Solve for x and y
(i) `[{:(,2,5),(,5,2):}] [{:(,x),(,y):}]=[{:(,-7),(,14):}]`
(ii) `[x+y, x-4] [{:(,-1,-2),(,2,2):}]=[-7, -11]`
(iii) `[{:(,-2,0),(,3,1):}] [{:(,-1),(,2x):}] +3[{:(,-2),(,1):}]=2 [{:(,y),(,3):}]`.

Answer

Let's solve the given equations step by step.

### Part (i)

Given the equation:
\[ \begin{pmatrix} 2 & 5 \\ 5 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -7 \\ 14 \end{pmatrix} \]

1. **Multiply the matrices**:
- The first row gives us: \(2x + 5y = -7\) (Equation 1)
- The second row gives us: \(5x + 2y = 14\) (Equation 2)

2. **Now we have the system of equations**:
- \(2x + 5y = -7\) (1)
- \(5x + 2y = 14\) (2)

3. **Multiply Equation 1 by 2**:
\[ 4x + 10y = -14 \quad \text{(Equation 3)} \]

4. **Multiply Equation 2 by 5**:
\[ 25x + 10y = 70 \quad \text{(Equation 4)} \]

5. **Subtract Equation 3 from Equation 4**:
\[ (25x + 10y) - (4x + 10y) = 70 - (-14) \]
\[ 21x = 84 \]
\[ x = 4 \]

6. **Substitute \(x = 4\) into Equation 1**:
\[ 2(4) + 5y = -7 \]
\[ 8 + 5y = -7 \]
\[ 5y = -15 \]
\[ y = -3 \]

**Final answer for part (i)**: \(x = 4\), \(y = -3\)

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### Part (ii)

Given the equation:
\[ \begin{pmatrix} x+y & x-4 \end{pmatrix} \begin{pmatrix} -1 & -2 \\ 2 & 2 \end{pmatrix} = \begin{pmatrix} -7 & -11 \end{pmatrix} \]

1. **Multiply the matrices**:
- First element: \((x+y)(-1) + (x-4)(2) = -7\)
- Second element: \((x+y)(-2) + (x-4)(2) = -11\)

2. **Set up the equations**:
- From the first element:
\[ -x - y + 2x - 8 = -7 \]
\[ x - y - 8 = -7 \quad \Rightarrow \quad x - y = 1 \quad \text{(Equation 5)} \]

- From the second element:
\[ -2x - 2y + 2x - 8 = -11 \]
\[ -2y - 8 = -11 \quad \Rightarrow \quad -2y = -3 \quad \Rightarrow \quad y = \frac{3}{2} \]

3. **Substitute \(y = \frac{3}{2}\) into Equation 5**:
\[ x - \frac{3}{2} = 1 \]
\[ x = 1 + \frac{3}{2} = \frac{5}{2} \]

**Final answer for part (ii)**: \(x = \frac{5}{2}\), \(y = \frac{3}{2}\)

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### Part (iii)

Given the equation:
\[ \begin{pmatrix} -2 & 0 \\ 3 & 1 \end{pmatrix} \begin{pmatrix} -1 \\ 2x \end{pmatrix} + 3 \begin{pmatrix} -2 \\ 1 \end{pmatrix} = 2 \begin{pmatrix} y \\ 3 \end{pmatrix} \]

1. **Multiply the matrices**:
- First term:
\[ -2(-1) + 0(2x) = 2 \]
\[ 3(-1) + 1(2x) = -3 + 2x \]

2. **Add the scalar multiplication**:
\[ 2 + 3(-2) = 2 - 6 = -4 \]
\[ -3 + 2x + 3(1) = -3 + 2x + 3 = 2x \]

3. **Set up the equations**:
- From the first term:
\[ 2 = 2y \quad \Rightarrow \quad y = 1 \]

- From the second term:
\[ 2x = 6 \quad \Rightarrow \quad x = 3 \]

**Final answer for part (iii)**: \(x = 3\), \(y = 1\)

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