Question

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

264

likes
1319 views

## Answer to a math question 4+168×10³×d1+36×10³×d2=-12 -10+36×10³×d1+72×10³×d2=0

4.7
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

Frequently asked questions $FAQs$
What is the equation of a hyperbola with center at $3, 2$, vertical transverse axis of length 8, and eccentricity 5/4?
+
What is the length of the altitude of an equilateral triangle whose side measures 10 units?
+
What is the scalar product of vector v with magnitude 3.5 and vector w with magnitude 2.8, given that the angle between them is 35 degrees?
+