# Question #91885

Feb 5, 2015

I am assuming the problem is $\frac{1}{2} x + \frac{2}{5} = 1 \frac{y}{6 y} + \frac{8}{5}$

The first step would be to simply $1 \frac{y}{6 y}$ to $1 \frac{1}{6}$

Then I would change the mixed number to a fraction.

$\frac{x}{2} + \frac{2}{5} = \frac{7}{6} + \frac{8}{5}$

To clear the fractions in this equation you will need to find the least common multiple for each of the fractions (in this cast 30) and multiply each term by it.

I would multiply each side of the equation by 30.

$30 \left(\frac{x}{2} + \frac{2}{5}\right) = 30 \left(\frac{7}{6} + \frac{8}{5}\right)$

Then I would use the distributive property and multiply each term and then simplify the expressions.

$\frac{30 x}{2} + \frac{60}{5} = \frac{210}{6} + \frac{240}{5}$

$15 x + 12 = 35 + 48$

$15 x + 12 = 83$

$15 x + 12 - 12 = 83 - 12$

$15 x = 71$

$x = \frac{71}{15}$ or $4 \frac{11}{15}$

and $y$ can be an number since it cancels out

Jul 15, 2016

Assuming that the two equations in the question are
$\frac{1}{2} x + \frac{2}{5} = y$ ......(1)
$\frac{1}{6} y + \frac{8}{5} = - x$ ......(2)
First step is to multiply each equation with the LCM of denominators of all the terms in that equation so as to get rid of the fractions.
Next step is to bring all unknowns to the LHS and constants on the RHS

For (1) we see that LCM of the denominators of its three terms, $2 , 5$ and $1$ is $10$. Equation (1) becomes
$10 \times \left(\frac{1}{2} x + \frac{2}{5} = y\right)$
$\implies 5 x + 4 = 10 y$
$\implies 5 x - 10 y = - 4$ .....(3)

Similarly for equation (2)
$30 \times \left(\frac{1}{6} y + \frac{8}{5} = - x\right)$
$\implies 5 y + 48 = - 30 x$
$\implies 30 x + 5 y = - 48$ .....(4)
Solve (3) and (4) for the unknown $x \mathmr{and} y$

To eliminate $y$ multiply (4) by $2$ and add to (3) we get
$5 x + 60 x = - 4 - 96$
$\implies 65 x = - 100$
$\implies x = - \frac{100}{65} = - \frac{20}{13}$
Inserting this value in (3)
$5 \times \left(- \frac{20}{13}\right) - 10 y = - 4$
$\implies - \frac{100}{13} - 10 y = - 4$
$\implies - 10 y = - 4 + \frac{100}{13}$
$\implies - 10 y = \frac{48}{13}$
$\implies y = - \frac{48}{130}$

Feb 22, 2018

$x = - 1 \mathmr{and} y = - \frac{1}{10}$

#### Explanation:

Luckily with equations, you can change the form into into one which suits you.

Do anything you like as long as you do the same thing to both sides. Multiply by the LCM to cancel the denominators

$\frac{1}{2} x + \frac{2}{5} = y \text{ } \leftarrow$multiply both sides by $10$

$\frac{{\cancel{10}}^{5} \times 1 x}{\cancel{2}} + \frac{{\cancel{10}}^{2} \times 2}{\cancel{5}} = 10 \times y$

$5 x + 4 = 10 y \text{ } \leftarrow$ there are no fractions.

$5 x - 10 y = - 4 \text{ } \leftarrow$ re-arranged

$6 y + \frac{8}{5} = - x \text{ } \leftarrow$ multiply both sides by $5$

$5 \times 6 y + \frac{\cancel{5} \times 8}{\cancel{5}} = 5 \times - x$

$30 y + 8 = - 5 x$

$5 x + 30 y = - 8 \text{ } \leftarrow$ re=arranged

Now you have two equations which are both in a better form and we can solve them simultaneously:

$\textcolor{w h i t e}{w w w w w} 5 x - 10 y = - 4 \textcolor{w h i t e}{w w w w w w w} A$
$\textcolor{w h i t e}{w w w w w} 5 x + 30 y = - 8 \textcolor{w h i t e}{w w w w . w . w} B$

$B - A : \textcolor{w h i t e}{w w . w} 40 y = - 4$
$\textcolor{w h i t e}{w w w w w w . w w .} y = - \frac{4}{40}$
$\textcolor{w h i t e}{w w w w w w . . w w} y = - \frac{1}{10}$

Substitute $y = - \frac{1}{10}$ in $A$

$\textcolor{w h i t e}{w w w w w w w} 5 x + 4 = 10 y$
$\textcolor{w h i t e}{w w w w w w w} 5 x + 4 = 10 \left(- \frac{1}{10}\right)$
$\textcolor{w h i t e}{w w w w w w w w . w} 5 x = - 1 - 4$
$\textcolor{w h i t e}{w w w w w w w w . w} 5 x = - 5$
$\textcolor{w h i t e}{w w w w w w w w . w} x = - 1$