Compute some scalars given to make given equations hold

Find $x,y \in \mathbb{R}$ such that $x(\mathbf{i} - \mathbf{j}) + y(\mathbf{i} + mathbf{j})$ is equal to

$\mathbf{i}$ ;
$\mathbf{j}$ ;
$3 \mathbf{i} - 5 \mathbf{j}$ ;
$7 \mathbf{i} + 5 \mathbf{j}$ ,

where $\mathbf{i}$ and $\mathbf{j}$ denote the unit coordinate vectors $(1,0)$ and $(0,1)$ , respectively.

We can use Theorem 12.9 (page 465 of Apostol) to compute the scalars. In this case we have $X = (1,0)$ corresponding to $\mathbf{i}$ , $A_1 = (1,-1)$ corresponding to $\mathbf{i} - \mathbf{j}$ and $A_2 = (1,1)$ corresponding to $\mathbf{i} + \mathbf{j}$ .
$\begin{align*} x &= \frac{(1,0) \cdot (1,-1)}{(1,-1) \cdot (1,-1)} = \frac{1}{2} \\ y &= \frac{(1,0) \cdot (1,1)}{(1,1) \cdot (1,1)} = \frac{1}{2}. \end{align*}$

Therefore, $x = y = \frac{1}{2}$ .
We can use Theorem 12.9 (page 465 of Apostol) to compute the scalars. In this case we have $X = (0,1)$ corresponding to $\mathbf{j}$ , $A_1 = (1,-1)$ corresponding to $\mathbf{i} - \mathbf{j}$ and $A_2 = (1,1)$ corresponding to $\mathbf{i} + \mathbf{j}$ .
$\begin{align*} x &= \frac{(0,1) \cdot (1,-1)}{(1,-1) \cdot (1,-1)} = -\frac{1}{2} \\ y &= \frac{(0,1) \cdot (1,1)}{(1,1) \cdot (1,1)} = \frac{1}{2}. \end{align*}$

Therefore, $x = -\frac{1}{2}$ and $y = \frac{1}{2}$ .
We can use Theorem 12.9 (page 465 of Apostol) to compute the scalars. In this case we have $X = (3,-5)$ corresponding to $3\mathbf{i}- 5 \mathbf{j}$ , $A_1 = (1,-1)$ corresponding to $\mathbf{i} - \mathbf{j}$ and $A_2 = (1,1)$ corresponding to $\mathbf{i} + \mathbf{j}$ .
$\begin{align*} x &= \frac{(3,-5) \cdot (1,-1)}{(1,-1) \cdot (1,-1)} = 4 \\ y &= \frac{(3,-5) \cdot (1,1)}{(1,1) \cdot (1,1)} = -1. \end{align*}$

Therefore, $x = 4$ and $y = -1$ .
We can use Theorem 12.9 (page 465 of Apostol) to compute the scalars. In this case we have $X = (7,5)$ corresponding to $7\mathbf{i} + 5 \mathbf{j}$ , $A_1 = (1,-1)$ corresponding to $\mathbf{i} - \mathbf{j}$ and $A_2 = (1,1)$ corresponding to $\mathbf{i} + \mathbf{j}$ .
$\begin{align*} x &= \frac{(7,5) \cdot (1,-1)}{(1,-1) \cdot (1,-1)} = 1 \\ y &= \frac{(7,5) \cdot (1,1)}{(1,1) \cdot (1,1)} = 6. \end{align*}$

Therefore, $x = 1$ and $y = 6$ .