diferira7c

2021-12-27

To solve: $\frac{{y}^{2}}{30}=\frac{y}{15}+\frac{1}{2}$

limacarp4

Calculation:
Multiply by 30 on both sides,
$30\cdot \frac{{y}^{2}}{30}=\left(\frac{y}{15}+\frac{1}{2}\right)30$
By distributive property,
$\left(\frac{y}{15}+\frac{1}{2}\right)30=\frac{y}{15}\cdot 30+\frac{1}{2}\cdot 30$
$\left(\frac{y}{15}+\frac{1}{2}\right)30=2y+15$
$30\cdot \frac{{y}^{2}}{30}={y}^{2}$
${y}^{2}=2y+15$
Subtract $-2y-15$ on both sides,
${y}^{2}-2y-15=2y+15-2y-15$
${y}^{2}-2y-15=0$
Split the middle term
Replace -2y with ,
${y}^{2}-5y+3y-15=0$
The above equation can be written as $y\cdot y-5\cdot y+3\cdot 3-3\cdot 5$
Apply distributive property for first two terms and last two terms,
$y\cdot y-5\cdot y+3\cdot y-3\cdot 5=y\left(y-5\right)+3\left(y-5\right)$
Again apply distributive property for $y\left(y-5\right)+3\left(y-5\right)$
$y\left(y-5\right)+3\left(y-5\right)=\left(y+3\right)\left(y-5\right)$
So the factors are
${y}^{2}-2y-15=\left(y+3\right)\left(y-5\right)$
To find the value of y,
Take $y-5=0$
$y-5+5=0+5$
$y=5$
Take $y+3=0$
Subtract 3 on both sides,
$y+3-3=0-3$
$y=-3$
The factors are
So, the values of y are -3 and 5.

sonorous9n

$\frac{{y}^{2}}{30}\cdot 30=\frac{y}{15}\cdot 30+\frac{1}{2}\cdot 30$
${y}^{2}=2y+15$
${y}^{2}-15=2y+15-15$
${y}^{2}-15=2y$
${y}^{2}-15-2y=2y-2y$
${y}^{2}-2y-15=0$
${y}_{1,2}=\frac{-\left(-2\right)±\sqrt{{\left(-2\right)}^{2}-4\cdot 1\cdot \left(-15\right)}}{2\cdot 1}$
${y}_{1,2}=\frac{-\left(-2\right)±8}{2\cdot 1}$
$y=\frac{-\left(-2\right)+8}{2\cdot 1},{y}_{2}=\frac{-\left(-2\right)-8}{2\cdot 1}$
$y=\frac{-\left(-2\right)+8}{2\cdot 1}:5$
$y=\frac{-\left(-2\right)-8}{2\cdot 1}:-3$
$y=5,y=-3$

karton

$\begin{array}{}\frac{{y}^{2}}{30}=\frac{y}{15}+\frac{1}{2}\\ {y}^{2}=2y+15\\ {y}^{2}-2y=15\\ {y}^{2}-2y-15=0\\ a+b=-2\\ ab=-15\\ 1,-15\\ 3,-5\\ 1-15=-14\\ 3-5=-2\\ a=-5\\ b=3\\ \left(y-5\right)\left(y+3\right)\\ y=5\\ y=-3\end{array}$