Solving Equations

**bearhug** · September 26th 2009, 12:15 PM

How do I solve this equation:
$\frac{x+4}{8}=1-\frac{x+3}{3}$

I know to get rid of the denominators I multiply
$3(x+4)=8(1-x+3)$

is that right so far or do I only multiply 'x+3' by 8 not 1?

**e^(i*pi)** · September 26th 2009, 12:21 PM

Originally Posted by bearhug

How do I solve this equation:
$\frac{x+4}{8}=1-\frac{x+3}{3}$

I know to get rid of the denominators I multiply
$3(x+4)=8(1-x+3)$

is that right so far or do I only multiply 'x+3' by 8 not 1?

Remember 1 can be expressed as a fraction with any non-zero denominator. In this case I would use:

$1 = \frac{3}{3}$

$\frac{x+4}{8} = \frac{3}{3} - \frac{x+3}{3} = \frac{3+x+3}{3} = \frac{x+6}{3}$

**skeeter** · September 26th 2009, 12:29 PM

correction ...

$\frac{3}{3} - \frac{x+3}{3} = \frac{3 -(x+3)}{3} = -\frac{x}{3}$

Thread: Solving Equations

LinkBack

Thread Tools

Display

Solving Equations

Similar Math Help Forum Discussions

Solving Linear Equations, Systems of Equations, and Functions.

Solving these equations...

Solving equations

Help solving equations please

solving equations

Search Tags