Given vectors $a = (1, 3)$ and $b = (4, -2)$, we need to express the given vectors $c$ and $d$ in the form $sa + tb$. (1) $c = (7, 7)$ (2) $d = (10, -12)$

1. Problem Description

Given vectors $a = (1, 3)$ and $b = (4, -2)$, we need to express the given vectors $c$ and $d$ in the form $sa + tb$.

(1) $c = (7, 7)$

(2) $d = (10, -12)$

2. Solution Steps

(1) We want to find scalars $s$ and $t$ such that $c = sa + tb$.

This means $(7, 7) = s(1, 3) + t(4, -2) = (s + 4t, 3s - 2t)$.

Equating the components, we have the following system of equations:

$s + 4t = 7$

$3s - 2t = 7$

Multiply the first equation by 2:

$2s + 8t = 14$

Multiply the second equation by 4:

$12s - 8t = 28$

Adding the two modified equations, we get:

$14s = 42$

$s = 3$

Substitute $s = 3$ into the first equation:

$3 + 4t = 7$

$4t = 4$

$t = 1$

Therefore, $c = 3a + b$.

(2) We want to find scalars $s$ and $t$ such that $d = sa + tb$.

This means $(10, -12) = s(1, 3) + t(4, -2) = (s + 4t, 3s - 2t)$.

Equating the components, we have the following system of equations:

$s + 4t = 10$

$3s - 2t = -12$

Multiply the first equation by 2:

$2s + 8t = 20$

Multiply the second equation by 4:

$12s - 8t = -48$

Adding the two modified equations, we get:

$14s = -28$

$s = -2$

Substitute $s = -2$ into the first equation:

$-2 + 4t = 10$

$4t = 12$

$t = 3$

Therefore, $d = -2a + 3b$.

3. Final Answer

(1) $c = 3a + b$

(2) $d = -2a + 3b$