4+168×10³×d1+36×10³×d2=-12 -10+36×10³×d1+72×10³×d2=0

To solve the system of equations:

\begin{align*}4+168\times 10^3\times d_1+36\times 10^3\times d_2 & = -12 \quad (1) \-10+36\times 10^3\times d_1+72\times 10^3\times d_2 & = 0 \quad (2)\end{align*}

We can use the method of substitution to eliminate one variable and solve for the other.

Let's solve equation (2) for d_1:

-10 + 36\times 10^3 \times d_1 + 72\times 10^3\times d_2 = 0

Rearranging the equation, we have:

36\times 10^3 \times d_1 = 10 + 72\times 10^3\times d_2

Dividing both sides by 36\times 10^3, we get:

d_1 = \frac{10 + 72\times 10^3\times d_2}{36\times 10^3} \quad (3)

Substituting equation (3) into equation (1), we have:

4 + 168\times 10^3\times \left(\frac{10 + 72\times 10^3\times d_2}{36\times 10^3}\right) + 36\times 10^3\times d_2 = -12

Simplifying the equation, we get:

4 + \frac{168}{36}\times (10 + 72\times 10^3\times d_2) + 36\times 10^3\times d_2 = -12

Simplifying further, we have:

4 + \frac{4}{3}\times (10 + 72\times 10^3\times d_2) + 36\times 10^3\times d_2 = -12

Expanding and simplifying, we get:

4 + \frac{40}{3} + \frac{288}{3}\times 10^3\times d_2 + 36\times 10^3\times d_2 = -12

Combining like terms, we have:

\frac{128}{3}\times 10^3\times d_2 + 36\times 10^3\times d_2 = -12 - \frac{40}{3}

Simplifying further, we get:

\frac{164}{3}\times 10^3\times d_2 = -\frac{76}{3}

Dividing both sides by \frac{164}{3}\times 10^3, we have:

d_2 = -\frac{76}{3} \div \frac{164}{3}\times 10^3

Simplifying the division, we get:

d_2 = -\frac{76}{164}\times 10^3

Simplifying the fraction, we have:

d_2 = -\frac{19}{41}\times 10^3

Finally, we can substitute the value of d_2 into equation (3) to solve for d_1:

d_1 = \frac{10 + 72\times 10^3\times \left(-\frac{19}{41}\times 10^3\right)}{36\times 10^3}

Simplifying the equation, we get:

d_1 = \frac{10 - \frac{72\times 19}{41}}{36}

Simplifying the fraction, we have:

d_1 = \frac{10 - \frac{1368}{41}}{36}

Calculating the numerator and denominator separately, we have:

d_1 = \frac{410 - 1368}{36} = \frac{-958}{36} = -\frac{479}{18} \quad (4)

Therefore, the solution to the system of equations is d_1 = -\frac{479}{18} and d_2 = -\frac{19}{41}\times 10^3.

\textbf{Answer:} d_1 = -\frac{479}{18}, d_2 = -\frac{19}{41}\times 10^3