All categories

3 years ago

In the system of equations below, I want to eliminate the variable y.

Mathematics

1 answer:

Mekhanik [1.2K]3 years ago

5 0

Answer:

(x,y)=(-1,1)

Step-by-step explanation:

Given that,

3x + 5y = 2 .....(1)

6x - 10y = -16 ....(2)

Finding the value of x from equation (1).

$x=\dfrac{2-5y}{3}$

Put the value of x in equation (2)

$6(\dfrac{2-5y}{3}) - 10y = -16\\\\2(2-5y)-10y=-16\\\\4-10y-10y=-16\\\\-20y=-16-4\\\\y=1$

Put the value of y in equation (1) :

$3x + 5(1) = 2\\\\3x=2-5\\\\3x=-3\\\\x=-1$

So, the solution of the equations is (-1,1).

You might be interested in

What would X is less than or equal to negative 7 look like as a graph

Anastaziya [24]

Answer:

Place a dot on 7 and do < underlined

8 0

3 years ago

2. Find the circumference of a circle if the area is 300 square centimeters. A. 84.74 cm B. 9.77 cm C. 61.38 cm D. 56.25 cm

Vadim26 [7]

The circumference of the circle is C. 61.38 cm

3 0

3 years ago

How many times does 9 go into 430?

sergejj [24]

9 goes into 430 47.777.... times repeateded. so 47.78

6 0

3 years ago

Write the equation in slope-intercept form. Do not use any spaces. Make sure your answer looks like y=mx+b. y+3=2(x+7) *

seraphim [82]

Answer:

y=2x+11

Step-by-step explanation:

You first have to distribute the 2 to (x+7) to get y+3=2x+14. after that all you have to do is subtract both sides of the equation by 3 to get y=2x+11

6 0

3 years ago

Using Rolle's theorem prove that the function has at most one root on the given interval <img src="https://tex.z-dn.net/?f=f%28x

Sophie [7]

Answer:

Show that if $f(a) = f(b) = 0$ for some $a,\, b \in [-2,\, 2]$ where $a \ne b$ , then by Rolle's Theorem $f^{\prime}(x) = 0$ for some $x \in (-2,\, 2)$ . However, no such $x$ exists since $f^{\prime}(x) < 0$ for all $x \in (-2,\, 2)\!$ .

Note that Rolle's Theorem alone does not give the exact value of the root. Neither does this theorem guarantee that a root exists in this interval.

Step-by-step explanation:

The function $f(x) = x^{3} - 12\, x + 11$ is continuous and differentiable over $[-2,\, 2]$ . By Rolle's Theorem. if $f(a) = f(b)$ for some $a,\, b \in [2,\, -2]$ where $a \ne b$ , then there would exist $x \in (a,\, b)$ such that $f^{\prime}(x) = 0$ .

Assume by contradiction $f(x)$ does have more than one roots over $[-2,\, 2]$ . Let $a$ and $b$ be (two of the) roots, such that $a \ne b$ . Notice that $f(a) = 0 = f(b)$ just as Rolle's Theorem requires. Thus- by Rolle's Theorem- there would exist $x \in (a,\, b)$ such that $f^{\prime}(x) = 0$ .

However, no such $x \in (a,\, b)$ could exist. Notice that $f^{\prime}(x) = 3\, x^{2} - 12$ , which is a parabola opening upwards. The only zeros of $f^{\prime}(x)$ are $x = (-2)$ and $x = 2$ .

However, neither $x = (-2)$ nor $x = 2$ are included in the open interval $(-2,\, 2)$ . Additionally, $a,\, b \in [-2,\, 2]$ , meaning that $(a,\, b)$ is a subset of the open interval $(-2,\, 2)$ . Thus, neither zero would be in the subset $(a,\, b)$ . In other words, there is no $x \in (a,\, b)$ such that $f^{\prime}(x) = 0$ . Contradiction.

Hence, $f(x) = x^{3} - 12\, x + 11$ has at most one root over the interval $[-2,\, 2]$ .

8 0

2 years ago