Chapter 3.2, Problem 47HP

To determine

To find: the solution of $\frac{2}{3}\left(x+3\right)=\frac{1}{2}\left(x+5\right)$ by graphing and verify the solution algebraically.

Answer to Problem 47HP

The solution will be $\left(x,y\right)=\left(3,4\right)$ .

Explanation of Solution

Given information:

Calculation:

Consider the equation,

  $\frac{2}{3}\left(x+3\right)=\frac{1}{2}\left(x+5\right)$

Once graphed, intersection points of these two lines need to be found.

The intersection point must be verified with algebra solution.

Given left hand equation $\frac{2}{3}(x+3)$

Points that lie on this equation will be

$x$	$y$	$\left(x,y\right)$
0	2	$\left(0,2\right)$
3	4	$\left(3,4\right)$
$-3$	0	$\left(-3,0\right)$

Given right hand equation $\frac{1}{2}(x+5)$

Points that lie on this equation will be,

$x$	$y$	$\left(x,y\right)$
1	3	$\left(1,3\right)$
3	4	$\left(3,4\right)$
$-1$	2	$\left(-1,2\right)$

Graph:

Given,

  $\frac{2}{3}\left(x+3\right)=\frac{1}{2}\left(x+5\right)$

The least common multiple of the denominators 2 & 3 is

  $2\times 3=6$

Multiplying with 6 both sides,

  $\begin{array}{c}6(\frac{2}{3}(x+3))=6(\frac{1}{2}(x+5))\\ 4\left(x+3\right)=3\left(x+5\right)\\ 4x+12=3x+15\end{array}$

Subtracting $3x$ and 12 both sides

  $\begin{array}{c}4x-3x+12-12=3x-3x+15-12\\ x=3\end{array}$

Therefore,

  $\begin{array}{c}y=\frac{2}{3}(x+3)\\ =\frac{2}{3}(3+3)\\ =\frac{2}{3}(6)\\ =4\end{array}$

Hence, the solution will be $\left(x,y\right)=\left(3,4\right)$ .