MHB Transpose Formula: Solve x=7-2y | Get Help Now

[fordy2707]

Hi,thanks in advance for your help
Ive been given the task to transpose

To make x the subject from y=7-2x

Which in my mind x= 7+y / 2

But my book is saying X=7-y / 2

Is the book correct if so where / why am I going wrong ?

 

[MarkFL]

We are given:

$$y=7-2x$$

and told to solve for $x$. First, let's add $2x-y$ to both sides, to get:

$$2x=7-y$$

Now, divide through by $2$:

$$x=\frac{7-y}{2}$$

This agrees with the book's result. :)

 

[fordy2707]

Thanks for that , I've been watching YouTube videos for the best part of today trying to get my head round transposing formulae .think I'm just about there now

 

[Deveno]

Here is another approach:

Suppose $y = 7 - 2x$

Let's examine the steps we take to get to $y$, starting with $x$.

1. First we multiply $x$ by $2$. Now we have $2x$.

2. Next we multiply by $-1$, so we have $-2x$.

3. Finally, we add $7$, so we have $7 - 2x$, and we have arrived at $y$.

To "undo" this, we do the "undoing" operation of each of our 3 steps, IN REVERSE ORDER.

First, we subtract $7$. this "undoes" the adding of $7$, so we have $y - 7$.

Next, we multiply by $-1$ (multiplying by $-1$ twice leaves us where we were originally, so multiplying by $-1$ "undoes itself"). This gives us $(-1)(y - 7) = -y - (-7) = -y + 7 = 7 - y$.

Finally, we multiply by $\frac{1}{2}$ which is what "undoes" multiplication by $2$:

$\frac{1}{2}(7 - y) = \dfrac{7-y}{2}$.

Since we "undid", everything we "did" to get from $x$ to $y$, we must now be back at $x$:

$x = \dfrac{7 - y}{2}$.

If we have $f(x) = 7 - 2x$, and $g(y) = \dfrac{7 - y}{2}$, as a final check, we verify that:

$g(f(x)) = x$, and $f(g(y)) = y$.

$g(f(x)) = \dfrac{7 - f(x)}{2} = \dfrac{7 - (7 - 2x)}{2} = \dfrac{7 - 7 + 2x}{2} = \dfrac{2x}{2} = x$

$f(g(y)) = 7 - 2(g(y)) = 7 - 2\left(\dfrac{7 - y}{2}\right) = 7 - (7 - y) = 7 - 7 + y = y$.