3(x-2)=18 I need help like asap

Mathematics

2 answers:

Naily [24]4 years ago

6 0

Answer: x=8

Step-by-step explanation:

Simplifying

3(x + -2) = 18

Reorder the terms:

3(-2 + x) = 18

(-2 * 3 + x * 3) = 18

(-6 + 3x) = 18

Solving

-6 + 3x = 18

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '6' to each side of the equation.

-6 + 6 + 3x = 18 + 6

Combine like terms: -6 + 6 = 0

0 + 3x = 18 + 6

3x = 18 + 6

Combine like terms: 18 + 6 = 24

3x = 24

Divide each side by '3'.

x = 8

Simplifying

x = 8

Aleks [24]4 years ago

5 0

The answer should be X=8

You might be interested in

Solve this problem by using variation of parameters method. y''-y=coshx.

Gekata [30.6K]

$y''-y=0\implies r^2-1=0\implies r=\pm1$
$\implies y_c=C_1e^x+C_2e^{-x}={C^*}_1\underbrace{\cosh x}_{y_1}+{C^*}_2\underbrace{\sinh x}_{y_2}$

For the nonhomogeneous ODE

$y''-y=\underbrace{\cosh x}_{f(x)}$

we're looking for a particular solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\int\frac{y_2(x)f(x)}{W(y_1(x),y_2(x))}\,\mathrm dx$
$u_2=\displaystyle\int\frac{y_1(x)f(x)}{W(y_1(x),y_2(x))}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the two fundamental solutions.

We have

$W(y_1,y_2)=\begin{vmatrix}\cosh x&\sinh x\\\sinh x&\cosh x\end{vmatrix}=\cosh^2x-\sinh^2x=1$

so we're left with

$u_1=-\displaystyle\int\sinh x\cosh x\,\mathrm dx=-\dfrac12\cosh^2x$
$u_2=\displaystyle\int\cosh^2x\,\mathrm dx=\dfrac12x+\dfrac14\sinh2x$

so that the particular solution is

$y_p=-\dfrac12\cosh^3x+\dfrac12x\sinh x+\dfrac14\sinh x\sinh2x$
$y_p=-\dfrac12\cosh x+\dfrac12x\sinh x$

As $y_1$ already accounts for the $\cosh x$ term in $y_p$ , we're left with the general solution

$y=C_1\cosh x+C_2\sinh x+\dfrac12x\sinh x$