-1
$\begingroup$

$$\frac{b}{a}y+2y=1$$ $$\Rightarrow y=\frac{a}{b+2a}$$

Can you help me with the steps to perform this calculation - how $y$ is found?

  • 0
    I am not sure why this under "optimization"? But start with factoring out $y$, do you know how to continue after that? – 2017-01-15
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    Just as you'd solve for $x$ in $$\frac 12 x + 2x = 1 \iff x\left(\frac 12 + 2 \right)= 1 \iff \left (\frac 52\right) x = \frac{1}{\frac 52} = \frac 25$$ – 2017-01-15
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    In this example, swap variable $x$ with $y$, and replace $2$ with $a$, and $1$ (in the numerator only) with $b$. $\frac ab = \frac 12$ – 2017-01-15

3 Answers 3

2

We have: $$\frac{b}{a}y+2y=1$$ This can be factorized to give: $$y\left(\frac{b}{a}+2\right)=1$$ $$y\left(\frac{b}{a}+\frac{2a}{a}\right)=1$$ $$y\left(\frac{b+2a}{a}\right)=1$$ Dividing both sides by $\frac{b+2a}{a}$: $$y=\frac{1}{\left(\frac{b+2a}{a}\right)}$$ $$\boxed{y=\frac{a}{b+2a}}$$

0

$$1=\frac ba y+2y=\left(\frac ba+2\right)y=\frac{2a+b}a\,y\implies y=\frac1{\frac{2a+b}a}=\frac a{2a+b}$$

0

Since the distributive property states that $a(b+c)=ab+ac$, the LHS becomes$$\dfrac bay+2y=y\left(\dfrac ba+2\right)$$ You can check that by redistributing the $y$ back into the parentheses. Now, your expression becomes$$\begin{align*}y\left(\dfrac ba+2\right) & =1\\y=\dfrac 1{\dfrac ba+2}\end{align*}$$ To simplify the ugly looking fraction, multiply both the numerator and the denominator by $a$ to get$$y=\dfrac {1\cdot\color{red}a}{\left(\dfrac ba+2\right)\color{red}a}=\dfrac a{b+2a}$$