Solving Equation Using Algebraic Method How to solve these equations using an algebraic method?
I need to show my working, don't you do something in reverse, like 7 multiplies by something. I haven't done it in class.
$$\dfrac{5(3y-4)}{2y}=7$$
 A: Multiply each side of the equation by $2y$ to get:
$$\dfrac{5(3y-4)}{2y}=7 \iff 5(3y - 4) = 7\cdot 2y = 14 y$$
Now, distribute, and then gather "like terms", and simplify:
$$
\begin{align} 5(3y - 4)  = 14 y & \iff 15y - 20 = 14y \\ \\
& \iff 15y - 14y = 20 \\ \\
& \iff y = 20.
\end{align}
$$
A: $$\begin{equation*}
\frac{5(3y-4)}{2y}=7.\tag{0}
\end{equation*}
$$

like 7 multiplies by something



*

*The given equation is only defined when the denominator of the left-hand side is different from $0$. So assume that $y\ne 0$. Then  you can multiply both sides of equation $(0)$ by $2y$ to obtain an equivalent one, i.e. the new equation has the same solution as the original.  $$
\begin{eqnarray*}
\frac{5(3y-4)}{2y}\times 2y &=&7\times 2y \tag{1$\mathrm{a}$} \\
\Leftrightarrow5(3y-4) &=&14y \tag{1$\mathrm{b}$} \\
\Leftrightarrow15y-20 &=&14y,\qquad\text{after expanding the LHS}.\tag{1$\mathrm{c}$}
\end{eqnarray*}$$

*You can subtract $14y$ from both sides. The new equation is equivalent to the previous  one:
$$
\begin{eqnarray*}
15y-20-14y &=&14y-14y \tag{2$\mathrm{a}$}\\
\Leftrightarrow y-20 &=&0.\tag{2$\mathrm{b}$}
\end{eqnarray*}
$$

*You can add $20$ to both sides. The new equation is equivalent to the previous  one:
$$
\begin{eqnarray*}
y-20+20 &=&0+20 \tag{3$\mathrm{a}$}\\
\Leftrightarrow y &=&20.\tag{3$\mathrm{b}$}
\end{eqnarray*}
$$

*Since the solution $y=20\neq 0$, the multiplication in 1 is valid.
Comment. In general to get an equivalent equation one can:


*

*multiply or divide both sides of a given equation by the same value, provided that it is different from $0$.

*add or subtract the same value to and from both sides.

*Simplify either side according to the algebraic rules as in $(1\mathrm{b})$ to $(1\mathrm{c})$.
A: We are given:
$\dfrac{5(3y-4)}{2y}=7$
We must next eliminate the $2y$ from the denominator. We can do this by multiplying $2y$ by the $7$ on the other side of the equation. What we have now is a straightforward solve for y question.
$\implies$ $5(3y-4)=7(2y)$
$\implies$ $15y-20=14y$
We must now isolate y.
$\implies$ $15y-20=14y\implies15y-14y=20\implies y=20$ 
We can now check our solution to see if we are correct by plugging the value of y (which is 20) back into the initial equation.
$\dfrac{5(3(20)-4)}{2(20)}=7$
$\dfrac{5(60-4)}{40}=7$
$\dfrac{280}{40}=7$
We find that 7 is indeed equal to 7 which proves our answer is correct.
$7=7$ 
