# (x+2)/5+(x+12)/10+2=x-2

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We have to solve (x+2)/5 + (x+12)/10 + 2 = x-2 for x.

Let's start with (x+2)/5 + (x+12)/10 + 2 = x-2

First eliminate the denominators in all the terms. To do this multiply all terms by 10.

=> 2*(x+2) + (x+12) + 2*10 = (x-2)*10

Open the brackets.

=> 2x + 4 + x + 12 + 20 = 10x - 20

Add up the similar terms.

=> 3x + 36 = 10x - 20

Bring all the terms with x to one side and the numerical terms to the other side.

=> 7x = 56

=> x = 56/7

=> x = 8

Therefore x = 8.

**The required value of x is 8.**

(x+ 2) / 5 + )x+12)/10 + 2 = x- 2

First let us subtract 2 from both sides:

==> (x+ 2)/5 + (x+ 12)/10 = x - 2 -2

==> (x+ 2)/5 + (x+12)/10 = x- 4

Now we will multiply all the terms by 10:

==> 10(x+2)/5 + 10(x+12)/10 = 10(x-4)

==> 2(x+ 2) + (x+12) = 10(x-4)

Now expand the brackets:

==> 2x + 4 + x + 12= 10x - 40

Now we will comine like terms:

==> 2x + x -10x + 4 + 12 + 40

==> -7x + 56 = 0

Now subtract 56 from both sides:

==> -7x = -56

now divide by -7:

==> x = -56/-7 = 8

**==> x = 8**

To solve the equation (x+2)/5 +(x+12)/10 +2 = x-2.

We multiply both sides by the LCM of the denominators.

LCM(5,10) = 10.

10(x+2)/5 +10(x+12)/10 +10*2 = 10(x-2)

2(x+2) +(x+12)+20 =10x-20

2x+4+x+12+20= 10x-20.

2x+x +4+12+20 = 10x-20

3x+36 = 10x-20.

We subtract 10x-36 from both sides:

3x-10x = -20-36

-7x = -56.

x= -56/-7 = 8.

Therefore x = 8 is the solution.