What is the solution to 3x-y=4 and 2y+x=11

Answers

Answer 1

Answer: 3x - 1y =   4 ⇒ 9x - 3y = 12
1x + 3y = 11 ⇒ 1x + 3y = 11
10x = 23
10   10
x = 2.3
  3x - y = 4
3(2.3) - y = 4
6.9 - y = 4
- 6.9   - 6.9
-y = -2.9
-1 -1
y = 2.9
  (x, y) = (2.3, 6.9)

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If the point (7, 3) is on the graph of an equation, which statement must be true?A. The values x = 3 and y = 7 make the equation true.
B. The values x = 7 and y = 3 are the only values that make the equation true.
C. There are solutions to the equation for the values x = 7 and x = 3.
D. The values x = 7 and y = 3 make the equation true.

Answers

Answer:

The value of x=7 and y=3 make the equation true.

D is correct.

Step-by-step explanation:

If the point (7, 3) is on the graph of an equation.

If point lie on any graph then the value of x and y satisfy the equation. It means equation would be true for the given passing point.

First value of point is x and second value is y

For given point: (7,3)

x=7 and y=3

If we put x=7 and y=3 in the graph of the equation then equation must be true.

because it makes equation true.

For true equation left side must be equal to right side.

Hence, The value of x=7 and y=3 make the equation true.

The best answer would be D. because:
1) x = 7 and y = 3 in that specific coordinate
2) if the point is on the graph and is labeled it is 1 of the MANY solutions that make that specific equation true

There is a chance for B. but if its infinate then (7,3) couldn't be the ONLY coordinate that makes it true

Solve for x: 2 over 3 (x - 4) = 2x.

Answers

2/3 (x - 4 ) = 2x

Simplify both sides of the equation.

2/3(x−4)=2x

Simplify:

2/3(x−4)=2x

(2/3)(x)+(2/3)(−4)=2x(Distribute)

2/3x+−8/3=2x

Subtract 2x from both sides.

2/3x+−8/3−2x=2x−2x

−4/3x+−8/3=0

Add 8/3 to both sides.

−4/3x+−8/3+8/3=0+8/3

−4/3x=8/3

Multiply both sides by 3/(-4).

(3/−4)*(−4/3x)=(3/−4)*(8/3)

x=−2

The correct answer for the value of for the expression $(2)/(3)(x - 4) = 2x$ is

-2.

To solve for x in the equation:

$(2)/(3)(x - 4) = 2x$

First, distribute the $(2)/(3)$ to the terms inside the parentheses:

$(2)/(3) x - (2)/(3)*4 = 2x$

$(2)/(3) x - (8)/(3) = 2x$

Next, Isolate the terms with x on one side of the equation.

Subtract $(2)/(3)x$ term from both side of the equation:

$(2)/(3) x - (2)/(3) x-(8)/(3) = 2x-(2)/(3) x$

$(-8)/(3) = (4)/(3) x$

Now, Solve for x by multiplying both sides of the equation by the reciprocal of $(4)/(3)$ , which is $(3)/(4)$ :

$(-8)/(3) * (3)/(4)= (4x)/(3) * (3)/(4)$

$x= (-8)/(4)$

Simplify further:

$x = -2$

The solution of the equation $(2)/(3)(x - 4) = 2x$ is $x = -2$ .

Learn more about Simplification here:

brainly.com/question/24140794

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What is 1/6 + (2/3 - 1/4) in simplest form

Answers

$(1)/(6)+((2)/(3)-(1)/(4))=\ \ \ \ | make\ a\ common\ denominator\n\n(2)/(12)+(8)/(12)-(3)/(12)=(2+8-3)/(12)=(7)/(12)\n\n The\ simplest\ form \ is\ (7)/(12).$

Which one is it , help me on this please please please please please please please

Answers

Answer:

Step-by-step explanation:

7^-2 = 1/7^2

The negative means you flip it

1/7^2 times 7^6 is 7^6/7^2 (just multiply).

using rules of exponents, 7^6/7^2 is 7^4 (just subtract 2 from 6)

D is the only one that equals 7^4, as 2 times 2 is 4.

Which of the following is not the same set of numbers?natural numbers
whole numbers
counting numbers

Answers

it is natural numbers

Find the second derivative for y=(x+2)/(x-3)

Answers

Let's find the first derivative.

Using quotient rule

$f(x)=(g(x))/(h(x))$

$f'(x)=(g'(x)*h(x)-g(x)*h'(x))/(h^2(x))$

then:

$y=(x+2)/(x-3)$
$f(x)=y$

$g(x)=x+2$

$g'(x)=1$

$h(x)=x-3$

$h'(x)=1$

Let's replace

$f'(x)=(g'(x)*h(x)-g(x)*h'(x))/(h^2(x))$

$y'=(1*(x-3)-(x+2)*1)/((x-3)^2)$

$y'=(x-3-x-2)/((x-3)^2)$

$\boxed{y'=(-5)/((x-3)^2)}$

the second derivative is the derivative of our first derivative.

$y'=(-5)/((x-3)^2)$

We can write this function as

$y'=-5*(x-3)^(-2)$

now we have to use the Chain rule's

$f(u)=-5u^(-2)$

and

$g(x)=x-3$

$f[g(x)]=-5*(x-3)^(-2)$

then

$f'[g(x)]=f'(u)*g'(x)$

$f'(u)=-5*(-2)*u^(-3)=10*u^(-3)$

$g'(x)=1$

$f'[g(x)]=f'(u)*g'(x)$

Let's replace

$f'[g(x)]=10*u^(-3)*1$

Let's change u by $(x-3)$

$f'[g(x)]=10*u^(-3)$

$f'[g(x)]=10*(x-3)^(-3)$

$\boxed{\boxed{y''=(10)/((x-3)^(3))}}$

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