## A Tutorial on Solving Equations Part 3

This is the third part of the series of tutorials on solving equations. In this part, we will solve more complicated equations especially those that contain fractions.  The first part and the second part of this series discuss 10 sample equations. We start with the 11th example.

Example 11: -5x – 3 = -4x + 12

This example deals with the question of what if $x$ is negative? Let us solve the equation. We want $x$ on the left and all the numbers on the right. So, we add 4x to both sides.

-5x – 3 + 4x = -4x + 4x + 12

x – 3 = 12

Next, we add 3 to both sides to eliminate -3 from the left hand side of the equation.

– 3 + 3 = 12 + 3

x = 15

You cannot have a final equation like this where there is a negative sign on x. To eliminate the negative sign on x, multiply both sides by -1.  That is

(-1)(-x) = (-1)(15)

So, x = -15 is the final answer.

Example 12: $2(x - 3) = 4(x + 8) - x$

This example highlights the distributive property. Notice that distributive property is also needed on equations with fractions. The idea is that if you have an expression that looks like $a(b+c)$; that is, a multiplied by the quantity $b + c$, you must “distribute $a$” over them. That is, $a(b + c) = ab + ac$ and $a(b-c) = ab - ac$.

Solving the equation above, we have $2(x) - 2(3) = 4(x) + 4(8) - x$

Notice on the right hand side that $4$ is not distributed to the second $x$ because the second $x$ is outside the parenthesis. We now simplify. $2x - 6 = 4x + 32 - x$.

Next, we simplify the expression $4x - x$ on the right hand side. $2x - 6 = 3x + 32$

Now, we want to put $x$ on the left and all the numbers on the right. We do this simultaneously. We subtract $3x$ from the right hand side and add 6 on the left hand side, so we add $6 - 3x$ to both sides of the equation. You can do this separately if you are confused. $2x - 6 + [6 - 3x] = 3x + 32 + [6 - 3x]$

On the left hand side: $-6 + 6 = 0$ and $2x - 3x = -x$. On the right hand side, $3x - 3x =0$ and $32 + 6 = 38$

This gives us $-x = 36$. Multiplying both sides by $-1$, as we have done in Example 11, we have $x = -38$

Example 13: $\frac{x}{2} + \frac{2x}{3} = \frac{3}{4}$.

This type of equation usually appears in work and motion problems which we will discuss later. Just like in solving fractions, all you have to do is get the least common denominator. Now, the least common denominator of 2, 3, and 4 is 12. So, all we have to do is to multiply everything with 12. That is $12(\frac{x}{2} + \frac{2x}{3}) = 12 (\frac{3}{4})$ $\frac{12x}{2} + \frac{12(2x)}{3} = \frac{12(3)}{4}$ $\frac{12x}{2} + \frac{24x}{3} = \frac{36}{4}$ $6x + 8x = 9$ $14x = 9$

Dividing both sides by $14$, we have $x = \frac{9}{14}$.

Example 14: $\frac{3x + 1}{5} = \frac{2x}{3}$

This is almost the same the above example. We get the least common denominator of $5$ and $3$ which is equal to $15$. Then, we multiply everything with $15$. That is $15 (\frac{3x + 1}{5}) = 15(\frac{2x}{3})$ $\frac{15(3x + 1)}{5} = \frac{15(2x)}{3}$

Now, on the left hand side, $15/5=3$ and on the right hand side $15/3 = 5$. This gives us $3(3x + 1) = 5(2x)$.

Simplifying the left hand side, we have $9x + 3 = 10x$

Now, $9x - 10x = - 3$ gives us $-x = - 3$. Multiplying both sides by $-1$ to make $x$ positive gives us the final answer $x = 3$.

Example 15: $\frac{3}{x} + \frac{1}{4} = \frac{2}{5}$

This example discusses the question “what if $x$ is in the denominator?” If $x$ is just in the denominator just like this example, the solution is quite similar to Example 13. However, if $x$ is both found in the numerator and denominator, this will result to a quadratic equation (something with $x^2$). This seldom comes out, and we will discuss this separately. For now, let us solve this example.

The strategy here is to get the least common denominator of the numbers and then include $x$ during the multiplication. In this example, we want to get the least common denominator of $4$ and $5$ which is $20$. Now, we include $x$ and the least common denominator of the equation above is $20x$. Now, we multiply everything with $20x$. That is, $20x(\frac{3}{x} + \frac{1}{4}) = 20x(\frac{2}{5})$ $\frac{(20x)(3)}{x} + \frac{(20x)(1)}{4} = \frac{(20x)(2)}{5}$ $\frac{60x}{x} + \frac{20x}{4} = \frac{40x}{5}$ $60 + 5x = 8x$ $60 = 3x$ $20 = x$. Therefore, the answer is $20$.

This ends the third part of this series, in the next part of this series (I am not sure if I will discuss this soon), we will discuss about dealing equations with radicals (square root and cube root).