Question

What is the simplified form of (x+9)/8 minus (x+3)/(x+2)?

Answer 1

The simplified form of (x+9)/8 - (x+3)/(x+2) is:

x^2+3x-6 over 8 (x+2)

Work:

1. (x+9)(x+2) - (x+3) x 8 over 8 (x+2)

2. (x+9)(x+2) - 8(x+3) over 8(x+2)

3. x^2 + 2x + 9x + 18 - 8x -24 over 8(x+2)

4. x^2 + (2x + 9x - 8x) + (18 - 24) over 8(x+2)

= x^2+3x-6 over 8 (x+2)

Answer 2

Here is the work to help you understand the problem.
$\frac{x+9}{8}-\frac{x+3}{x+2}$

$\mathrm{Find\:the\:least\:common\:denominator\:} \ \textgreater \ 8\left(x+2\right)$

$\mathrm{Adjust\:Fractions\:based\:on\:the\:LCD} \ \textgreater \ \frac{\left(x+9\right)\left(x+2\right)}{8\left(x+2\right)}-\frac{\left(x+3\right)\cdot \:8}{8\left(x+2\right)}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}: \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$\frac{\left(x+9\right)\left(x+2\right)-8\left(x+3\right)}{8\left(x+2\right)}$

$Expand \left(x+9\right)\left(x+2\right)-8\left(x+3\right)$

$\left(x+9\right)\left(x+2\right)$
$\mathrm{Distribute\:parentheses\:using}: \left(a+b\right)\left(c+d\right)=ac+ad+bc+bd$
$Where\; a=x,\:b=9,\:c=x,\:d=2$

$x\cdot \:x+x\cdot \:2+9\cdot \:x+9\cdot \:2 \ \textgreater \ \mathrm{Add\:similar\:elements:}\:2x+9x=11x$

$xx+11x+2\cdot \:9 \ \textgreater \ \mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c}$

$xx=\:x^{1+1}=\:x^2 \ \textgreater \ x^2+11x+2\cdot \:9 \ \textgreater \ \mathrm{Multiply\:the\:numbers:}\:9\cdot \:2=18$

$x^2+11x+18 \ \textgreater \ x^2+11x+18-8\left(x+3\right)$

$-8\left(x+3\right) \ \textgreater \ \mathrm{Distribute\:parentheses\:using}: \:a\left(b+c\right)=ab+ac$
$Where\;a=-8,\:b=x,\:c=3$

$-8\cdot \:x-8\cdot \:3 \ \textgreater \ \mathrm{Multiply\:the\:numbers:}\:8\cdot \:3=24 \ \textgreater \ -8x-24$

$x^2+11x+18-8x-24 \ \textgreater \ \mathrm{Group\:like\:terms} \ \textgreater \ x^2+11x-8x+18-24$

$\mathrm{Add\:similar\:elements:}\:11x-8x=3x \ \textgreater \ x^2+3x+18-24$

$\mathrm{Add/Subtract\:the\:numbers:}\:18-24=-6 \ \textgreater \ x^2+3x-6$

$\frac{x^2+3x-6}{8\left(x+2\right)}$

Hope this helps!