All categories

2 years ago

-6x-14y=16 and -2x+7y=17 solve by elimination show the work

Mathematics

1 answer:

professor190 [17]2 years ago

6 0

Answer:

(- 5, 1 )

Step-by-step explanation:

- 6x - 14y = 16 → (1)

- 2x + 7y = 17 → (2)

Multiplying (2) by - 3 and adding to (1) will eliminate the x- term

6x - 21y = - 51 → (3)

Add (1) and (3) term by term to eliminate x

0 - 35y = - 35

- 35y = - 35 ( divide both sides by - 35 )

y = 1

Substitute y = 1 into either of the 2 equations and solve for x

Substituting into (1)

- 6x - 14(1) = 16

- 6x - 14 = 16 ( add 14 to both sides )

- 6x = 30 ( divide both sides by - 6 )

x = - 5

solution is (- 5, 1 )

You might be interested in

Select the correct answer.

FinnZ [79.3K]

Answer:

ghytmujtjy46hgqmwneymymwrj25rhq

5 0

2 years ago

What is the simplified form of the following expression?a. (34)3b. 7[62 ÷ (52 − 42) + 8]

Natasha_Volkova [10]

Answer:

a.) 102 b.) 99.4

Step-by-step explanation:

a. 34 x 3 = 102

b. 7[62 ÷ (52 − 42) + 8] = $\frac{497}{5}$ or $99\frac{2}{5}$ or 99.4

6 0

2 years ago

Read 2 more answers

Find the area of the SHADED region.<br> Round to the nearest TENTH if necessary.

lisov135 [29]

The area of the shaded region is 49.2.

6 0

2 years ago

Find an exact value.

Westkost [7]

Answer:

$\displaystyle \cos\left(-\frac{7\,\pi}{12}\right) = \frac{\sqrt{2} - \sqrt{6}}{4}$ .

Step-by-step explanation:

Convert the angle $\displaystyle \left(-\frac{7\, \pi}{12}\right)$ to degrees:

$\displaystyle \left(-\frac{7\, \pi}{12}\right) = \left(-\frac{7\, \pi}{12}\right) \times \frac{180^\circ}{\pi} = -105^\circ$ .

Note, that $\left(-105^\circ\right)$ is the sum of two common angles: $\left(-45^\circ\right)$ and $\left(-60^\circ\right)$ .

$\displaystyle \cos\left(-45^\circ\right) = \cos\left(45^\circ\right) = \frac{\sqrt{2}}{2}$ .
$\displaystyle \cos\left(-60^\circ\right) = \cos\left(60^\circ\right) = \frac{1}{2}$ .

$\displaystyle \sin\left(-45^\circ\right) = -\sin\left(45^\circ\right) = -\frac{\sqrt{2}}{2}$ .
$\displaystyle \sin\left(-60^\circ\right) = -\sin\left(60^\circ\right) = -\frac{\sqrt{3}}{2}$ .

By the sum-angle identity of cosine:

$\cos(A + B) = \cos(A)\cdot \cos(B) - \sin(A) \cdot \sin(B)$ .

Apply the sum formula for cosine to find the exact value of $\cos\left(-105^\circ \right)$ .

$\begin{aligned}\cos\left(-105^\circ \right) &= \cos\left(\left(-45^\circ\right) + \left(-60^\circ\right)\right) \\ &= \cos\left(-45^\circ\right) \cdot \cos\left(-60^\circ\right)\right) - \sin\left(-45^\circ\right) \cdot \sin\left(-60^\circ\right)\right) \\ &= \frac{\sqrt{2}}{2} \times \frac{1}{2} - \left(-\frac{\sqrt{2}}{2}\right)\times \left(-\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{2} - \sqrt{6}}{4}\end{aligned}$ .

$\displaystyle \left(-\frac{7\, \pi}{12}\right) = \left(-\frac{7\, \pi}{12}\right) \times \frac{180^\circ}{\pi} = -105^\circ$ . In other words, $\displaystyle \left(-\frac{7\, \pi}{12}\right)$ and $\left(-105^\circ\right)$ correspond to the same angle. Therefore, the cosine of $\displaystyle \left(-\frac{7\, \pi}{12}\right)\!$ would be equal to the cosine of $\left(-105^\circ\right)\!$ .