# Solve the system : 3x+3>2-x 5x-4=Solve the system : 3x+3>2-x 5x-4=<2x+8

### 4 Answers | Add Yours

3x + 3 > 2-x

5x-4 =< 2x + 8

First we will solve each inequality and then the solution would be the intersection of both solutions.

3x + 3 > 2- x

Group similar terms:

==> 4x + 1 > 0

==> 4x > -1

==> x > -1/4

**Then x = (-1/4 , inf) ...........(1)**

** **

5x-4 = < 2x + 8

==> 5x =< 2x + 12

subtrtact 2x:

==> 3x =< 12

Divide by 3:

==> x =< 12/3

==> x =< 4

**Then x = (-inf, 4 ].........(2)**

**Then the solution is:**

**x= (-1/4, inf) AND (-inf, 4]**

**==> x= (-1/4, 4]**

The given in-equations are 3x+3>2-x and 5x-4=<2x+8

From the two we can get a range of values that x can take for which both the in-equations will be valid.

We can write 3x+3>2-x as 4x>-1 => x>(-1/4)

We can write 5x-4=<2x+8 as 3x=<12 => x=< 4

Therefore from the two in-equations we arrive at the conclusion that x has to be greater than (-1/4) and x has to be less than or equal to 4.

**Any value greater than (-1/4) and less than or equal to 4 can be a solution for the given system of in-equations.**

To solve 3x+3>2-x

5x-4 = <2x+8.

Solution:

1st inequality 3x+3 > 2-x . Add x

3x+x+4 <>2. Subtract 4.

4x > 2-4 = -2. Divide by 4,

x > -2/4 = -1/2.............(1)

2nd inequality: 5x-4 <= 2x+8. Add 4 to both sides.

5x <= 2x+8+4 = 2x+12

5x <= 2x+12. Subtract 2x.

5x-2x <= 12.

3x <= 12. Divide by 3.

x < 12/3 = 4.

x < = 4.................(2).

Combining the inequalities (1) and (2), we get:

- (1/2 < x < = 4.

We'll note the inequations as:

3x+3>2-x (1)

5x-4=<2x+8 (2)

We'll solve the inequation (1):

3x+3>2-x

We'll subtract 3 both sides:

(3x+3)-3>(2-x)-3

We'll remove the brackets and eliminate like terms:

3x>-x-1

We'll add x both sides:

3x + x>-x-1+x

We'll eliminate like terms:

4x>-1

We'll divide by 4:

x>-1/4

**x belongs to the interval (-1/4 , +inf.)**

Now, we'll solve the second inequality:

5x-4=<2x+8

We'll add 4 both sides;

5x=<2x+8+4

We'll subtract 2x both sides:

3x=<12

We'll divide by 3:

x=<4

**x belongs to the interval (-inf., 4]**

**The common solution of the system is (-1/4 , 4].**