& y &=& 35 \\[3pt] -~x &-& y &=& -5 \\[3pt] \hline 6x & & &=& 30 \end{array} \Rightarrow x=5 \]Step 3 — Find \(y\)
\[
5+y=5 \;\Rightarrow\; y=0
\]
Step 4 — Find \(z\)
\[
\begin{array}{rcrcrl}
5 &+& 2(0) &+& z &= 6\\[3pt]
5 &+& 0 &+& z &= 6\\[3pt]
\hline
& & & & z &= 1
\end{array}
\]
✅ Final Answer
\[
(x,y,z)=(5,0,1)
\]
Problem Solving with Simultaneous Equations
An opera was attended by 240 people. Two ticket prices, €31 and €16, were available.
If the total takings on the night were €5595, find, using this data:
(i) two linear equations connecting the two types of tickets sold
(ii) the number of €31 tickets sold
(iii) the number of €16 tickets sold.
Problem Solving with Simultaneous Equations — Coins
Fifty, twenty and ten cent coins are collected from a coin machine and counted.
The total value of the coins is €32. When counting, the cashier noted that twice
the number of twenty cent coins, added to the number of ten cent coins, equalled
three times the number of fifty cent coins. She then noticed that four times the
number of fifty cent coins, added to the number of ten cent coins, equalled six
times the number of twenty cent coins.
Find the number of each type of coin in the machine.
Simultaneous Equations: Practice Questions
From \(2x-y=1\Rightarrow y=2x-1\). Substitute into \(3x+2y=12\):
\(3x+2(2x-1)=12\Rightarrow 3x+4x-2=12\Rightarrow 7x=14\Rightarrow x=2\).
Then \(y=2(2)-1=3\). So \(\boxed{(x,y)=(2,3)}\).
\(3x+2(2x-1)=12\Rightarrow 3x+4x-2=12\Rightarrow 7x=14\Rightarrow x=2\).
Then \(y=2(2)-1=3\). So \(\boxed{(x,y)=(2,3)}\).
Subtract first from second: \(x-2y= -1\). Subtract first from third: \(y-2z= -2\Rightarrow y=2z-2\).
Then \(x= -1+2y= -1+2(2z-2)=4z-5\). Substitute in \(x+y+z=6\):
\((4z-5)+(2z-2)+z=6\Rightarrow 7z-7=6\Rightarrow z=\tfrac{13}{7}\).
\(y=2z-2=\tfrac{26}{7}-\tfrac{14}{7}=\tfrac{12}{7}\).
\(x=4z-5=\tfrac{52}{7}-\tfrac{35}{7}=\tfrac{17}{7}\).
Hence \(\boxed{\left(x,y,z\right)=\left(\tfrac{17}{7},\,\tfrac{12}{7},\,\tfrac{13}{7}\right)}\).
Then \(x= -1+2y= -1+2(2z-2)=4z-5\). Substitute in \(x+y+z=6\):
\((4z-5)+(2z-2)+z=6\Rightarrow 7z-7=6\Rightarrow z=\tfrac{13}{7}\).
\(y=2z-2=\tfrac{26}{7}-\tfrac{14}{7}=\tfrac{12}{7}\).
\(x=4z-5=\tfrac{52}{7}-\tfrac{35}{7}=\tfrac{17}{7}\).
Hence \(\boxed{\left(x,y,z\right)=\left(\tfrac{17}{7},\,\tfrac{12}{7},\,\tfrac{13}{7}\right)}\).
