Question

In each of the following ,make y the subject and hence find the value of y when a=2, b=3 ,and c=4.

Answer 1

Part i

$\dfrac{2a-7}{6y} = \dfrac{3b-5}{2c}$

$y = \dfrac{ 2c(2a-7)}{6(3b-5)} = \dfrac{c(2a-7)}{9b-15}$

Substituting,

$y = \dfrac{4(2(2)-7)}{9(3)-15} = \dfrac{ -12}{12} = -1$

Answer: -1

Part ii

I'm not sure that one's typed in correctly but I'll solve it as written.

$3-34y + 2a = \dfrac{3b-5}{2c + 1}$

$34y = 3+2a-\dfrac{3b-5}{2c + 1}$

$y = \frac 1 {34}\left(3+2a-\dfrac{3b-5}{2c + 1} \right)$

We're not asked to simplify it so I wont. Substituting,

$y = \frac 1 {34}\left(3+2(2)-\dfrac{3(3)-5}{2(4) + 1} \right) = \frac 1 {34}(7-4/9) = \dfrac{59}{306}$

Answer: 59/306