# Question #d2c06

Sep 7, 2017

1. $x = 4$
2. $x > 6.25$
3. $x = - 10$
4. ...
5. $- 2 {x}^{2} + 3 x + 9$
6. ..

#### Explanation:

• $3 \cdot x + 5 = 17$
To solve these of equations, the unknown, $x$, has to be isolated one side of the equation,
First, we subtract 5 from both side of the equation to get rid of the 5 on the left side.
$3 \cdot x + 5 - 5 = 17 - 5$
$3 \cdot x = 12$

Then divide by 3 both sides
$\frac{3 \cdot x}{3} = \frac{12}{3}$
$x = 4$
(3 by 3 cancel out, and 12/3=4)

• $4 \cdot x - 5 > 20$

Same as before, add 5 to both sides of the equation
$4 \cdot x - 5 + 5 > 20 + 5$
$4 \cdot x > 25$

divide by 4 both sides of the equation
$\frac{4 \cdot x}{4} > \frac{25}{4}$
$x > 6 \frac{1}{4}$
$x > 6.25$

• $2 - x = 12$
subtract 2 from both sides of the equation
$2 - x - 2 = 12 - 2$
$- x = 10$
since x is negative, we have to multiply both sides by -1
$\left(- x\right) \left(- 1\right) = \left(10\right) \left(- 1\right)$
$x = - 10$

• 4th try it yourself

• $\left(2 x + 3\right) \left(3 - x\right)$
in these kind of equations, you have to multiply each term of the first with the each terms of the second,
it is written as $\left(a x + b\right) \cdot \left(c x + d\right)$
and it is equal to
$\left(a x\right) \left(c x\right) + \left(b\right) \left(c x\right) + \left(a x\right) \left(d\right) + \left(b\right) \left(d\right)$

in the one in your example,
$\left(2 x + 3\right) \left(3 - x\right)$
is equal to
$\left(2 x\right) \left(- x\right) + \left(3\right) \left(- x\right) + \left(2 x\right) \left(3\right) + \left(3\right) \left(3\right) = - 2 {x}^{2} + \left(- 3 x\right) + 6 x + 9$
simplifying the answer by adding $- 3 x + 6 x = 3 x$
=$- 2 {x}^{2} + 3 x + 9$