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
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.
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: .$\displaystyle x \:=\:-\frac{5}{12}$
$\displaystyle \frac{x+3}{2}+\frac{x+4}{3}=2$
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.