# Math Help - Finding a value for x

1. ## Finding a value for x

Just seem to be struggling to understand what I am missing here?

I have;

x - 8 = 3x + 20
7

This is what I have done to try and find the value for x

7(x - 8) = 7(3x + 20)
7

56 - 7x = 3x + 20

-7x -3x = 20 - 56

-10x = -36
-10 -10

x = 3.6

Substitute value into x

3.6 - 8 = 3 x 3.6 + 20
7

-4.4 = 4.4

My problem to me is that -4.4 is not equal to 4.4?

The equation should not have two routes?

Please advise if there is an alterative technique to solve this type of problem?

P.S. the 7 above (denominator) should be divided by the numberators on the RHS of the equals sign. This software program seens to move the integer to the LHS

2. Originally Posted by David Green
7(x - 8) = 7(3x + 20)
7

56 - 7x = 3x + 20
Why did the '7' magically turn negative on the left-hand side?

3. Originally Posted by TKHunny
Why did the '7' magically turn negative on the left-hand side?
Thank you for your help, but there must be another method to solve this equation which I am not aware of?

4. The basic principle is there, the algebraic manipulation is tripping you up.

$\frac{x-8}{7}=3x+20$

This is what it looks like to me. I assume (from your working) that you meant to present:

$x-8=\frac{3x+20}{7}$

Which of these represents your question?

5. Originally Posted by Quacky
The basic principle is there, the algebraic manipulation is tripping you up.

$\frac{x-8}{7}=3x+20$

This is what it looks like to me. I assume (from your working) that you meant to present:

$x-8=\frac{3x+20}{7}$

Which of these represents your question?
Sorry the equation you wrote is the wrong way round. The way I presented it was incorrectly positioned on this forum because the forum moves all intergers placed as denominators to the far left hand side for some reason?

I'll do it again;

x - 8 is on the left hand side and 3x + 20 divided by 7 is on the right hand side.

so x - 8 = 3x + 20 / 7.

multiplying both sides of the equation by 7 gives;

7(x - 8) = 3x + 20 multiply out the LHS.

7x - 56 = 3x + 20 Move the -56 to the RHS
7x = 3x + 20 + 56 Now move the 3x to the LHS
7x - 3x = 76 Subtract LHS
4x = 76 Divide both sides by 4
x = 19

The original equation; x - 8 = 3x + 20 / 7

Putting in the x value; 19 - 8 = 3 \cdot x + 20 / 7

the like values are; 11 = 11

Whatever the reason I couldn't see it before is just a state of mind?

6. Originally Posted by David Green
Sorry the equation you wrote is the wrong way round. The way I presented it was incorrectly positioned on this forum because the forum moves all intergers placed as denominators to the far left hand side for some reason?

I'll do it again;

x - 8 is on the left hand side and 3x + 20 divided by 7 is on the right hand side.

so x - 8 = 3x + 20 / 7.

multiplying both sides of the equation by 7 gives;

7(x - 8) = 3x + 20 multiply out the LHS.

7x - 56 = 3x + 20 Move the -56 to the RHS
7x = 3x + 20 + 56 Now move the 3x to the LHS
7x - 3x = 76 Subtract LHS
4x = 76 Divide both sides by 4
x = 19

The original equation; x - 8 = 3x + 20 / 7

Putting in the x value; 19 - 8 = 3 \cdot x + 20 / 7

the like values are; 11 = 11

Whatever the reason I couldn't see it before is just a state of mind?