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.


  • $\begingroup$ $\dfrac{5(3y-4)}{2y}=7 \implies$ $5(3y-4)=14y$, bring all the like factors together. And you get the value of $y$. $\endgroup$ – Inceptio May 18 '13 at 13:50
  • $\begingroup$ @Inceptio please can you edit it for me, i'm not sure how $\endgroup$ – user61406 May 18 '13 at 13:51
  • $\begingroup$ HonkyHanka: Enclose \frac{5(3y-4)}{2y}=7 in dollar signs. When you have a little time, and for reference, see this nice mathjax tutorial. You can start simply, formatting equations like $x^2 + y^2 = 1$, e.g.: $x^2 + y^2 = 1$ $\endgroup$ – Namaste May 18 '13 at 13:53
  • $\begingroup$ Ok thanks but any answers lol? $\endgroup$ – user61406 May 18 '13 at 14:00

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:

$$ \begin{align} 5(3y - 4) = 14 y & \iff 15y - 20 = 14y \\ \\ & \iff 15y - 14y = 20 \\ \\ & \iff y = 20. \end{align} $$

  • $\begingroup$ Gets a thumbs up! +1 $\endgroup$ – Amzoti May 19 '13 at 0:42

$$\begin{equation*} \frac{5(3y-4)}{2y}=7.\tag{0} \end{equation*} $$

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 $(1\mathrm{c})$.


We are given:


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.




We find that 7 is indeed equal to 7 which proves our answer is correct. $7=7$


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