diferira7c

Answered

2021-12-27

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

Answer & Explanation

limacarp4

Expert

2021-12-28Added 39 answers

Calculation:

Multiply by 30 on both sides,

$30\cdot \frac{{y}^{2}}{30}=(\frac{y}{15}+\frac{1}{2})30$

By distributive property,

$(\frac{y}{15}+\frac{1}{2})30=\frac{y}{15}\cdot 30+\frac{1}{2}\cdot 30$

$(\frac{y}{15}+\frac{1}{2})30=2y+15$

$30\cdot \frac{{y}^{2}}{30}={y}^{2}$

${y}^{2}=2y+15$

Subtract$-2y-15$ on both sides,

${y}^{2}-2y-15=2y+15-2y-15$

${y}^{2}-2y-15=0$

Split the middle term$-2y\text{}\text{as}\text{}-5y\text{}\text{and}\text{}+3y$

Replace -2y with$-5y+3y\text{}\in \text{}{y}^{2}-2y-15=0$ ,

${y}^{2}-5y+3y-15=0$

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,

$y\cdot y-5\cdot y+3\cdot y-3\cdot 5=y(y-5)+3(y-5)$

Again apply distributive property for$y(y-5)+3(y-5)$

$y(y-5)+3(y-5)=(y+3)(y-5)$

So the factors are$(y+3)\text{}\text{and}\text{}(y-5)$

${y}^{2}-2y-15=(y+3)(y-5)$

To find the value of y,

Take$y-5=0$

Add 5 on both sides,

$y-5+5=0+5$

Add the like terms,

$y=5$

Take$y+3=0$

Subtract 3 on both sides,

$y+3-3=0-3$

Add the like terms,

$y=-3$

The factors are$y=-3\text{}\text{and}\text{}y=5$

So, the values of y are -3 and 5.

Multiply by 30 on both sides,

By distributive property,

Subtract

Split the middle term

Replace -2y with

The above equation can be written as

Apply distributive property for first two terms and last two terms,

Again apply distributive property for

So the factors are

To find the value of y,

Take

Add 5 on both sides,

Add the like terms,

Take

Subtract 3 on both sides,

Add the like terms,

The factors are

So, the values of y are -3 and 5.

sonorous9n

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2021-12-29Added 34 answers

karton

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