All categories

3 years ago

Find the solution to the equation 3x + 1 = 6x + 1

Mathematics

2 answers:

I am Lyosha [343]3 years ago

5 0

X= 0 there’s no other way that equation would work

Musya8 [376]3 years ago

3 0

Answer:

x=0

Step-by-step explanation:

First, start by getting rid of your X values. You must take your lowest X value and find a way to move it to the other side. Our lowest X value is 3x, and it's currently being added to 1.

In order to move it, we must subtract it on both sides.

3x + 1 = 6x + 1

-3x -3x

1 = 3x + 1

Now we must get the X value by itself, which means doing everything reverse. For example, you would subtract the 1 instead of adding it.

1 = 3x + 1

-1 -1

3x = 0

divide by 3 on both sides

x = 0

Let's check

3(0) + 1 = 6(0) + 1

1 = 1

This is a true statement

You might be interested in

Find the particular solution of the differential equation that satisfies the initial condition(s). f ''(x) = x−3/2, f '(4) = 1,

sweet [91]

Answer:

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

Step-by-step explanation:

This differential equation has separable variable and can be solved by integration. First derivative is now obtained:

$f'' = x - \frac{3}{2}$

$f' = \int {\left(x-\frac{3}{2}\right) } \, dx$

$f' = \int {x} \, dx -\frac{3}{2}\int \, dx$

$f' = \frac{1}{2}\cdot x^{2} - \frac{3}{2}\cdot x + C$ , where C is the integration constant.

The integration constant can be found by using the initial condition for the first derivative ( $f'(4) = 1$ ):

$1 = \frac{1}{2}\cdot 4^{2} - \frac{3}{2}\cdot (4) + C$

$C = 1 - \frac{1}{2}\cdot 4^{2} + \frac{3}{2}\cdot (4)$

$C = -1$

The first derivative is $y' = \frac{1}{2}\cdot x^{2}- \frac{3}{2}\cdot x - 1$ , and the particular solution is found by integrating one more time and using the initial condition ( $f(0) = 0$ ):

$y = \int {\left(\frac{1}{2}\cdot x^{2}-\frac{3}{2}\cdot x -1 \right)} \, dx$

$y = \frac{1}{2}\int {x^{2}} \, dx - \frac{3}{2}\int {x} \, dx - \int \, dx$

$y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x + C$

$C = 0 - \frac{1}{6}\cdot 0^{3} + \frac{3}{4}\cdot 0^{2} + 0$

$C = 0$

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

5 0

4 years ago

Solve for the variable: x + 9 > -4

Marina CMI [18]

$Hi \\ \\ x+9\ \textgreater \ -4 \\ \\ x\ \textgreater \ -4-9 \\ \\ \boxed{x\ \textgreater \ -13} \\ \\ S=\{x\in~\mathbb{R} |x\ \textgreater \ -13\} \\ \\ \\$