# Finding a value for x

• May 4th 2011, 03:27 AM
David Green
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(Itwasntme)
• May 4th 2011, 04:03 AM
TKHunny
Quote:

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?
• May 4th 2011, 06:03 AM
David Green
Quote:

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?

• May 4th 2011, 06:52 AM
Quacky
The basic principle is there, the algebraic manipulation is tripping you up.

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

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

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

Which of these represents your question?
• May 4th 2011, 10:03 AM
David Green
Quote:

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

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

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

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

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?

• May 4th 2011, 10:06 AM
Quacky
Quote:

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?