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3 years ago

How do you solve 15-2x=3x

Mathematics

2 answers:

sergeinik [125]3 years ago

8 0

HEY There,

15-2x=3x

bring 2x to the right hand side

15 = 3x + 2x

15 = 5x

then, bring 5 to the denominator

15/5 = x

3 = x

HOPE YOU UNDERSTAND!!!!!!!!!!!!!

bekas [8.4K]3 years ago

6 0

3) 15 - 2x = 3x

Add 2x to both sides so only one side contains the variable x.

15 = 5x

Divide both sides by 5.

x = 3; if you want to check your solution then plug x back into the original equation.

15 - 2(3) = 3(3)

15 - 6 = 9 ⇒ 9 = 9 so this would be true, your answer is correct.

____________________________________________________________

15) 10(2y + 2) - y = 2(8y - 8)

Distribute 10 and 2 inside their respective parentheses.

(20y + 20) - y = (16y - 16)

Combine like terms on the left side of the equation.

19y + 20 = 16y - 16

Subtract 16y from both sides.

3y + 20 = -16

Subtract 20 from both sides.

3y = -36

Divide both sides by 3.

y = -12

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Find two vectors in R2 with Euclidian Norm 1<br> whoseEuclidian inner product with (3,1) is zero.

alina1380 [7]

Answer:

$v_1=(\frac{1}{10},-\frac{3}{10})$

$v_2=(-\frac{1}{10},\frac{3}{10})$

Step-by-step explanation:

First we define two generic vectors in our $\mathbb{R}^2$ space:

$v_1 = (x_1,y_1)$
$v_2 = (x_2,y_2)$

By definition we know that Euclidean norm on an 2-dimensional Euclidean space $\mathbb{R}^2$ is:

$\left \| v \right \|= \sqrt{x^2+y^2}$

Also we know that the inner product in $\mathbb{R}^2$ space is defined as:

$v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2$

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

$\left \| v_1 \right \|= \sqrt{x^2+y^2}=1$

and

$\left \| v_2 \right \|= \sqrt{x^2+y^2}=1$

As second condition we have that:

$v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0$

$v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0$

Which is the same:

$y_1=-3x_1\\y_2=-3x_2$

Replacing the second condition on the first condition we have:

$\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}$

Since $x_1^2= \frac{1}{10}$ we have two posible solutions, $x_1=\frac{1}{10}$ or $x_1=-\frac{1}{10}$ . If we choose $x_1=\frac{1}{10}$ , we can choose next the other solution for $x_2$ .

Remembering,

$y_1=-3x_1\\y_2=-3x_2$

The two vectors we are looking for are:

$v_1=(\frac{1}{10},-\frac{3}{10})\\v_2=(-\frac{1}{10},\frac{3}{10})$

5 0

3 years ago