Solve the inequality x-5#3x+10+

x-5\leq +3x+10

Go!
1
2
3
4
5
6
7
8
9
0
x
y
(◻)
◻/◻
2

e
π
ln
log
lim
d/dx
d/dx
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

Answer

$x\geq \frac{1}{2}-\frac{15}{2}$

Step by step solution

Problem

$x-5\leq +3x+10$
1

Grouping terms

$-3x-5+x\leq 10+$
2

Adding $-3x$ and $x$

$-2x-5\leq 10+$
3

Moving the term $-5$ to the other side of the inequation with opposite sign

$-2x\leq 5+10+$
4

Add the values $10$ and $5$

$-2x\leq 15$
5

Multiply both sides of the inequality by $-1$, reversing the sign

$2x\geq \left(15\right)\left(-1\right)$
6

Multiply $\left(15+\right)$ by $-1$

$2x\geq -15-1$
7

Divide both sides of the inequation by $2$

$x\geq \frac{-15-1}{2}$
8

Split the fraction $\frac{-15+-1}{2}$ in two terms with same denominator

$x\geq \frac{-1}{2}+\frac{-15}{2}$
9

Divide $-15$ by $2$

$x\geq \frac{-1}{2}-\frac{15}{2}$
10

Apply the formula: $\frac{b\cdot a}{c}$$=b\frac{a}{c}$, where $a=-1$, $b=$ and $c=2$

$x\geq -\frac{15}{2}-1\cdot \left(-\frac{1}{2}\right)$
11

Multiply $-\frac{1}{2}$ times $-1$

$x\geq \frac{1}{2}-\frac{15}{2}$

Answer

$x\geq \frac{1}{2}-\frac{15}{2}$

Problem Analysis

Main topic:

Polynomials

Time to solve it:

0.23 seconds

Views:

74