CPhill is correct - He's rarely wrong. I just want to look at it myself.
Think about this for a moment
$$|a|=|t|$$
This would be true if x=t and it would also be true if x=-t
The same thing applies to your question:
$$\begin{array}{rlll}
|3y+6|&=&|3y+3|\\
(3y+6)&=&\pm (3y+3)\quad &\mbox{now factorise both sides}\\
3(y+2)&=&\pm 3(y+1)\quad &\mbox{next divide both sides by 3}\\
y+2&=&\pm (y+1)
\end{array}$$
Now we seperate this into 2 equations
$$\begin{array}{rlllrl}
y+2&=&+(y+1)\quad &\mbox{and}\qquad y+2&=&-(y+1)\\
y+2&=&y+1\quad &\mbox{and}\qquad y+2&=&-y-1\\
y&=&y-1\quad &\mbox{and}\qquad 2y&=&-3\\
y&=&\mbox{no soln}\quad &\mbox{and}\qquad y&=&\frac{-3}{2}\\
\end{array}$$
So the answer is y=-1.5
I'm going to change the "y's" to "x's"
So we have
l3x+6l = l3x+3l
Factor a 3 out of each and we get
lx+2l = lx+1l
Take both things out of the absolute value bars. So we have
x+2 = x+1 and there's no solution to this
Now set x+2 = -(x+1) .......So we have
x+2 = -x -1 Add x to both sides
2x + 2 = -1 Subtract 2 from both sides
2x = -3 Divide by 2 on both sides
x = -3/2 = -1.5
Here's a graph of the solution:
P.S. ... You can back-substitute "y" for "x" if you'd like....
The leftmost function on the graph is l3x+6l and the rightmost is l3x+3l
You can back-substitute "y" for "x" if you'd like......
CPhill is correct - He's rarely wrong. I just want to look at it myself.
Think about this for a moment
$$|a|=|t|$$
This would be true if x=t and it would also be true if x=-t
The same thing applies to your question:
$$\begin{array}{rlll}
|3y+6|&=&|3y+3|\\
(3y+6)&=&\pm (3y+3)\quad &\mbox{now factorise both sides}\\
3(y+2)&=&\pm 3(y+1)\quad &\mbox{next divide both sides by 3}\\
y+2&=&\pm (y+1)
\end{array}$$
Now we seperate this into 2 equations
$$\begin{array}{rlllrl}
y+2&=&+(y+1)\quad &\mbox{and}\qquad y+2&=&-(y+1)\\
y+2&=&y+1\quad &\mbox{and}\qquad y+2&=&-y-1\\
y&=&y-1\quad &\mbox{and}\qquad 2y&=&-3\\
y&=&\mbox{no soln}\quad &\mbox{and}\qquad y&=&\frac{-3}{2}\\
\end{array}$$
So the answer is y=-1.5