# 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.

$$\dfrac{5(3y-4)}{2y}=7$$

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$\dfrac{5(3y-4)}{2y}=7 \implies$ $5(3y-4)=14y$, bring all the like factors together. And you get the value of $y$. –  Inceptio May 18 '13 at 13:50
@Inceptio please can you edit it for me, i'm not sure how –  user61406 May 18 '13 at 13:51
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.: $x^2 + y^2 = 1$ –  amWhy May 18 '13 at 13:53
Ok thanks but any answers lol? –  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}

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Gets a thumbs up! +1 –  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 $(1\mathrm{c})$.

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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. $7=7$

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