# How do you solve the following linear system:  x=1-y , x-y=5 ?

Nov 13, 2015

$\textcolor{b l u e}{x = 3}$
A large amount of detail given for the method on purpose!
Using my methods you should be able to find the value of y$\to \textcolor{b r o w n}{\text{Practice makes it much faster}}$

#### Explanation:

$\textcolor{b r o w n}{\text{You will get used to this and be able to do simple ones in your head}}$

Given:

$x = 1 - y \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(1\right)$
$x - y = 5. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(2\right)$

In each case make y the dependant variable (answer) then equate to each other through y to find x

$\textcolor{b l u e}{\text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$
$\textcolor{red}{\text{For (1)}}$

Add y to both sides giving:

$x + y = 1 - y + y \textcolor{w h i t e}{\times x} \text{Note that" -y +y=0 " so disappears!}$
$x + y = 1$

Subtract $x$ from both sides
$x - x + y = 1 - x$
$y = 1 - x \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left({1}_{a}\right)$

$\textcolor{b l u e}{\text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$
$\textcolor{red}{\text{For (2)}}$

Add y to both sides giving:
$x - y + y = 5 + y$
$x = 5 + y$

Subtract 5 from bot sides giving:
$x - 5 = 5 - 5 + y$
$x - 5 = y$

Reverse so that y is on the left
$y = x - 5. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left({2}_{a}\right)$

$\textcolor{b l u e}{\text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$
$\text{Equation "(1_a) = color(blue)(y) = "Equation } \left({2}_{a}\right)$

$\textcolor{b r o w n}{1 - x} = \textcolor{b l u e}{y} = \textcolor{b r o w n}{x - 5}$
$\textcolor{b r o w n}{1 - x} = \textcolor{b r o w n}{x - 5}$

Add $x$ to both sides

$1 - x + x = x + x - 5$
$1 = 2 x - 5$

$1 + 5 = 2 x - 5 + 5$

$2 x = 6$

$\textcolor{b r o w n}{\underline{\text{DIVIDE}}}$ BOTH SIDES BY 2

$\frac{2}{2} x = \frac{6}{2}$

## $\textcolor{b l u e}{x = 3}$

$\textcolor{b l u e}{\text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$

Using the methods I have shown you first substitute $x = 3$ in equations #(1_a) "or " (2_a) then manipulate to find y