How do I solve this equation: $\displaystyle \frac{x+4}{8}=1-\frac{x+3}{3}$ I know to get rid of the denominators I multiply $\displaystyle 3(x+4)=8(1-x+3)$ is that right so far or do I only multiply 'x+3' by 8 not 1?
Follow Math Help Forum on Facebook and Google+
Originally Posted by bearhug How do I solve this equation: $\displaystyle \frac{x+4}{8}=1-\frac{x+3}{3}$ I know to get rid of the denominators I multiply $\displaystyle 3(x+4)=8(1-x+3)$ 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: $\displaystyle 1 = \frac{3}{3}$ $\displaystyle \frac{x+4}{8} = \frac{3}{3} - \frac{x+3}{3} = \frac{3+x+3}{3} = \frac{x+6}{3}$
correction ... $\displaystyle \frac{3}{3} - \frac{x+3}{3} = \frac{3 -(x+3)}{3} = -\frac{x}{3}$
View Tag Cloud