# How do you solve the following system: x - 3y = -1 , 8x-y=19 ?

Feb 26, 2016

If you start to use decimals you will introduce errors. keep to fractions!
$\textcolor{b l u e}{x = \frac{58}{23} \text{ } y = 1 \frac{58}{69}}$

#### Explanation:

The objective when solving an equation is to have only 1 unknown and some numeric values. This is then solvable.

For a system to be solvable you have to have the same count of equations as the count of unknowns.

You have two unknowns and two equations so this system is solvable.
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$\textcolor{b l u e}{\text{Initial approach}}$

Change one of the equations so that there is the same count of the variable $y$ in each of them. Then by subtraction remove this variable. Hence solve for $x$

$\textcolor{b l u e}{\text{Solving for } x}$

Given:
$\text{ "x-3y=-1" }$..................................(1)
$\text{ "8x-y=19" }$.....................................(2)

Multiply equation (2) by 3 so that we have the same number of $y ' s$ in each.

$\text{ "x-3y=-1" }$..................................(1)
" "underline(24x-3y=57" ")..........................(2_a)

I do not like it this way round for subtraction so reverse order

$\text{ "24x-3y=57" } \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left({2}_{a}\right)$
" "underline(color(white)(..)x-3y=-1" ")..................................(1)
Subtraction$\text{ } 23 x + 0 \textcolor{w h i t e}{.} = 58$

Divide both sides by 23 giving

$\text{ } \textcolor{b l u e}{x = \frac{58}{23}}$
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$\textcolor{b l u e}{\text{Solving for } y}$

Substitute $\textcolor{b l u e}{x = \frac{58}{23}}$ into equation (1)
I chose equation (1) to make the calculation easier!

$\textcolor{b r o w n}{x - 3 y = - 1 \text{ "->" } \textcolor{b l u e}{\left(\frac{58}{23}\right)} - 3 y = - 1}$

$\text{ } 3 y = \frac{58}{23} + 1$

$y = \frac{58}{23 \times 3} + \frac{1}{3}$

$\textcolor{b l u e}{y = \frac{58 + 69}{69} = 1 \frac{58}{69}}$
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