# How do you solve the following system: 3x+y=8, 2x +5y = 5 ?

Jul 26, 2018

The more formal layout approach.

This $\underline{\text{clearly}}$ shows the thinking processes and is easy to follow for teacher marking.

I have determined the value of $x$ to start you off.

#### Explanation:

Given:
$3 x + y = 8 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(1\right)$
$2 x + 5 y = 5 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(2\right)$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{This bit is not part of the formal approach.}}$

We need one equation with just one unknown.

I am using method by example: known that color(green)(2=2) color(red)(larr" True"

Multiply both sides by $\textcolor{p u r p \le}{3}$

color(green)(ubrace([2color(purple)(xx3)]) = ubrace([2color(purple)(xx3)])
$\textcolor{g r e e n}{\textcolor{w h i t e}{\text{.d") darr color(white)("dddd.d}} \downarrow}$
$\textcolor{g r e e n}{\textcolor{w h i t e}{\text{dd") 6color(white)("dd") =color(white)("dd") 6 color(red)(larr" Still true}}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Back to the solution.}}$

Multiply both sides of $E q n \left(1\right)$ by 5 $\textcolor{w h i t e}{\text{d}}$ ( 'to get rid' of the $y$)

$15 x + 5 y = 40 \text{ } \ldots \ldots . . E q u a t i o n \left({1}_{a}\right)$

$E q n \left({1}_{a}\right) - E q n \left(2\right)$

$15 x + 5 y = 40$
$\underline{\textcolor{w h i t e}{1} 2 x + 5 y = \textcolor{w h i t e}{4} 5 \leftarrow \text{ Subtract}}$
$13 x + 0 \textcolor{w h i t e}{\text{d}} = 35$

Divide both sides by 13

$\frac{13 x}{13} = \frac{35}{13}$

$x = \frac{35}{13}$

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

Substitute this into either $E q n \left(1\right) \mathmr{and} E q n \left(2\right)$ to determine the value of $y$

I will let you do that