# Question 2d385

Jun 23, 2016

First part in a lot of detail to demonstrate algebraic manipulation.
Second part not so.

Point common to both equations is:
$\left(x , y\right) \to \left(3 , 2\right)$

#### Explanation:

Given:$\text{ } 7 x + y = 23$ ........................Equation (1)
$\text{ } 3 x = 4 y + 1$.........................Equation (2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Consider equation(1)

Subtract $7 x$ from both sides

$y = - 7 x + 23 \text{ ............................Equation} \left({1}_{a}\right)$

$\textcolor{g r e e n}{\text{Substitute for "color(blue)(y)" in equation (2) using } \left({1}_{a}\right)}$

$\textcolor{b r o w n}{3 x = 4 \textcolor{b l u e}{y} + 1 \text{ "->" } 3 x = 4 \textcolor{b l u e}{\left(- 7 x + 23\right)} + 1}$

$\textcolor{w h i t e}{}$

color(brown)(3x=-28x+92+1)" "color(green)(larr" multiplied out the bracket")
$\textcolor{w h i t e}{}$

$\textcolor{b r o w n}{3 x \textcolor{b l u e}{+ 28 x} = - 28 x \textcolor{b l u e}{+ 28 x} + 93} \textcolor{g r e e n}{\leftarrow \text{ add "color(blue)(28x)" to both sides}}$
$\textcolor{w h i t e}{}$

$\textcolor{b r o w n}{31 x = 0 + 93}$
$\textcolor{w h i t e}{}$

$\textcolor{b r o w n}{\frac{31}{\textcolor{b l u e}{31}} x = \frac{93}{\textcolor{b l u e}{31}}} \textcolor{g r e e n}{\leftarrow \text{ divide both sides by } \textcolor{b l u e}{31}}$

$\textcolor{w h i t e}{}$

color(brown)(x=3)" "color(green)(larr31/31=1" and "93/31 = 3)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute for $x$ in (1)

$7 x + y = 23 \text{ " =>" } y = 2$