More complex equations with fractions

• Jan 1st 2009, 12:05 PM
bronaghxo
More complex equations with fractions
Can someone help explain this to me please..

x+3/2 + x+4/3 = 2

y+5/6 + y-1/2 = 10

z+1/2 - z-1/4 = 1
d+12/3 + d+3/6 = 3d + 11/2
• Jan 1st 2009, 12:23 PM
Soroban
Hello, bronaghxo!

I assume it's the same old complaint: "I hate fractions!"
Unless you plan to count apples or sheep all your life,
. . you better learn to handle them.

Quote:

Solve for $\displaystyle x\!:\;\;x+\frac{3}{2} + x + \frac{4}{3} \:=\: 2$

Well, there's one blessing . . . With an equation, we can eliminate the fractions.

We have: .$\displaystyle x + \frac{3}{2} + x + \frac{4}{3} \;=\;2$

Multiply both sides by the LCD (6):

. . $\displaystyle 6\,\bigg[x + \frac{3}{2} + x + \frac{4}{3}\bigg] \;=\;6\,\bigg[2\bigg]$

. . $\displaystyle 6(x) + 6\left(\frac{3}{2}\right) + 6(x) + 6\left(\frac{4}{3}\right) \:=\:6(2)$

. . $\displaystyle 6x + 9 + 6x + 8 \:=\:12 \quad\Rightarrow\quad 12x + 17 \:=\:12 \quad\Rightarrow\quad 12x \:=\:-5$

Therefore: .$\displaystyle x \:=\:-\frac{5}{12}$

• Jan 1st 2009, 12:24 PM
red_dog
$\displaystyle \frac{x+3}{2}+\frac{x+4}{3}=2$
The common denominator is 6. So amplify the first fraction with 3 and the second with 2 and the right hand member with 6

$\displaystyle 3(x+3)+2(x+4)=12\Leftrightarrow 5x=-5\Leftrightarrow x=-1$

Try to solve the others equations in the same manner.