Solve the simultaneous equations 4a + 12b = 16 and 2a + 3b = 7, and find the positive value of K if (a+b) = K2.

Given equations:

$$4a + 12b = 16 \quad (1)$$ $$2a + 3b = 7 \quad (2)$$

Multiply equation (2) by 2 to eliminate $a$:

$$2(2a + 3b) = 2(7)$$ $$4a + 6b = 14 \quad (3)$$

Subtract equation (3) from equation (1):

$$(4a + 12b) - (4a + 6b) = 16 - 14$$ $$6b = 2$$ $$b = \frac{2}{6}$$ $$b = \frac{1}{3}$$

$$2a + 3\left(\frac{1}{3}\right) = 7$$ $$2a + 1 = 7$$ $$2a = 7 - 1$$ $$2a = 6$$ $$a = \frac{6}{2}$$ $$a = 3$$

Given that $(a+b) = K^2$. Substitute the values of $a$ and $b$:

$$\left(3 + \frac{1}{3}\right) = K^2$$ $$\left(\frac{9}{3} + \frac{1}{3}\right) = K^2$$ $$\frac{10}{3} = K^2$$

Take the square root of both sides. Since we need the positive value of K:

$$K = \sqrt{\frac{10}{3}}$$

Rationalize the denominator:

$$K = \frac{\sqrt{10}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$$ $$K = \frac{\sqrt{30}}{3}$$

The positive value of K is $\frac{\sqrt{30}}{3}$.